Download Week 06

Integration by parts (Sect. 8.1)
I
Integral form of the product rule.
I
Exponential and logarithms.
I
Trigonometric functions.
I
Definite integrals.
I
Substitution and integration by parts.
Integral form of the product rule
Remark: The integration by parts formula is an integral form of
the product rule for derivatives:
Integral form of the product rule
Remark: The integration by parts formula is an integral form of
the product rule for derivatives: (fg )0 = f 0 g + f g 0 .
Integral form of the product rule
Remark: The integration by parts formula is an integral form of
the product rule for derivatives: (fg )0 = f 0 g + f g 0 .
Theorem
For all differentiable functions g , f : R → R holds
Z
Z
f (x) g 0 (x) dx = f (x) g (x) − f 0 (x) g (x) dx.
Integral form of the product rule
Remark: The integration by parts formula is an integral form of
the product rule for derivatives: (fg )0 = f 0 g + f g 0 .
Theorem
For all differentiable functions g , f : R → R holds
Z
Z
f (x) g 0 (x) dx = f (x) g (x) − f 0 (x) g (x) dx.
Proof: Integrate the product rule f g 0 = (fg )0 − f 0 g ,
Integral form of the product rule
Remark: The integration by parts formula is an integral form of
the product rule for derivatives: (fg )0 = f 0 g + f g 0 .
Theorem
For all differentiable functions g , f : R → R holds
Z
Z
f (x) g 0 (x) dx = f (x) g (x) − f 0 (x) g (x) dx.
Proof: Integrate the product rule f g 0Z= (fg )0 − f 0 g , and use the
Fundamental Theorem of Calculus in
(fg )0 dx = fg .
Integral form of the product rule
Remark: The integration by parts formula is an integral form of
the product rule for derivatives: (fg )0 = f 0 g + f g 0 .
Theorem
For all differentiable functions g , f : R → R holds
Z
Z
f (x) g 0 (x) dx = f (x) g (x) − f 0 (x) g (x) dx.
Proof: Integrate the product rule f g 0Z= (fg )0 − f 0 g , and use the
Fundamental Theorem of Calculus in
(fg )0 dx = fg .
Z
Notation: It is common to write
Z
u dv = uv −
v du,
Integral form of the product rule
Remark: The integration by parts formula is an integral form of
the product rule for derivatives: (fg )0 = f 0 g + f g 0 .
Theorem
For all differentiable functions g , f : R → R holds
Z
Z
f (x) g 0 (x) dx = f (x) g (x) − f 0 (x) g (x) dx.
Proof: Integrate the product rule f g 0Z= (fg )0 − f 0 g , and use the
Fundamental Theorem of Calculus in
(fg )0 dx = fg .
Z
Notation: It is common to write
u = f (x),
dv = g 0 (x) dx,
Z
u dv = uv −
and v = g (x),
v du, where
du = f 0 (x) dx.
Integration by parts (Sect. 8.1)
I
Integral form of the product rule.
I
Exponential and logarithms.
I
Trigonometric functions.
I
Definite integrals.
I
Substitution and integration by parts.
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
0
f (x) g (x) dx = f (x) g (x) −
Z
f 0 (x) g (x) dx.
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
0
f (x) g (x) dx = f (x) g (x) −
We need to choose the functions f and g .
Z
f 0 (x) g (x) dx.
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
0
f (x) g (x) dx = f (x) g (x) −
Z
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side.
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
0
f (x) g (x) dx = f (x) g (x) −
Z
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side.
f (x) = x,
g 0 (x) = e 2x
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
Z
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side.
f (x) = x,
g 0 (x) = e 2x
⇒
f 0 (x) = 1,
g (x) =
e 2x
.
2
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
Z
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side.
f (x) = x,
Z
xe
2x
g 0 (x) = e 2x
x e 2x
dx =
−
2
⇒
Z
f 0 (x) = 1,
e 2x
dx
2
g (x) =
e 2x
.
2
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
Z
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side.
f (x) = x,
Z
xe
2x
g 0 (x) = e 2x
x e 2x
dx =
−
2
⇒
Z
f 0 (x) = 1,
g (x) =
e 2x
.
2
e 2x
e 2x
x e 2x
dx =
−
+ c.
2
2
4
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Z
0
f (x) g (x) dx = f (x) g (x) −
Solution: Recall:
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side.
f (x) = x,
Z
xe
2x
g 0 (x) = e 2x
x e 2x
dx =
−
2
We conclude I =
⇒
Z
f 0 (x) = 1,
g (x) =
e 2x
.
2
e 2x
e 2x
x e 2x
dx =
−
+ c.
2
2
4
e 2x
(2x − 1) + c.
4
C
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
0
f (x) g (x) dx = f (x) g (x) −
Z
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side.
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
0
f (x) g (x) dx = f (x) g (x) −
Z
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side. Wrong choice:
f (x) = e 2x ,
g 0 (x) = x
Exponentials and logarithms
Example
Z
Evaluate I =
x e 2x dx.
Z
Solution: Recall:
Z
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side. Wrong choice:
f (x) = e 2x ,
g 0 (x) = x
⇒
f 0 (x) = 2 e 2x ,
g (x) =
x2
.
2
Exponentials and logarithms
Example
Z
x e 2x dx.
Evaluate I =
Z
Solution: Recall:
Z
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side. Wrong choice:
g 0 (x) = x
f (x) = e 2x ,
Z
xe
2x
⇒
f 0 (x) = 2 e 2x ,
x 2 e 2x
dx =
−
2
Z
x 2 e 2x dx.
g (x) =
x2
.
2
Exponentials and logarithms
Example
Z
x e 2x dx.
Evaluate I =
Z
Solution: Recall:
Z
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
We need to choose the functions f and g .
They should be chosen in a way that the right-hand side above is
simpler to integrate than the original left-hand side. Wrong choice:
g 0 (x) = x
f (x) = e 2x ,
Z
xe
2x
⇒
f 0 (x) = 2 e 2x ,
x 2 e 2x
dx =
−
2
This is the wrong choice.
Z
g (x) =
x2
.
2
x 2 e 2x dx.
C
Exponentials and logarithms
Z
Remark: We use now the
Z
u dv = uv −
v du notation.
Exponentials and logarithms
Z
Remark: We use now the
Example
Z
Evaluate I =
ln(x) dx.
Z
u dv = uv −
v du notation.
Exponentials and logarithms
Z
Remark: We use now the
Z
u dv = uv −
Example
Z
Evaluate I =
ln(x) dx.
Solution: We need to choose u and v :
v du notation.
Exponentials and logarithms
Z
Remark: We use now the
Z
u dv = uv −
Example
Z
Evaluate I =
ln(x) dx.
Solution: We need to choose u and v :
u = ln(x),
dv = dx
v du notation.
Exponentials and logarithms
Z
Remark: We use now the
Z
u dv = uv −
v du notation.
Example
Z
Evaluate I =
ln(x) dx.
Solution: We need to choose u and v :
u = ln(x),
dv = dx
⇒
du =
dx
,
x
v = x.
Exponentials and logarithms
Z
Remark: We use now the
Z
u dv = uv −
v du notation.
Example
Z
Evaluate I =
ln(x) dx.
Solution: We need to choose u and v :
u = ln(x),
⇒
dv = dx
Z
Z
ln(x) dx = x ln(x) −
x
du =
dx
x
dx
,
x
v = x.
Exponentials and logarithms
Z
Remark: We use now the
Z
u dv = uv −
v du notation.
Example
Z
Evaluate I =
ln(x) dx.
Solution: We need to choose u and v :
u = ln(x),
⇒
dv = dx
Z
Z
ln(x) dx = x ln(x) −
du =
dx
,
x
v = x.
dx
= x ln(x) −
x
x
Z
dx.
Exponentials and logarithms
Z
Remark: We use now the
Z
u dv = uv −
v du notation.
Example
Z
Evaluate I =
ln(x) dx.
Solution: We need to choose u and v :
u = ln(x),
⇒
dv = dx
Z
Z
ln(x) dx = x ln(x) −
du =
dx
,
x
v = x.
dx
= x ln(x) −
x
x
We conclude that I = x ln(x) − x + c.
Z
dx.
C
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Solution: u = x 2 ,
dv = e x dx
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Solution: u = x 2 ,
dv = e x dx
⇒
du = 2x dx,
v = ex .
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Solution: u = x 2 , dv = e x dx ⇒ du = 2x dx,
Z
Z
x 2 e x dx = x 2 e x − 2 x e x dx
v = ex .
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Solution: u = x 2 , dv = e x dx ⇒ du = 2x dx,
Z
Z
x 2 e x dx = x 2 e x − 2 x e x dx
We integrate by parts one more time.
v = ex .
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Solution: u = x 2 , dv = e x dx ⇒ du = 2x dx,
Z
Z
x 2 e x dx = x 2 e x − 2 x e x dx
We integrate by parts one more time.
u = x,
dv = e x dx
v = ex .
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Solution: u = x 2 , dv = e x dx ⇒ du = 2x dx,
Z
Z
x 2 e x dx = x 2 e x − 2 x e x dx
v = ex .
We integrate by parts one more time.
u = x,
dv = e x dx
⇒
du = dx,
v = ex .
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Solution: u = x 2 , dv = e x dx ⇒ du = 2x dx,
Z
Z
x 2 e x dx = x 2 e x − 2 x e x dx
v = ex .
We integrate by parts one more time.
u = x,
Z
du = dx, v = e x .
Z
h
i
x 2 e x dx = x 2 e x − 2 x e x − e x dx
dv = e x dx
⇒
Exponentials and logarithms
Remark: Integration by parts can be used more than once.
Example
Z
Evaluate I =
x 2 e x dx.
