Download Aside from the most basic relations such as tanx = sin(θ) cos(θ) and

Aside from the most basic relations such as
sin(θ)
1
tan x =
and sec x =
, you should
cos(θ)
cos(θ)
know the following trig identities:
Aside from the most basic relations such as
sin(θ)
1
tan x =
and sec x =
, you should
cos(θ)
cos(θ)
know the following trig identities:
cos2 (θ) + sin2 (θ) = 1.
sec2 (θ) − tan2 (θ) = 1.
1 + cos(2θ)
cos2 (θ) =
2
1 − cos(2θ)
sin2 (θ) =
2
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2 (θ)−sin2 (θ) = 2 cos2 (θ)−1 = 1−2 sin2 (θ)
The following 8 problems are from the Spring
2007 Midterm 1.
The following 8 problems are from the Spring
2007 Midterm 1.
Problem
1 Evaluate: R √
x + e x + 3 sec2 x dx
The following 8 problems are from the Spring
2007 Midterm 1.
Problem
1 Evaluate: R √
x + e x + 3 sec2 x dx
Solution Simplify and then evaluate.
The following 8 problems are from the Spring
2007 Midterm 1.
Problem
1 Evaluate: R √
x + e x + 3 sec2 x dx
Solution Simplify and then evaluate.
Z
√
x + e x + 3 sec2 x dx
The following 8 problems are from the Spring
2007 Midterm 1.
Problem
1 Evaluate: R √
x + e x + 3 sec2 x dx
Solution Simplify and then evaluate.
Z
√
x + e x + 3 sec2 x dx
Z
=
1
2
x dx +
Z
x
e dx + 3
Z
sec2 x dx
The following 8 problems are from the Spring
2007 Midterm 1.
Problem
1 Evaluate: R √
x + e x + 3 sec2 x dx
Solution Simplify and then evaluate.
Z
√
x + e x + 3 sec2 x dx
Z
=
1
2
x dx +
Z
x
e dx + 3
Z
sec2 x dx
2 3
= x 2 + e x + 3 tan x + C .
3
Problem 2 Evaluate:
Z
sin x
dx
1 + cos2 x
Problem 2 Evaluate:
Z
sin x
dx
1 + cos2 x
Solution Make the substitution u = cos x,
du = − sin x dx.
Problem 2 Evaluate:
Z
sin x
dx
1 + cos2 x
Solution Make the substitution u = cos x,
du = − sin x dx.
Z
sin x
dx
1 + cos2 x
Problem 2 Evaluate:
Z
sin x
dx
1 + cos2 x
Solution Make the substitution u = cos x,
du = − sin x dx.
Z
Z
sin x
1
dx
du
=
−
1 + cos2 x
1 + u2
Problem 2 Evaluate:
Z
sin x
dx
1 + cos2 x
Solution Make the substitution u = cos x,
du = − sin x dx.
Z
Z
sin x
1
dx
du
=
−
1 + cos2 x
1 + u2
= − tan−1 (u) + C
Problem 2 Evaluate:
Z
sin x
dx
1 + cos2 x
Solution Make the substitution u = cos x,
du = − sin x dx.
Z
Z
sin x
1
dx
du
=
−
1 + cos2 x
1 + u2
= − tan−1 (u) + C = − tan−1 (cos x) + C .
Problem 3 Evaluate:
R
ln(ln x)
x
dx
R
Problem 3 Evaluate: ln(lnx x) dx
Solution Make the substitution w = ln x,
dw = 1x dx.
R
Problem 3 Evaluate: ln(lnx x) dx
Solution Make the substitution w = ln x,
dw = 1x dx.
Z
Z
ln(ln x)
dx = ln w dw .
x
R
Problem 3 Evaluate: ln(lnx x) dx
Solution Make the substitution w = ln x,
dw = 1x dx.
Z
Z
ln(ln x)
dx = ln w dw .
x
Now perform integration by parts.