Solution: u = x 2 , dv = e x dx ⇒ du = 2x dx,
Z
Z
x 2 e x dx = x 2 e x − 2 x e x dx
v = ex .
We integrate by parts one more time.
u = x,
Z
du = dx, v = e x .
Z
h
i
x 2 e x dx = x 2 e x − 2 x e x − e x dx
dv = e x dx
⇒
We conclude that I = x 2 e x − 2x e x + 2e x + c.
C
Integration by parts (Sect. 8.1)
I
Integral form of the product rule.
I
Exponential and logarithms.
I
Trigonometric functions.
I
Definite integrals.
I
Substitution and integration by parts.
Trigonometric functions
Example
Z
Evaluate I =
x sin(x) dx.
Trigonometric functions
Example
Z
Evaluate I =
x sin(x) dx.
Solution: We choose u, v so that we simplify the integral:
Trigonometric functions
Example
Z
Evaluate I =
x sin(x) dx.
Solution: We choose u, v so that we simplify the integral:
u = x,
dv = sin(x) dx,
Trigonometric functions
Example
Z
Evaluate I =
x sin(x) dx.
Solution: We choose u, v so that we simplify the integral:
u = x,
dv = sin(x) dx,
⇒
du = dx,
v = − cos(x).
Trigonometric functions
Example
Z
Evaluate I =
x sin(x) dx.
Solution: We choose u, v so that we simplify the integral:
u = x,
dv = sin(x) dx,
⇒
du = dx,
v = − cos(x).
Z
I = −x cos(x) −
[− cos(x)] dx.
Trigonometric functions
Example
Z
Evaluate I =
x sin(x) dx.
Solution: We choose u, v so that we simplify the integral:
u = x,
dv = sin(x) dx,
⇒
du = dx,
v = − cos(x).
Z
I = −x cos(x) −
[− cos(x)] dx.
Z
I = −x cos(x) + cos(x) dx.
Trigonometric functions
Example
Z
Evaluate I =
x sin(x) dx.
Solution: We choose u, v so that we simplify the integral:
u = x,
dv = sin(x) dx,
⇒
du = dx,
v = − cos(x).
Z
I = −x cos(x) −
[− cos(x)] dx.
Z
I = −x cos(x) + cos(x) dx.
We conclude that I = −x cos(x) + sin(x) + c.
C
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: In this case we need to integrate by parts twice.
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: In this case we need to integrate by parts twice.
u = e ax ,
dv = sin(x) dx
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: In this case we need to integrate by parts twice.
u = e ax ,
dv = sin(x) dx
⇒
du = a e ax dx,
v = − cos(x).
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: In this case we need to integrate by parts twice.
u = e ax ,
du = a e ax dx, v = − cos(x).
Z
ax
I = −e cos(x) + a e ax cos(x) dx.
dv = sin(x) dx
⇒
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: In this case we need to integrate by parts twice.
u = e ax ,
u = e ax ,
du = a e ax dx, v = − cos(x).
Z
ax
I = −e cos(x) + a e ax cos(x) dx.
dv = sin(x) dx
dv = cos(x) dx
⇒
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: In this case we need to integrate by parts twice.
u = e ax ,
u = e ax ,
du = a e ax dx, v = − cos(x).
Z
ax
I = −e cos(x) + a e ax cos(x) dx.
dv = sin(x) dx
dv = cos(x) dx
⇒
⇒
du = a e ax dx,
v = sin(x).
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: In this case we need to integrate by parts twice.
u = e ax ,
du = a e ax dx, v = − cos(x).
Z
ax
I = −e cos(x) + a e ax cos(x) dx.
dv = sin(x) dx
⇒
dv = cos(x) dx ⇒ du = a e ax dx, v = sin(x).
Z
h
i
I = −e ax cos(x) + a e ax sin(x) − a e ax sin(x) dx .
u = e ax ,
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: In this case we need to integrate by parts twice.
u = e ax ,
du = a e ax dx, v = − cos(x).
Z
ax
I = −e cos(x) + a e ax cos(x) dx.
dv = sin(x) dx
⇒
dv = cos(x) dx ⇒ du = a e ax dx, v = sin(x).
Z
h
i
I = −e ax cos(x) + a e ax sin(x) − a e ax sin(x) dx .
u = e ax ,
Z
e
ax
sin(x) dx = −e
ax
cos(x) + a e
ax
sin(x) − a
2
Z
e ax sin(x) dx.
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: Recall:
Z
Z
ax
ax
ax
2
e sin(x) dx = −e cos(x) + a e sin(x) − a
e ax sin(x) dx.
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: Recall:
Z
Z
ax
ax
ax
2
e sin(x) dx = −e cos(x) + a e sin(x) − a
e ax sin(x) dx.
Remark: The last term on the right-hand side is proportional to
the negative of the left-hand side.
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: Recall:
Z
Z
ax
ax
ax
2
e sin(x) dx = −e cos(x) + a e sin(x) − a
e ax sin(x) dx.
Remark: The last term on the right-hand side is proportional to
the negative of the left-hand side. So, for all a 6= 0 holds
Z
2
(1 + a ) e ax sin(x) dx = −e ax cos(x) + a e ax sin(x).
Trigonometric functions
Example
Z
Evaluate I =
e ax sin(x) dx.
Solution: Recall:
Z
Z
ax
ax
ax
2
e sin(x) dx = −e cos(x) + a e sin(x) − a
e ax sin(x) dx.
Remark: The last term on the right-hand side is proportional to
the negative of the left-hand side. So, for all a 6= 0 holds
Z
2
(1 + a ) e ax sin(x) dx = −e ax cos(x) + a e ax sin(x).
We then conclude that
Z
e ax sin(x) dx =
e ax − cos(x) + a sin(x) .
2
(1 + a )
C
Integration by parts (Sect. 8.1)
I
Integral form of the product rule.
I
Exponential and logarithms.
I
Trigonometric functions.
I
Definite integrals.
I
Substitution and integration by parts.
Definite integrals
Remark: Integration by parts can be used with definite integrals.
Definite integrals
Remark: Integration by parts can be used with definite integrals.
Theorem
For all differentiable functions f , g : R → R holds
Z b
Z b
b
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
a
a
a
Definite integrals
Remark: Integration by parts can be used with definite integrals.
Theorem
For all differentiable functions f , g : R → R holds
Z b
Z b
b
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
a
a
Example
Z
Evaluate I =
0
π
e ax sin(x) dx.
a
Definite integrals
Remark: Integration by parts can be used with definite integrals.
Theorem
For all differentiable functions f , g : R → R holds
Z b
Z b
b
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
a
a
Example
Z
Evaluate I =
a
π
e ax sin(x) dx.
0
Solution: Use integrations by parts and evaluate the result:
Definite integrals
Remark: Integration by parts can be used with definite integrals.
Theorem
For all differentiable functions f , g : R → R holds
Z b
Z b
b
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
a
a
Example
Z
Evaluate I =
a
π
e ax sin(x) dx.
0
Solution: Use integrations by parts and evaluate the result:
Z π
h e ax iπ
e ax sin(x) dx =
−
cos(x)
+
a
sin(x)
(1 + a2 )
0
0
Definite integrals
Remark: Integration by parts can be used with definite integrals.
Theorem
For all differentiable functions f , g : R → R holds
Z b
Z b
b
0
f (x) g (x) dx = f (x) g (x) −
f 0 (x) g (x) dx.
a
a
Example
Z
Evaluate I =
a
π
e ax sin(x) dx.
0
Solution: Use integrations by parts and evaluate the result:
Z π
h e ax iπ
e ax sin(x) dx =
−
cos(x)
+
a
sin(x)
(1 + a2 )
0
0
Z π
(e aπ + 1)
.
e ax sin(x) dx =
(1 + a2 )
0
C
Integration by parts (Sect. 8.1)
I
Integral form of the product rule.
I
Exponential and logarithms.
I
Trigonometric functions.
I
Definite integrals.
I
Substitution and integration by parts.
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Solution: We start with the substitution y = ln(x),
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Solution: We start with the substitution y = ln(x), and dy =
dx
.
x
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Solution: We start with the substitution y = ln(x), and dy =
dx = x dy
dx
.
x
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Solution: We start with the substitution y = ln(x), and dy =
dx = x dy = e y dy .
dx
.
x
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Solution: We start with the substitution y = ln(x), and dy =
dx = x dy = e y dy .
Z
The integral is I =
cos(y ) e y dy ,
dx
.
x
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Solution: We start with the substitution y = ln(x), and dy =
dx
.
x
dx = x dy = e y dy .
Z
The integral is I =
cos(y ) e y dy , and we integrate by parts.
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Solution: We start with the substitution y = ln(x), and dy =
dx
.
x
dx = x dy = e y dy .
Z
The integral is I =
cos(y ) e y dy , and we integrate by parts.
If u = e y , dv = cos(y ) dy , then du = e y dy , v = sin(y ),
Substitution and integration by parts
Remark: Substitution and integration by parts can be used on the
same integral.
Example
Z
Evaluate I =
cos ln(x) dx.
Solution: We start with the substitution y = ln(x), and dy =
dx
.
x
dx = x dy = e y dy .
Z
The integral is I =
cos(y ) e y dy , and we integrate by parts.
If u = e y , dv = cos(y ) dy , then du = e y dy , v = sin(y ),
Z
Z
y
y
e cos(y ) dy = e sin(y ) − e y sin(y ) dy .
Substitution and integration by parts
Example
Z
cos ln(x) dx.
Z
Z
e y cos(y ) dy = e y sin(y ) − e y sin(y ) dy .
Solution: Recall:
Evaluate I =
Substitution and integration by parts
Example
Z
cos ln(x) dx.
Z
Z
e y cos(y ) dy = e y sin(y ) − e y sin(y ) dy .
Solution: Recall:
Evaluate I =
One more integration by parts,
Substitution and integration by parts
Example
Z
cos ln(x) dx.
Z
Z
e y cos(y ) dy = e y sin(y ) − e y sin(y ) dy .
Solution: Recall:
Evaluate I =
One more integration by parts,
u = ey ,
dv = sin(y ) dy ,
Substitution and integration by parts
Example
Z
cos ln(x) dx.
Z
Z
e y cos(y ) dy = e y sin(y ) − e y sin(y ) dy .
Solution: Recall:
Evaluate I =
One more integration by parts,
u = ey ,
dv = sin(y ) dy ,
⇒
du = e y dy ,
v = − cos(y ).
Substitution and integration by parts
Example
Z
cos ln(x) dx.
Z
Z
e y cos(y ) dy = e y sin(y ) − e y sin(y ) dy .
Solution: Recall:
Evaluate I =
One more integration by parts,
u = e y , dv = sin(y ) dy , ⇒ du = e y dy , v = − cos(y ).
Z
Z
h
i
y
y
y
e cos(y ) dy = e sin(y ) − −e cos(y ) + e y cos(y ) dy .
Substitution and integration by parts
Example
Z
cos ln(x) dx.
Z
Z
e y cos(y ) dy = e y sin(y ) − e y sin(y ) dy .
Solution: Recall:
Evaluate I =
One more integration by parts,
u = e y , dv = sin(y ) dy , ⇒ du = e y dy , v = − cos(y ).
Z
Z
h
i
y
y
y
e cos(y ) dy = e sin(y ) − −e cos(y ) + e y cos(y ) dy .
Z
2
e y cos(y ) dy = e y sin(y ) + e y cos(y )
Substitution and integration by parts
Example
Z
cos ln(x) dx.
Z
Z
e y cos(y ) dy = e y sin(y ) − e y sin(y ) dy .
Solution: Recall:
Evaluate I =
One more integration by parts,
u = e y , dv = sin(y ) dy , ⇒ du = e y dy , v = − cos(y ).
Z
Z
h
i
y
y
y
e cos(y ) dy = e sin(y ) − −e cos(y ) + e y cos(y ) dy .
Z
2
Z
e y cos(y ) dy = e y sin(y ) + e y cos(y )
e y cos(y ) =
ey sin(y ) + cos(y ) .
2
Substitution and integration by parts
Example
Z
cos ln(x) dx.
Z
Z
e y cos(y ) dy = e y sin(y ) − e y sin(y ) dy .
Solution: Recall:
Evaluate I =
One more integration by parts,
u = e y , dv = sin(y ) dy , ⇒ du = e y dy , v = − cos(y ).
Z
Z
h
i
y
y
y
e cos(y ) dy = e sin(y ) − −e cos(y ) + e y cos(y ) dy .
Z
2
e y cos(y ) dy = e y sin(y ) + e y cos(y )
ey sin(y ) + cos(y ) .
2
Z
x
We conclude:
cos ln(x) dx = sin(ln(x))) + cos(ln(x)) . C
2
Z
e y cos(y ) =
Trigonometric integrals (Sect. 8.2)
I
Product of sines and cosines.
I
Eliminating square roots.
I
Integrals of tangents and secants.
I
Products of sines and cosines.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
(a) If m = 2k + 1, (odd),
sinm (x) cosn (x) dx.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(a) If m = 2k + 1, (odd), then sin(2k+1) (x) =
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(a) If m = 2k + 1, (odd), then sin(2k+1) (x) = sin2 (x) sin(x);
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(a) If m = 2k + 1, (odd), then sin(2k+1) (x) = sin2 (x) sin(x);
Z
k
I =
1 − cos2 (x) cosn (x) sin(x) dx.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(a) If m = 2k + 1, (odd), then sin(2k+1) (x) = sin2 (x) sin(x);
Z
k
I =
1 − cos2 (x) cosn (x) sin(x) dx.
Substitute u = cos(x), so
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(a) If m = 2k + 1, (odd), then sin(2k+1) (x) = sin2 (x) sin(x);
Z
k
I =
1 − cos2 (x) cosn (x) sin(x) dx.
Substitute u = cos(x), so du = − sin(x) dx,
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(a) If m = 2k + 1, (odd), then sin(2k+1) (x) = sin2 (x) sin(x);
Z
k
I =
1 − cos2 (x) cosn (x) sin(x) dx.
Substitute u = cos(x), so du = − sin(x) dx, hence
Z
I = − (1 − u 2 )k u n du.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(a) If m = 2k + 1, (odd), then sin(2k+1) (x) = sin2 (x) sin(x);
Z
k
I =
1 − cos2 (x) cosn (x) sin(x) dx.
Substitute u = cos(x), so du = − sin(x) dx, hence
Z
I = − (1 − u 2 )k u n du.
We now need to integrate a polynomial.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
(b) If n = 2k + 1, (odd),
sinm (x) cosn (x) dx.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(b) If n = 2k + 1, (odd), then cos(2k+1) (x) =
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(b) If n = 2k + 1, (odd), then cos(2k+1) (x) = cos2 (x) cos(x);
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(b) If n = 2k + 1, (odd), then cos(2k+1) (x) = cos2 (x) cos(x);
Z
k
I = sinm (x) 1 − sin2 (x) cos(x) dx.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(b) If n = 2k + 1, (odd), then cos(2k+1) (x) = cos2 (x) cos(x);
Z
k
I = sinm (x) 1 − sin2 (x) cos(x) dx.
Substitute u = sin(x),
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(b) If n = 2k + 1, (odd), then cos(2k+1) (x) = cos2 (x) cos(x);
Z
k
I = sinm (x) 1 − sin2 (x) cos(x) dx.
Substitute u = sin(x), so du = cos(x) dx,
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(b) If n = 2k + 1, (odd), then cos(2k+1) (x) = cos2 (x) cos(x);
Z
k
I = sinm (x) 1 − sin2 (x) cos(x) dx.
Substitute u = sin(x), so du = cos(x) dx, hence
Z
I = u m (1 − u 2 )k du.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
k
(b) If n = 2k + 1, (odd), then cos(2k+1) (x) = cos2 (x) cos(x);
Z
k
I = sinm (x) 1 − sin2 (x) cos(x) dx.
Substitute u = sin(x), so du = cos(x) dx, hence
Z
I = u m (1 − u 2 )k du.
Again, we now need to integrate a polynomial.
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even,
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even, say m = 2k and n = 2`,
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even, say m = 2k and n = 2`, then
Z
I = sin2k (x) cos2` (x) dx
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even, say m = 2k and n = 2`, then
Z
Z
k
`
2k
2`
sin2 (x)
cos2 (x) dx.
I = sin (x) cos (x) dx =
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even, say m = 2k and n = 2`, then
Z
Z
k
`
2k
2`
sin2 (x)
cos2 (x) dx.
I = sin (x) cos (x) dx =
Now use the identities
sin2 (x) =
1
1 − cos(2x) ,
2
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even, say m = 2k and n = 2`, then
Z
Z
k
`
2k
2`
sin2 (x)
cos2 (x) dx.
I = sin (x) cos (x) dx =
Now use the identities
sin2 (x) =
1
1 − cos(2x) ,
2
cos2 (x) =
1
1 + cos(2x) .
2
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even, say m = 2k and n = 2`, then
Z
Z
k
`
2k
2`
sin2 (x)
cos2 (x) dx.
I = sin (x) cos (x) dx =
Now use the identities
sin2 (x) =
1
1 − cos(2x) ,
2
cos2 (x) =
Depending whether k or ` are odd,
1
1 + cos(2x) .
2
Product of sines and cosines
Remark: There is a procedure to compute integrals of the form
Z
I =
sinm (x) cosn (x) dx.
(c) If both m and n are even, say m = 2k and n = 2`, then
Z
Z
k
`
2k
2`
sin2 (x)
cos2 (x) dx.
I = sin (x) cos (x) dx =
Now use the identities
sin2 (x) =
1
1 − cos(2x) ,
2
cos2 (x) =
1
1 + cos(2x) .
2
Depending whether k or ` are odd, repeat (a), (b) or (c).
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd,
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
I = sin4+1 (x) dx
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Z
2
I =
1 − cos2 (x) sin(x) dx.
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Z
2
I =
1 − cos2 (x) sin(x) dx.
Introduce the substitution
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Z
2
I =
1 − cos2 (x) sin(x) dx.
Introduce the substitution u = cos(x),
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Z
2
I =
1 − cos2 (x) sin(x) dx.
Introduce the substitution u = cos(x), then du = − sin(x) dx,
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Z
2
I =
1 − cos2 (x) sin(x) dx.
Introduce the substitution u = cos(x), then du = − sin(x) dx,
Z
I =−
1 − u 2 )2 du
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Z
2
I =
1 − cos2 (x) sin(x) dx.
Introduce the substitution u = cos(x), then du = − sin(x) dx,
Z
Z
I =−
1 − u 2 )2 du = − (1 − 2u 2 + u 4 ) du.
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Z
2
I =
1 − cos2 (x) sin(x) dx.
Introduce the substitution u = cos(x), then du = − sin(x) dx,
Z
Z
I =−
1 − u 2 )2 du = − (1 − 2u 2 + u 4 ) du.
I = −u + 2
u3 u5
−
+ c.
3
5
Product of sines and cosines
Example
Z
Evaluate I =
sin5 (x) dx.
Solution: Since m = 5 is odd, we write it as m = 4 + 1,
Z
Z
2
I = sin4+1 (x) dx =
sin2 (x) sin(x) dx
Z
2
I =
1 − cos2 (x) sin(x) dx.