R
Problem 3 Evaluate: ln(lnx x) dx
Solution Make the substitution w = ln x,
dw = 1x dx.
Z
Z
ln(ln x)
dx = ln w dw .
x
Now perform integration by parts.
u = ln w
dv = dw
1
du = dw
v = w.
w
R
Problem 3 Evaluate: ln(lnx x) dx
Solution Make the substitution w = ln x,
dw = 1x dx.
Z
Z
ln(ln x)
dx = ln w dw .
x
Now perform integration by parts.
u = ln w
dv = dw
1
du = dw
v = w.
w
Z
ln w dw
R
Problem 3 Evaluate: ln(lnx x) dx
Solution Make the substitution w = ln x,
dw = 1x dx.
Z
Z
ln(ln x)
dx = ln w dw .
x
Now perform integration by parts.
Z
u = ln w
dv = dw
1
du = dw
v = w.
w Z
ln w dw = w ln w −
dw
R
Problem 3 Evaluate: ln(lnx x) dx
Solution Make the substitution w = ln x,
dw = 1x dx.
Z
Z
ln(ln x)
dx = ln w dw .
x
Now perform integration by parts.
Z
u = ln w
dv = dw
1
du = dw
v = w.
w Z
ln w dw = w ln w −
dw = w ln w − w + C .
R
Problem 3 Evaluate: ln(lnx x) dx
Solution Make the substitution w = ln x,
dw = 1x dx.
Z
Z
ln(ln x)
dx = ln w dw .
x
Now perform integration by parts.
Z
u = ln w
dv = dw
1
du = dw
v = w.
w Z
ln w dw = w ln w −
dw = w ln w − w + C .
So the answer is ln x · ln(ln x) − ln x + C.
Z
Problem 4 Evaluate:
x 2 cos x dx
Z
Problem 4 Evaluate:
x 2 cos x dx
Solution Do integration by parts twice.
Z
Problem 4 Evaluate:
x 2 cos x dx
Solution Do integration by parts twice.
u = x2
dv = cos x dx
du = 2x dx
v = sin x
Z
Problem 4 Evaluate:
x 2 cos x dx
Solution Do integration by parts twice.
u = x2
dv = cos x dx
du = 2x dx
Z
v = sin x
Z
x 2 cos x dx = x 2 sin x − 2 x sin x dx.
Z
Problem 4 Evaluate:
x 2 cos x dx
Solution Do integration by parts twice.
u = x2
dv = cos x dx
du = 2x dx
v = sin x
Z
x 2 cos x dx = x 2 sin x − 2 x sin x dx.
R
To evaluate x sin x dx, we let u = x du = dx
dv = sin x dx v = − cos x.
Z
Z
Problem 4 Evaluate:
x 2 cos x dx
Solution Do integration by parts twice.
u = x2
dv = cos x dx
du = 2x dx
v = sin x
Z
x 2 cos x dx = x 2 sin x − 2 x sin x dx.
R
To evaluate x sin x dx, we let u = x du = dx
dv = sin
Z x dx v = − cos x.
Z
Z
x sin x dx = −x cos x +
cos x dx
Z
Problem 4 Evaluate:
x 2 cos x dx
Solution Do integration by parts twice.
u = x2
dv = cos x dx
du = 2x dx
v = sin x
Z
x 2 cos x dx = x 2 sin x − 2 x sin x dx.
R
To evaluate x sin x dx, we let u = x du = dx
dv = sin
Z x dx v = − cos x.
Z
Z
x sin x dx = −x cos x +
cos x dx
= −x cos x + sin x + C .
Z
Problem 4 Evaluate:
x 2 cos x dx
Solution Do integration by parts twice.
u = x2
dv = cos x dx
du = 2x dx
v = sin x
Z
x 2 cos x dx = x 2 sin x − 2 x sin x dx.
R
To evaluate x sin x dx, we let u = x du = dx
dv = sin
Z x dx v = − cos x.
Z
Z
x sin x dx = −x cos x +
cos x dx
= −x cos x + sin x + C .