Introduce the substitution u = cos(x), then du = − sin(x) dx,
Z
Z
I =−
1 − u 2 )2 du = − (1 − 2u 2 + u 4 ) du.
u3 u5
−
+ c.
3
5
2
1
We conclude I = − cos(x) + cos3 (x) − cos5 (x) + c.
3
5
I = −u + 2
C
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even,
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even, we write it as m = 2(3),
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even, we write it as m = 2(3),
Z
3
I =
sin2 (x) dx
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even, we write it as m = 2(3),
Z Z
3
3
1
[1 − cos(2x)] dx
I =
sin2 (x) dx =
2
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even, we write it as m = 2(3),
Z Z
3
3
1
[1 − cos(2x)] dx
I =
sin2 (x) dx =
2
I =
1
8
Z
1 − 3 cos(2x) + 3 cos2 (2x) − cos3 (2x) dx.
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even, we write it as m = 2(3),
Z Z
3
3
1
[1 − cos(2x)] dx
I =
sin2 (x) dx =
2
I =
1
8
Z
1 − 3 cos(2x) + 3 cos2 (2x) − cos3 (2x) dx.
Z
The first two terms are:
(1 − 3 cos(2x)) dx = x −
3
sin(2x).
2
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even, we write it as m = 2(3),
Z Z
3
3
1
[1 − cos(2x)] dx
I =
sin2 (x) dx =
2
I =
1
8
Z
1 − 3 cos(2x) + 3 cos2 (2x) − cos3 (2x) dx.
Z
The first two terms are:
(1 − 3 cos(2x)) dx = x −
The third term can be integrated as follows,
Z
Z
1
2
(1 + cos(4x)) dx
3 cos (2x) dx = 3
2
3
sin(2x).
2
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: Since m = 6 is even, we write it as m = 2(3),
Z Z
3
3
1
[1 − cos(2x)] dx
I =
sin2 (x) dx =
2
I =
1
8
Z
1 − 3 cos(2x) + 3 cos2 (2x) − cos3 (2x) dx.
Z
The first two terms are:
(1 − 3 cos(2x)) dx = x −
3
sin(2x).
2
The third term can be integrated as follows,
Z
Z
1
3
1
2
(1 + cos(4x)) dx =
3 cos (2x) dx = 3
x + sin(4x) .
2
2
4
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
J = cos2 (2x) cos(2x) dx
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
Z
J = cos2 (2x) cos(2x) dx =
1 − sin2 (2x) cos(2x) dx.
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
Z
J = cos2 (2x) cos(2x) dx =
1 − sin2 (2x) cos(2x) dx.
Introduce the substitution u = sin(2x),
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
Z
J = cos2 (2x) cos(2x) dx =
1 − sin2 (2x) cos(2x) dx.
Introduce the substitution u = sin(2x), then du = 2 cos(2x) dx.
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
Z
J = cos2 (2x) cos(2x) dx =
1 − sin2 (2x) cos(2x) dx.
Introduce the substitution u = sin(2x), then du = 2 cos(2x) dx.
Z
1
J=
(1 − u 2 ) du
2
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
Z
J = cos2 (2x) cos(2x) dx =
1 − sin2 (2x) cos(2x) dx.
Introduce the substitution u = sin(2x), then du = 2 cos(2x) dx.
Z
1
1
u3 J=
(1 − u 2 ) du =
u−
2
2
3
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
Z
J = cos2 (2x) cos(2x) dx =
1 − sin2 (2x) cos(2x) dx.
Introduce the substitution u = sin(2x), then du = 2 cos(2x) dx.
Z
1
1
u3 1
1
J=
(1 − u 2 ) du =
u−
= sin(2x) − sin3 (2x).
2
2
3
2
6
Product of sines and cosines
Example
Z
Evaluate I =
sin6 (x) dx.
Solution: So far we have found that
i 1 Z
1h
3
3
1
I = x − sin(2x) +
x + sin(4x) −
cos3 (2x) dx.
8
2
2
4
8
Z
The last term J = cos3 (2x) dx can we computed as follows,
Z
Z
J = cos2 (2x) cos(2x) dx =
1 − sin2 (2x) cos(2x) dx.
Introduce the substitution u = sin(2x), then du = 2 cos(2x) dx.
Z
1
1
u3 1
1
J=
(1 − u 2 ) du =
u−
= sin(2x) − sin3 (2x).
2
2
3
2
6
h
i
1
3
3
1
1
3
I = x − sin(2x) + x + sin(4x) − sin(2x) + sin3 (2x) + c.
8
2
2
8
2
6
Trigonometric integrals (Sect. 8.2)
I
Product of sines and cosines.
I
Eliminating square roots.
I
Integrals of tangents and secants.
I
Products of sines and cosines.
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
These identities can be used to simplify certain square roots.
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
I
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
These identities can be used to simplify certain square roots.
The same holds for Pythagoras Theorem,
sin2 (θ) = 1 − cos2 (θ),
cos2 (θ) = 1 − sin2 (θ).
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
I
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
These identities can be used to simplify certain square roots.
The same holds for Pythagoras Theorem,
sin2 (θ) = 1 − cos2 (θ),
Example
Z
Evaluate I =
0
π/8 p
1 + cos(8x) dx.
cos2 (θ) = 1 − sin2 (θ).
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
I
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
These identities can be used to simplify certain square roots.
The same holds for Pythagoras Theorem,
sin2 (θ) = 1 − cos2 (θ),
cos2 (θ) = 1 − sin2 (θ).
Example
Z
Evaluate I =
π/8 p
1 + cos(8x) dx.
0
Solution: Use that : 1 + cos(8x) = 2 cos2 (4x).
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
I
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
These identities can be used to simplify certain square roots.
The same holds for Pythagoras Theorem,
sin2 (θ) = 1 − cos2 (θ),
cos2 (θ) = 1 − sin2 (θ).
Example
Z
Evaluate I =
π/8 p
1 + cos(8x) dx.
0
Solution: Use that : 1 + cos(8x) = 2 cos2 (4x). Hence,
√ Z π/8
I = 2
cos(4x) dx
0
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
I
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
These identities can be used to simplify certain square roots.
The same holds for Pythagoras Theorem,
sin2 (θ) = 1 − cos2 (θ),
cos2 (θ) = 1 − sin2 (θ).
Example
Z
Evaluate I =
π/8 p
1 + cos(8x) dx.
0
Solution: Use that : 1 + cos(8x) = 2 cos2 (4x). Hence,
√
π/8
√ Z π/8
2
sin(4x)
I = 2
cos(4x) dx =
4
0
0
Eliminating square roots
Remarks:
I
Recall the double angle identities:
sin2 (θ) =
I
1
1 − cos(2θ) ,
2
cos2 (θ) =
1
1 + cos(2θ) .
2
These identities can be used to simplify certain square roots.
The same holds for Pythagoras Theorem,
sin2 (θ) = 1 − cos2 (θ),
cos2 (θ) = 1 − sin2 (θ).
Example
Z
Evaluate I =
π/8 p
1 + cos(8x) dx.
0
Solution: Use that : 1 + cos(8x) = 2 cos2 (4x). Hence,
√
√
π/8
√ Z π/8
2
2
sin(4x)
⇒ I =
.
I = 2
cos(4x) dx =
4
4
0
0
Trigonometric integrals (Sect. 8.2)
I
Product of sines and cosines.
I
Eliminating square roots.
I
Integrals of tangents and secants.
I
Products of sines and cosines.
Integrals of tangents and secants
Remark: Recall the identities:
tan0 (x) = sec2 (x) = tan2 (x) + 1.
Integrals of tangents and secants
Remark: Recall the identities:
tan0 (x) = sec2 (x) = tan2 (x) + 1.
First equation comes from quotient rule,
Integrals of tangents and secants
Remark: Recall the identities:
tan0 (x) = sec2 (x) = tan2 (x) + 1.
First equation comes from quotient rule, the second from
Pythagoras Theorem.
Integrals of tangents and secants
Remark: Recall the identities:
tan0 (x) = sec2 (x) = tan2 (x) + 1.
First equation comes from quotient rule, the second from
Pythagoras Theorem. These identities can be used to compute
Z
I = tan2k (x) dx,
k ∈ N.
Integrals of tangents and secants
Remark: Recall the identities:
tan0 (x) = sec2 (x) = tan2 (x) + 1.
First equation comes from quotient rule, the second from
Pythagoras Theorem. These identities can be used to compute
Z
I = tan2k (x) dx,
k ∈ N.
Example
Z
Evaluate I =
tan2 (x) dx.
Integrals of tangents and secants
Remark: Recall the identities:
tan0 (x) = sec2 (x) = tan2 (x) + 1.
First equation comes from quotient rule, the second from
Pythagoras Theorem. These identities can be used to compute
Z
I = tan2k (x) dx,
k ∈ N.
Example
Z
Evaluate I =
tan2 (x) dx.
Solution: The identity on the far left above implies
Z
I =
tan0 (x) − 1 dx
Integrals of tangents and secants
Remark: Recall the identities:
tan0 (x) = sec2 (x) = tan2 (x) + 1.
First equation comes from quotient rule, the second from
Pythagoras Theorem. These identities can be used to compute
Z
I = tan2k (x) dx,
k ∈ N.
Example
Z
Evaluate I =
tan2 (x) dx.
Solution: The identity on the far left above implies
Z
I =
tan0 (x) − 1 dx ⇒ I = tan(x) − x + c.
C
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
I = tan(2k−2) (x) tan2 (x) dx
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
(2k−2)
2
I = tan
(x) tan (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
(2k−2)
2
I = tan
(x) tan (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Z
I =
(2k−2)
tan
0
(x) tan (x) dx −
Z
tan(2k−2) (x) dx.
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
(2k−2)
2
I = tan
(x) tan (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Z
I =
(2k−2)
tan
0
(x) tan (x) dx −
In the first term on the right,
Z
tan(2k−2) (x) dx.
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
(2k−2)
2
I = tan
(x) tan (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Z
I =
(2k−2)
tan
0
Z
(x) tan (x) dx −
In the first term on the right, u = tan(x),
tan(2k−2) (x) dx.
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
(2k−2)
2
I = tan
(x) tan (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Z
I =
(2k−2)
tan
0
(x) tan (x) dx −
Z
tan(2k−2) (x) dx.
In the first term on the right, u = tan(x), then du = tan0 (x) dx,
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
(2k−2)
2
I = tan
(x) tan (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Z
I =
(2k−2)
tan
0
(x) tan (x) dx −
Z
tan(2k−2) (x) dx.
In the first term on the right, u = tan(x), then du = tan0 (x) dx,
Z
Z
tan(2k−2) (x) tan0 (x) dx = u (2k−2) du
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
(2k−2)
2
I = tan
(x) tan (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Z
I =
(2k−2)
tan
0
(x) tan (x) dx −
Z
tan(2k−2) (x) dx.
In the first term on the right, u = tan(x), then du = tan0 (x) dx,
Z
Z
u (2k−1)
tan(2k−2) (x) tan0 (x) dx = u (2k−2) du =
.
(2k − 1)
Integrals of tangents and secants
Example
Z
Find a recurrence formula to compute I =
tan2k (x) dx, k ∈ N.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Z
Z
(2k−2)
2
I = tan
(x) tan (x) dx = tan(2k−2) (x) tan0 (x) − 1 dx
Z
I =
(2k−2)
tan
0
(x) tan (x) dx −
Z
tan(2k−2) (x) dx.
In the first term on the right, u = tan(x), then du = tan0 (x) dx,
Z
Z
u (2k−1)
tan(2k−2) (x) tan0 (x) dx = u (2k−2) du =
.
(2k − 1)
Z
1
(2k−1)
C
I =
tan
(x) − tan2(k−1) (x) dx.
(2k − 1)
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Rewrite the integral as follows,
Z
I = sec(x) sec2 (x) dx
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Rewrite the integral as follows,
Z
Z
2
I = sec(x) sec (x) dx = sec(x) tan0 (x) dx.
Where we used that sec2 (x) = tan0 (x).
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Rewrite the integral as follows,
Z
Z
2
I = sec(x) sec (x) dx = sec(x) tan0 (x) dx.
Where we used that sec2 (x) = tan0 (x). Integrate by parts,
u = sec(x),
dv = tan0 (x) dx
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Rewrite the integral as follows,
Z
Z
2
I = sec(x) sec (x) dx = sec(x) tan0 (x) dx.
Where we used that sec2 (x) = tan0 (x). Integrate by parts,
u = sec(x),
dv = tan0 (x) dx ⇒ du = sec0 (x) dx,
v = tan(x).
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Rewrite the integral as follows,
Z
Z
2
I = sec(x) sec (x) dx = sec(x) tan0 (x) dx.
Where we used that sec2 (x) = tan0 (x). Integrate by parts,
u = sec(x),
dv = tan0 (x) dx ⇒ du = sec0 (x) dx, v = tan(x).
Z
I = sec(x) tan(x) − tan(x) sec0 (x) dx.
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Rewrite the integral as follows,
Z
Z
2
I = sec(x) sec (x) dx = sec(x) tan0 (x) dx.
Where we used that sec2 (x) = tan0 (x). Integrate by parts,
u = sec(x),
dv = tan0 (x) dx ⇒ du = sec0 (x) dx, v = tan(x).
Z
I = sec(x) tan(x) − tan(x) sec0 (x) dx.
Recall: sec0 (x) =
sin(x)
cos2 (x)
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Solution: Recall: tan0 (x) = sec2 (x) = tan2 (x) + 1.
Rewrite the integral as follows,
Z
Z
2
I = sec(x) sec (x) dx = sec(x) tan0 (x) dx.
Where we used that sec2 (x) = tan0 (x). Integrate by parts,
u = sec(x),
dv = tan0 (x) dx ⇒ du = sec0 (x) dx, v = tan(x).
Z
I = sec(x) tan(x) − tan(x) sec0 (x) dx.
Recall: sec0 (x) =
sin(x)
= sec(x) tan(x).
cos2 (x)
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Z
Solution: I = sec(x) tan(x) −
know sec0 (x) = sec(x) tan(x).
tan(x) sec0 (x) dx, and we also
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Z
Solution: I = sec(x) tan(x) −
tan(x) sec0 (x) dx, and we also
know sec0 (x) = sec(x) tan(x).
Z
I = sec(x) tan(x) −
sec(x) tan2 (x) dx
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Z
Solution: I = sec(x) tan(x) −
tan(x) sec0 (x) dx, and we also
know sec0 (x) = sec(x) tan(x).
Z
I = sec(x) tan(x) − sec(x) tan2 (x) dx
Z
I = sec(x) tan(x) − sec(x) (sec2 (x) − 1) dx
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Z
Solution: I = sec(x) tan(x) −
tan(x) sec0 (x) dx, and we also
know sec0 (x) = sec(x) tan(x).
Z
I = sec(x) tan(x) − sec(x) tan2 (x) dx
Z
I = sec(x) tan(x) − sec(x) (sec2 (x) − 1) dx
Z
3
sec (x) dx = sec(x) tan(x) +
Z
Z
sec(x) dx −
sec3 (x) dx.
Integrals of tangents and secants
Example
Z
Evaluate I =
sec3 (x) dx.
Z
Solution: I = sec(x) tan(x) −
tan(x) sec0 (x) dx, and we also
know sec0 (x) = sec(x) tan(x).
Z
I = sec(x) tan(x) − sec(x) tan2 (x) dx
Z
I = sec(x) tan(x) − sec(x) (sec2 (x) − 1) dx
Z
3
Z
sec (x) dx = sec(x) tan(x) +
Z
sec3 (x) dx =
Z
sec(x) dx −
sec3 (x) dx.
1
1
sec(x) tan(x) + ln sec(x) + tan(x) + c.
2
2
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
Proof:
Z
I =
sec(x) dx
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
Proof:
Z
I =
Z
sec(x) dx =
1
dx
cos(x)
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
Proof:
Z
I =
Z
I =
Z
sec(x) dx =
1
dx
cos(x)
[1 + sin(x)] cos(x)
1
dx
cos(x)
cos(x) [1 + sin(x)]
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
Proof:
Z
I =
Z
Z
sec(x) dx =
1
dx
cos(x)
[1 + sin(x)] cos(x)
1
dx
cos(x)
cos(x) [1 + sin(x)]
Z
[1 + sin(x)]
1
I =
dx
cos2 (x) [1 + sin(x)] cos(x)
Z 1
[1 + sin(x)] 0
I =
[1 + sin(x)] dx
cos(x)
cos(x)
I =
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
I =
Z [1 + sin(x)] 0
1
[1 + sin(x)] dx
cos(x)
cos(x)
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
I =
Z
I =
Z [1 + sin(x)] 0
1
[1 + sin(x)] dx
cos(x)
cos(x)
0
sec(x) + tan(x)
Substitute u = sec(x) + tan(x), then
1
dx
sec(x) + tan(x)
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
I =
Z
I =
Z [1 + sin(x)] 0
1
[1 + sin(x)] dx
cos(x)
cos(x)
0
sec(x) + tan(x)
Substitute u = sec(x) + tan(x), then
Z
du
I =
u
1
dx
sec(x) + tan(x)
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
I =
Z
I =
Z [1 + sin(x)] 0
1
[1 + sin(x)] dx
cos(x)
cos(x)
0
sec(x) + tan(x)
1
dx
sec(x) + tan(x)
Substitute u = sec(x) + tan(x), then
Z
du
I =
= ln(u) + c.
u
Integrals of tangents and secants
Z
Recall:
sec(x) dx = ln sec(x) + tan(x) + c.
I =
Z
I =
Z [1 + sin(x)] 0
1
[1 + sin(x)] dx
cos(x)
cos(x)
0
sec(x) + tan(x)
1
dx
sec(x) + tan(x)
Substitute u = sec(x) + tan(x), then
Z
du
I =
= ln(u) + c.
u
So we obtain the formula,
Z
sec(x) dx = ln sec(x) + tan(x) + c.
Trigonometric integrals (Sect. 8.2)
I
Product of sines and cosines.
I
Eliminating square roots.
I
Integrals of tangents and secants.
I
Products of sines and cosines.
Products of sines and cosines
Remark: The identities
1
cos(θ − φ) − cos(θ + φ)
2
1
sin(θ) cos(φ) = sin(θ − φ) + sin(θ + φ)
2
1
cos(θ) cos(φ) = cos(θ − φ) + cos(θ + φ) .
2
sin(θ) sin(φ) =
can be used to compute integrals of the form
Z
Z
sin(mx) sin(nx) dx,
sin(mx) cos(nx) dx,
Z
cos(mx) cos(nx) dx.
Products of sines and cosines
Example
Z
Evaluate: I =
sin(3x) cos(4x) dx.
Products of sines and cosines
Example
Z
Evaluate: I =
sin(3x) cos(4x) dx.
Solution: Recall: sin(θ) cos(φ) =
1
sin(θ − φ) + sin(θ + φ) .
2
Products of sines and cosines
Example
Z
Evaluate: I =
sin(3x) cos(4x) dx.
1
Solution: Recall: sin(θ) cos(φ) = sin(θ − φ) + sin(θ + φ) .
2
The formula above implies,
Z
1 I =
sin((3 − 4)x) + sin((3 + 4)x) dx,
2
Products of sines and cosines
Example
Z
Evaluate: I =
sin(3x) cos(4x) dx.
1
Solution: Recall: sin(θ) cos(φ) = sin(θ − φ) + sin(θ + φ) .
2
The formula above implies,
Z
1 I =
sin((3 − 4)x) + sin((3 + 4)x) dx,
2
that is,
1
I =
2
Z
− sin(x) + sin(7x) dx.
Products of sines and cosines
Example
Z
Evaluate: I =
sin(3x) cos(4x) dx.
1
Solution: Recall: sin(θ) cos(φ) = sin(θ − φ) + sin(θ + φ) .
2
The formula above implies,
Z
1 I =
sin((3 − 4)x) + sin((3 + 4)x) dx,
2
that is,
1
I =
2
Z
− sin(x) + sin(7x) dx.
This integral is simple to do,
Products of sines and cosines
Example
Z
Evaluate: I =
sin(3x) cos(4x) dx.
1
Solution: Recall: sin(θ) cos(φ) = sin(θ − φ) + sin(θ + φ) .
2
The formula above implies,
Z
1 I =
sin((3 − 4)x) + sin((3 + 4)x) dx,
2
that is,
1
I =
2
Z
− sin(x) + sin(7x) dx.
This integral is simple to do,
I =
i
1h
1
cos(x) − cos(7x) + c.
2
7
C
Trigonometric substitutions (Sect. 8.3)
I
I
I
I
Substitutions to cancel the square root
√
Integrals involving a2 − x 2 : Use x = a sin(θ).
√
Integrals involving a2 + x 2 : Use x = a tan(θ).
√
Integrals involving x 2 − a2 : Use x = a sec(θ).
Substitutions to cancel the square root
Remark: Integrals involving
substitution x = a sin(θ).
√
a2 − x 2 can be found with the
Substitutions to cancel the square root
√
a2 − x 2 can be found with the
substitution x = a sin(θ). Indeed,
q
p
2
2
a − x = a2 − a2 sin2 (θ)
Remark: Integrals involving
Substitutions to cancel the square root
√
a2 − x 2 can be found with the
substitution x = a sin(θ). Indeed,
q
q
p
2
2
2
2
2
a − x = a − a sin (θ) = |a| 1 − sin2 (θ)
Remark: Integrals involving
Substitutions to cancel the square root
√
a2 − x 2 can be found with the
substitution x = a sin(θ). Indeed,
q
q
p
2
2
2
2
2
a − x = a − a sin (θ) = |a| 1 − sin2 (θ) = |a| | cos(θ)|.
Remark: Integrals involving
Substitutions to cancel the square root
√
a2 − x 2 can be found with the
substitution x = a sin(θ). Indeed,
q
q
p
2
2
2
2
2
a − x = a − a sin (θ) = |a| 1 − sin2 (θ) = |a| | cos(θ)|.
Remark: Integrals involving
We conclude that
√
a2 − x 2 = |a| | cos(θ)|.
Substitutions to cancel the square root
√
a2 − x 2 can be found with the
substitution x = a sin(θ). Indeed,
q
q
p
2
2
2
2
2
a − x = a − a sin (θ) = |a| 1 − sin2 (θ) = |a| | cos(θ)|.
Remark: Integrals involving
We conclude that
√
a2 − x 2 = |a| | cos(θ)|.
Notice: The substitution x = a cos(θ) also works.
Substitutions to cancel the square root
√
a2 − x 2 can be found with the
substitution x = a sin(θ). Indeed,
q
q
p
2
2
2
2
2
a − x = a − a sin (θ) = |a| 1 − sin2 (θ) = |a| | cos(θ)|.
Remark: Integrals involving
We conclude that
√
a2 − x 2 = |a| | cos(θ)|.
Notice: The substitution x = a cos(θ) also works.
Remark: We have used Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1.
Substitutions to cancel the square root
√
a2 − x 2 can be found with the
substitution x = a sin(θ). Indeed,
q
q
p
2
2
2
2
2
a − x = a − a sin (θ) = |a| 1 − sin2 (θ) = |a| | cos(θ)|.
Remark: Integrals involving
We conclude that
√
a2 − x 2 = |a| | cos(θ)|.
Notice: The substitution x = a cos(θ) also works.
Remark: We have used Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1.
To compute dx we will need the following derivatives:
sin0 (θ) = cos(θ),
cos0 (θ) = − sin(θ).
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ),
sec0 (θ) = sec(θ) tan(θ).
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ).
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
p
2
2
a + x = a2 + a2 tan2 (θ)
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ)
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
√
Hence, a2 + x 2 = |a| | sec(θ)|,
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
√
Hence, a2 + x 2 = |a| | sec(θ)|, and dx = a sec2 (θ) dθ.
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
√
Hence, a2 + x 2 = |a| | sec(θ)|, and dx = a sec2 (θ) dθ.
√
Remark: Integrals involving x 2 − a2 can be found with the
substitution x = a sec(θ).
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
√
Hence, a2 + x 2 = |a| | sec(θ)|, and dx = a sec2 (θ) dθ.
√
Remark: Integrals involving x 2 − a2 can be found with the
substitution x = a sec(θ). Indeed,
q
p
2
2
x − a = a2 sec2 (θ) − a2
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
√
Hence, a2 + x 2 = |a| | sec(θ)|, and dx = a sec2 (θ) dθ.
√
Remark: Integrals involving x 2 − a2 can be found with the
substitution x = a sec(θ). Indeed,
q
q
p
2
2
2
2
2
x − a = a sec (θ) − a = |a| sec2 (θ) − 1
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
√
Hence, a2 + x 2 = |a| | sec(θ)|, and dx = a sec2 (θ) dθ.
√
Remark: Integrals involving x 2 − a2 can be found with the
substitution x = a sec(θ). Indeed,
q
q
p
2
2
2
2
2
x − a = a sec (θ) − a = |a| sec2 (θ) − 1 = |a| | tan(θ)|.
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
√
Hence, a2 + x 2 = |a| | sec(θ)|, and dx = a sec2 (θ) dθ.
√
Remark: Integrals involving x 2 − a2 can be found with the
substitution x = a sec(θ). Indeed,
q
q
p
2
2
2
2
2
x − a = a sec (θ) − a = |a| sec2 (θ) − 1 = |a| | tan(θ)|.
√
Hence, x 2 − a2 = |a| | tan(θ)|,
Substitutions to cancel the square root
Recall:
sec2 (θ) = tan2 (θ) + 1 = tan0 (θ), sec0 (θ) = sec(θ) tan(θ).
√
Remark: Integrals involving a2 + x 2 can be found with the
substitution x = a tan(θ). Indeed,
q
q
p
2
2
2
2
2
a + x = a + a tan (θ) = |a| 1 + tan2 (θ) = |a| | sec(θ)|.
√
Hence, a2 + x 2 = |a| | sec(θ)|, and dx = a sec2 (θ) dθ.
√
Remark: Integrals involving x 2 − a2 can be found with the
substitution x = a sec(θ). Indeed,
q
q
p
2
2
2
2
2
x − a = a sec (θ) − a = |a| sec2 (θ) − 1 = |a| | tan(θ)|.
√
Hence, x 2 − a2 = |a| | tan(θ)|, and dx = a sec(θ) tan(θ) dθ.
Trigonometric substitutions (Sect. 8.3)
I
I
I
I
Substitutions to cancel the square root
√
Integrals involving a2 − x 2 : Use x = a sin(θ).
√
Integrals involving a2 + x 2 : Use x = a tan(θ).
√
Integrals involving x 2 − a2 : Use x = a sec(θ).
Integrals involving
√
a2 − x 2
Example
Z
5
Evaluate I =
−5
p
25 − x 2 dx.
Integrals involving
√
a2 − x 2
Example
Z
5
Evaluate I =
p
25 − x 2 dx.
−5
Solution: Recall Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1
Integrals involving
√
a2 − x 2
Example
Z
5
Evaluate I =
p
25 − x 2 dx.
−5
Solution: Recall Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1
⇒
| cos(θ)| =
q
1 − sin2 (θ).
Integrals involving
√
a2 − x 2
Example
Z
5
Evaluate I =
p
25 − x 2 dx.
−5
Solution: Recall Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1
Substitute: x = 5 sin(θ),
⇒
| cos(θ)| =
q
1 − sin2 (θ).
Integrals involving
√
a2 − x 2
Example
Z
5
Evaluate I =
p
25 − x 2 dx.
−5
Solution: Recall Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1
⇒
| cos(θ)| =
q
Substitute: x = 5 sin(θ), then dx = 5 cos(θ) dθ.
1 − sin2 (θ).
Integrals involving
√
a2 − x 2
Example
Z
5
Evaluate I =
p
25 − x 2 dx.
−5
Solution: Recall Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1
⇒
| cos(θ)| =
q
1 − sin2 (θ).
Substitute: x = 5 sin(θ), then dx = 5 cos(θ) dθ. Hence,
Z π/2 q
I =
25 − 25 sin2 (θ) 5 cos(θ) dθ ,
−π/2
Integrals involving
√
a2 − x 2
Example
Z
5
Evaluate I =
p
25 − x 2 dx.
−5
Solution: Recall Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1
⇒
| cos(θ)| =
q
1 − sin2 (θ).
Substitute: x = 5 sin(θ), then dx = 5 cos(θ) dθ. Hence,
Z π/2 q
I =
25 − 25 sin2 (θ) 5 cos(θ) dθ ,
−π/2
Z
π/2
I =5
−π/2
q
25 1 − sin2 (θ) cos(θ) dθ,
Integrals involving
√
a2 − x 2
Example
Z
5
p
25 − x 2 dx.
Evaluate I =
−5
Solution: Recall Pythagoras Theorem:
sin2 (θ) + cos2 (θ) = 1
⇒
| cos(θ)| =
q
1 − sin2 (θ).
Substitute: x = 5 sin(θ), then dx = 5 cos(θ) dθ. Hence,
Z π/2 q
I =
25 − 25 sin2 (θ) 5 cos(θ) dθ ,
−π/2
Z
π/2
I =5
q
25 1 − sin2 (θ) cos(θ) dθ,
−π/2
I = 52
Z
π/2
−π/2
q
1 − sin2 (θ) cos(θ) dθ.
Integrals involving
√
a2 − x 2
Example
Z
π/2
Evaluate I =
p
25 − x 2 dx.
−π/2
Solution: Recall: I = 52
Z
π/2
−π/2
q
1 − sin2 (θ) cos(θ) dθ.
Integrals involving
√
a2 − x 2
Example
Z
π/2
Evaluate I =
p
25 − x 2 dx.
−π/2
Solution: Recall: I = 52
Z
π/2
q
1 − sin2 (θ) cos(θ) dθ.
−π/2
2
Z
π/2
| cos(θ)| cos(θ) dx
I =5
−π/2
Integrals involving
√
a2 − x 2
Example
Z
π/2
Evaluate I =
p
25 − x 2 dx.
−π/2
Solution: Recall: I = 52
Z
π/2
q
1 − sin2 (θ) cos(θ) dθ.
−π/2
2
Z
π/2
2
Z
π/2
| cos(θ)| cos(θ) dx = 5
I =5
−π/2
−π/2
cos2 (θ) dθ.
Integrals involving
√
a2 − x 2
Example
Z
π/2
Evaluate I =
p
25 − x 2 dx.
−π/2
Solution: Recall: I = 52
Z
π/2
q
1 − sin2 (θ) cos(θ) dθ.
−π/2
2
Z
π/2
2
π/2
| cos(θ)| cos(θ) dx = 5
I =5
−π/2
−π/2
25
I =
2
Z
Z
π/2 −π/2
1 + cos(2θ) dθ
cos2 (θ) dθ.
Integrals involving
√
a2 − x 2
Example
Z
π/2
Evaluate I =
p
25 − x 2 dx.
−π/2
Solution: Recall: I = 52
Z
π/2
q
1 − sin2 (θ) cos(θ) dθ.
−π/2
2
Z
π/2
2
π/2
| cos(θ)| cos(θ) dx = 5
I =5
Z
π/2 −π/2
cos2 (θ) dθ.
−π/2
−π/2
25
I =
2
Z
25 sin(2θ) π/2
1 + cos(2θ) dθ =
θ+
.
2
2
−π/2
Integrals involving
√
a2 − x 2
Example
Z
π/2
Evaluate I =
p
25 − x 2 dx.
−π/2
Solution: Recall: I = 52
Z
π/2
q
1 − sin2 (θ) cos(θ) dθ.
−π/2
2
Z
π/2
2
π/2
| cos(θ)| cos(θ) dx = 5
I =5
Z
cos2 (θ) dθ.
−π/2
−π/2
25
I =
2
Z
π/2 −π/2
25 sin(2θ) π/2
1 + cos(2θ) dθ =
θ+
.
2
2
−π/2
We conclude that I =
25π
.
2
C
Trigonometric substitutions (Sect. 8.3)
I
I
I
I
Substitutions to cancel the square root
√
Integrals involving a2 − x 2 : Use x = a sin(θ).
√
Integrals involving a2 + x 2 : Use x = a tan(θ).
√
Integrals involving x 2 − a2 : Use x = a sec(θ).
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Hence the substitution x = 2 tan(θ),
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Hence the substitution x = 2 tan(θ), for θ ∈ (−π/2, π/2).
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Hence the substitution x = 2 tan(θ), for θ ∈ (−π/2, π/2).
Then, dx = 2 tan0 (θ) dθ,
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Hence the substitution x = 2 tan(θ), for θ ∈ (−π/2, π/2).
Then, dx = 2 tan0 (θ) dθ, that is dx = 2 sec2 (θ) dθ.
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Hence the substitution x = 2 tan(θ), for θ ∈ (−π/2, π/2).
Then, dx = 2 tan0 (θ) dθ, that is dx = 2 sec2 (θ) dθ.
23 tan3 (θ)
Z
I =
p
4 tan2 (θ) + 4
2 sec2 (θ) dθ
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Hence the substitution x = 2 tan(θ), for θ ∈ (−π/2, π/2).
Then, dx = 2 tan0 (θ) dθ, that is dx = 2 sec2 (θ) dθ.
23 tan3 (θ)
Z
I =
p
4 tan2 (θ) + 4
2
2 sec (θ) dθ = 16
Z
tan3 (θ) sec2 (θ)
p
dθ.
4 [tan2 (θ) + 1]
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Hence the substitution x = 2 tan(θ), for θ ∈ (−π/2, π/2).
Then, dx = 2 tan0 (θ) dθ, that is dx = 2 sec2 (θ) dθ.
23 tan3 (θ)
Z
I =
p
4 tan2 (θ) + 4
16
I =
2
Z
2
2 sec (θ) dθ = 16
tan3 (θ) sec2 (θ)
dθ
| sec(θ)|
Z
tan3 (θ) sec2 (θ)
p
dθ.
4 [tan2 (θ) + 1]
Integrals involving
√
a2 + x 2
Example
Z
Evaluate I =
x 3 dx
√
.
x2 + 4
Solution: The only thing that matters to choose the substitution is
the argument of the square root.
In this case we need to recall 1 + tan2 (θ) = sec2 (θ).
Hence the substitution x = 2 tan(θ), for θ ∈ (−π/2, π/2).
Then, dx = 2 tan0 (θ) dθ, that is dx = 2 sec2 (θ) dθ.
23 tan3 (θ)
Z
I =
p
4 tan2 (θ) + 4
16
I =
2
Z
Z
2
2 sec (θ) dθ = 16
tan3 (θ) sec2 (θ)
dθ = 8
| sec(θ)|
Z
tan3 (θ) sec2 (θ)
p
dθ.
4 [tan2 (θ) + 1]
tan3 (θ) sec2 (θ)
dθ.
sec(θ)
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
tan3 (θ) sec2 (θ)
Solution: Recall: I = 8
dθ.
sec(θ)
Z
Evaluate I =
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
tan3 (θ) sec2 (θ)
Solution: Recall: I = 8
dθ.
sec(θ)
Z
I = 8 tan3 (θ) sec(θ) dθ
Z
Evaluate I =
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
tan3 (θ) sec2 (θ)
Solution: Recall: I = 8
dθ.
sec(θ)
Z
Z
I = 8 tan3 (θ) sec(θ) dθ = 8 tan2 (θ) tan(θ) sec(θ) dθ.
Z
Evaluate I =
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
tan3 (θ) sec2 (θ)
Solution: Recall: I = 8
dθ.
sec(θ)
Z
Z
I = 8 tan3 (θ) sec(θ) dθ = 8 tan2 (θ) tan(θ) sec(θ) dθ.
Z
Evaluate I =
Now recall tan(θ) sec(θ) = sec0 (θ),
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
tan3 (θ) sec2 (θ)
Solution: Recall: I = 8
dθ.
sec(θ)
Z
Z
I = 8 tan3 (θ) sec(θ) dθ = 8 tan2 (θ) tan(θ) sec(θ) dθ.
Z
Evaluate I =
Now recall tan(θ) sec(θ) = sec0 (θ), and that tan2 (θ) = sec2 (θ) − 1.
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
tan3 (θ) sec2 (θ)
Solution: Recall: I = 8
dθ.
sec(θ)
Z
Z
I = 8 tan3 (θ) sec(θ) dθ = 8 tan2 (θ) tan(θ) sec(θ) dθ.
Z
Evaluate I =
Now recall tan(θ) sec(θ) = sec0 (θ), and that tan2 (θ) = sec2 (θ) − 1.
Z
2
I =8
sec (θ) − 1 sec0 (θ) dθ.
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
tan3 (θ) sec2 (θ)
Solution: Recall: I = 8
dθ.
sec(θ)
Z
Z
I = 8 tan3 (θ) sec(θ) dθ = 8 tan2 (θ) tan(θ) sec(θ) dθ.
Z
Evaluate I =
Now recall tan(θ) sec(θ) = sec0 (θ), and that tan2 (θ) = sec2 (θ) − 1.
Z
2
I =8
sec (θ) − 1 sec0 (θ) dθ.
We now do a substitution: u = sec(θ),
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
tan3 (θ) sec2 (θ)
Solution: Recall: I = 8
dθ.
sec(θ)
Z
Z
I = 8 tan3 (θ) sec(θ) dθ = 8 tan2 (θ) tan(θ) sec(θ) dθ.
Z
Evaluate I =
Now recall tan(θ) sec(θ) = sec0 (θ), and that tan2 (θ) = sec2 (θ) − 1.
Z
2
I =8
sec (θ) − 1 sec0 (θ) dθ.
We now do a substitution: u = sec(θ), hence du = sec0 (θ) dθ.
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
2
Solution: Recall: I = 8
sec (θ) − 1 sec0 (θ) dθ together with
Z
Evaluate I =
the substitution: u = sec(θ), so du = sec0 (θ) dθ.
Z
I = 8 (u 2 − 1) du
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
2
Solution: Recall: I = 8
sec (θ) − 1 sec0 (θ) dθ together with
Z
Evaluate I =
the substitution: u = sec(θ), so du = sec0 (θ) dθ.
Z
u3
−u+c .
I = 8 (u 2 − 1) du = 8
3
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
2
Solution: Recall: I = 8
sec (θ) − 1 sec0 (θ) dθ together with
Z
Evaluate I =
the substitution: u = sec(θ), so du = sec0 (θ) dθ.
Z
u3
−u+c .
I = 8 (u 2 − 1) du = 8
3
Substitute back u = sec(θ),
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
2
Solution: Recall: I = 8
sec (θ) − 1 sec0 (θ) dθ together with
Z
Evaluate I =
the substitution: u = sec(θ), so du = sec0 (θ) dθ.
Z
u3
−u+c .
I = 8 (u 2 − 1) du = 8
3
Substitute back u = sec(θ), that is,
8
I = sec3 (θ) − 8 sec(θ) + c.
3
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
2
Solution: Recall: I = 8
sec (θ) − 1 sec0 (θ) dθ together with
Z
Evaluate I =
the substitution: u = sec(θ), so du = sec0 (θ) dθ.
Z
u3
−u+c .
I = 8 (u 2 − 1) du = 8
3
Substitute back u = sec(θ), that is,
8
I = sec3 (θ) − 8 sec(θ) + c.
3
Still substitute back x = 2 tan(θ),
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
2
Solution: Recall: I = 8
sec (θ) − 1 sec0 (θ) dθ together with
Z
Evaluate I =
the substitution: u = sec(θ), so du = sec0 (θ) dθ.
Z
u3
−u+c .
I = 8 (u 2 − 1) du = 8
3
Substitute back u = sec(θ), that is,
8
I = sec3 (θ) − 8 sec(θ) + c.
3
Still substitute back x = 2 tan(θ), that is θ = arctan(x/2),
Integrals involving
√
a2 + x 2
Example
x 3 dx
√
.
x2 + 4
Z
2
Solution: Recall: I = 8
sec (θ) − 1 sec0 (θ) dθ together with
Z
Evaluate I =
the substitution: u = sec(θ), so du = sec0 (θ) dθ.
Z
u3
−u+c .
I = 8 (u 2 − 1) du = 8
3
Substitute back u = sec(θ), that is,
8
I = sec3 (θ) − 8 sec(θ) + c.
3
Still substitute back x = 2 tan(θ), that is θ = arctan(x/2), hence
8
I = sec3 arctan(x/2) − 8 sec(arctan(x/2) + c.
C
3
Trigonometric substitutions (Sect. 8.3)
I
I
I
I
Substitutions to cancel the square root
√
Integrals involving a2 − x 2 : Use x = a sin(θ).
√
Integrals involving a2 + x 2 : Use x = a tan(θ).
√
Integrals involving x 2 − a2 : Use x = a sec(θ).
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1,
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1, so sec2 (θ) − 1 = tan2 (θ).
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1, so sec2 (θ) − 1 = tan2 (θ).
The substitution is x = 3 sec(θ),
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1, so sec2 (θ) − 1 = tan2 (θ).
The substitution is x = 3 sec(θ), hence dx = 3 sec0 (θ) dθ.
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1, so sec2 (θ) − 1 = tan2 (θ).
The substitution is x = 3 sec(θ), hence dx = 3 sec0 (θ) dθ.
Z
2
p
I =
3 sec0 (θ) dθ.
2
2
9 sec (θ) 9 sec (θ) − 9
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1, so sec2 (θ) − 1 = tan2 (θ).
The substitution is x = 3 sec(θ), hence dx = 3 sec0 (θ) dθ.
Z
2
p
I =
3 sec0 (θ) dθ.
2
2
9 sec (θ) 9 sec (θ) − 9
2
I =
9
Z
sec0 (θ) dθ
p
sec2 (θ) sec2 (θ) − 1
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1, so sec2 (θ) − 1 = tan2 (θ).
The substitution is x = 3 sec(θ), hence dx = 3 sec0 (θ) dθ.
Z
2
p
I =
3 sec0 (θ) dθ.
2
2
9 sec (θ) 9 sec (θ) − 9
2
I =
9
Z
sec0 (θ) dθ
2
p
=
2
2
9
sec (θ) sec (θ) − 1
Z
sec0 (θ) dθ
.
sec2 (θ)| tan(θ)|
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1, so sec2 (θ) − 1 = tan2 (θ).
The substitution is x = 3 sec(θ), hence dx = 3 sec0 (θ) dθ.
Z
2
p
I =
3 sec0 (θ) dθ.
2
2
9 sec (θ) 9 sec (θ) − 9
2
I =
9
Z
sec0 (θ) dθ
2
p
=
2
2
9
sec (θ) sec (θ) − 1
Recall: sec0 (θ) = sec(θ) tan(θ),
Z
sec0 (θ) dθ
.
sec2 (θ)| tan(θ)|
Integrals involving
√
x 2 − a2
Example
Z
Evaluate I =
x2
2 dx
√
, for x > 0.
x2 − 9
Solution: Once again, it only matters what is in the square root.
We need to recall sec2 (θ) = tan2 (θ) + 1, so sec2 (θ) − 1 = tan2 (θ).
The substitution is x = 3 sec(θ), hence dx = 3 sec0 (θ) dθ.
Z
2
p
I =
3 sec0 (θ) dθ.
2
2
9 sec (θ) 9 sec (θ) − 9
2
I =
9
Z
sec0 (θ) dθ
2
p
=
2
2
9
sec (θ) sec (θ) − 1
Z
sec0 (θ) dθ
.
sec2 (θ)| tan(θ)|
Recall: sec0 (θ) = sec(θ) tan(θ), and x > 0 implies tan(θ) > 0.
Integrals involving
√
x 2 − a2
Example
Z
2 dx
√
, for x > 0.
x2 − 9
Z
2
sec0 (θ) dθ
Solution: So, I =
, and sec0 (θ) = sec(θ) tan(θ).
9
sec2 (θ) tan(θ)
Evaluate I =
x2
Integrals involving
√
x 2 − a2
Example
Z
2 dx
√
, for x > 0.
x2 − 9
Z
2
sec0 (θ) dθ
Solution: So, I =
, and sec0 (θ) = sec(θ) tan(θ).
9
sec2 (θ) tan(θ)
Evaluate I =
I =
2
9
Z
x2
sec(θ) tan(θ) dθ
sec2 (θ) tan(θ)
Integrals involving
√
x 2 − a2
Example
Z
2 dx
√
, for x > 0.
x2 − 9
Z
2
sec0 (θ) dθ
Solution: So, I =
, and sec0 (θ) = sec(θ) tan(θ).
9
sec2 (θ) tan(θ)
Evaluate I =
I =
2
9
Z
x2
sec(θ) tan(θ) dθ
2
=
sec2 (θ) tan(θ)
9
Z
dθ
sec(θ)
Integrals involving
√
x 2 − a2
Example
Z
2 dx
√
, for x > 0.
x2 − 9
Z
2
sec0 (θ) dθ
Solution: So, I =
, and sec0 (θ) = sec(θ) tan(θ).
9
sec2 (θ) tan(θ)
Evaluate I =
I =
2
9
Z
x2
sec(θ) tan(θ) dθ
2
=
sec2 (θ) tan(θ)
9
Z
2
dθ
=
sec(θ)
9
Z
cos(θ) dθ.
Integrals involving
√
x 2 − a2
Example
Z
2 dx
√
, for x > 0.
x2 − 9
Z
2
sec0 (θ) dθ
Solution: So, I =
, and sec0 (θ) = sec(θ) tan(θ).
9
sec2 (θ) tan(θ)
Evaluate I =
I =
2
9
Z
x2
sec(θ) tan(θ) dθ
2
=
sec2 (θ) tan(θ)
9
I =
Z
2
dθ
=
sec(θ)
9
2
sin(θ) + c.
9
Z
cos(θ) dθ.
Integrals involving
√
x 2 − a2
Example
Z
2 dx
√
, for x > 0.
x2 − 9
Z
2
sec0 (θ) dθ
Solution: So, I =
, and sec0 (θ) = sec(θ) tan(θ).
9
sec2 (θ) tan(θ)
Evaluate I =
I =
2
9
Z
x2
sec(θ) tan(θ) dθ
2
=
sec2 (θ) tan(θ)
9
I =
Substitute back x = 3 sec(θ),
Z
2
dθ
=
sec(θ)
9
2
sin(θ) + c.
9
Z
cos(θ) dθ.
Integrals involving
√
x 2 − a2
Example
Z
2 dx
√
, for x > 0.
x2 − 9
Z
2
sec0 (θ) dθ
Solution: So, I =
, and sec0 (θ) = sec(θ) tan(θ).
9
sec2 (θ) tan(θ)
Evaluate I =
I =
2
9
Z
x2
sec(θ) tan(θ) dθ
2
=
sec2 (θ) tan(θ)
9
I =
Z
2
dθ
=
sec(θ)
9
Z
cos(θ) dθ.
2
sin(θ) + c.
9
Substitute back x = 3 sec(θ), that is, θ = arcsec(x/3).
Integrals involving
√
x 2 − a2
Example
Z
2 dx
√
, for x > 0.
x2 − 9
Z
2
sec0 (θ) dθ
Solution: So, I =
, and sec0 (θ) = sec(θ) tan(θ).
9
sec2 (θ) tan(θ)
Evaluate I =
I =
2
9
Z
x2
sec(θ) tan(θ) dθ
2
=
sec2 (θ) tan(θ)
9
I =
Z
2
dθ
=
sec(θ)
9
Z
cos(θ) dθ.
2
sin(θ) + c.
9
Substitute back x = 3 sec(θ), that is, θ = arcsec(x/3).
I =
2
sin arcsec(x/3) + c.
9
C