So the answer is x2 sin x − 2[−x cos x + sin x] + C.
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Solution Make the trig substitution x = 3 sin θ,
dx = 3 cos θ dθ.
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Solution Make the trig substitution x = 3 sin θ,
3
.
dx = 3 cos θ dθ. Note that cos1 θ = √9−x
2
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Solution Make the trig substitution x = 3 sin θ,
1
√ 1
dx = 3 cos θ dθ. Note that 3 cos
θ = 9−x 2 .
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Solution Make the trig substitution x = 3 sin θ,
1
1
dx = 3 cos θ dθ. Note that 3 cos
= √9−x
.
2
θ
R 3x−1
R
1
√
dx = (3x − 1) √9−x 2 dx
9−x 2
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Solution Make the trig substitution x = 3 sin θ,
1
1
dx = 3 cos θ dθ. Note that 3 cos
= √9−x
.
2
θ
R 3x−1
R
1
√
dx = (3x − 1) √9−x 2 dx
R 9−x 2
1
= [3(3 sin θ) − 1] 3 cos
θ · 3 cos θ dθ
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Solution Make the trig substitution x = 3 sin θ,
1
1
dx = 3 cos θ dθ. Note that 3 cos
= √9−x
.
2
θ
R 3x−1
R
1
√
dx = (3x − 1) √9−x 2 dx
R 9−x 2
1
= R [3(3 sin θ) − R1] 3 cos
θ · 3 cos θ dθ
= 9 sin θ dθ − dθ
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Solution Make the trig substitution x = 3 sin θ,
1
1
dx = 3 cos θ dθ. Note that 3 cos
= √9−x
.
2
θ
R 3x−1
R
1
√
dx = (3x − 1) √9−x 2 dx
R 9−x 2
1
= R [3(3 sin θ) − R1] 3 cos
θ · 3 cos θ dθ
= 9 sin θ dθ − dθ
= −9 cos θ − θ + C
Problem 5 Evaluate:
R
√3x−1
9−x 2
dx
Solution Make the trig substitution x = 3 sin θ,
1
1
dx = 3 cos θ dθ. Note that 3 cos
= √9−x
.
2
θ
R 3x−1
R
1
√
dx = (3x − 1) √9−x 2 dx
R 9−x 2
1
= R [3(3 sin θ) − R1] 3 cos
θ · 3 cos θ dθ
= 9 sin θ dθ − dθ
√
= −9 cos θ − θ + C = −3 9 − x 2 − sin−1 ( x3 ) + C .
Z
Problem 6 Evaluate:
tan3 x
dx
cos x
Z
Problem 6 Evaluate:
tan3 x
dx
cos x
Solution Simplify using tan2 x = sec2 x − 1 and
0
sec x = sec x tan x and make the substitution
u = sec x, du = sec x tan x dx.
Z
Problem 6 Evaluate:
tan3 x
dx
cos x
Solution Simplify using tan2 x = sec2 x − 1 and
0
sec x = sec x tan x and make the substitution
u = sec x, du = sec x tan x dx.
Z
tan3 x
dx
cos x
Z
Problem 6 Evaluate:
tan3 x
dx
cos x
Solution Simplify using tan2 x = sec2 x − 1 and
0
sec x = sec x tan x and make the substitution
u = sec x, du = sec x tan x dx.
Z
Z
tan3 x
dx = tan2 x (sec x tan x) dx
cos x
Z
Problem 6 Evaluate:
tan3 x
dx
cos x
Solution Simplify using tan2 x = sec2 x − 1 and
0
sec x = sec x tan x and make the substitution
u = sec x, du = sec x tan x dx.
Z
Z
tan3 x
dx = tan2 x (sec x tan x) dx
cos x
Z
= (sec2 x − 1)(sec x tan x) dx
Z
Problem 6 Evaluate:
tan3 x
dx
cos x
Solution Simplify using tan2 x = sec2 x − 1 and
0
sec x = sec x tan x and make the substitution
u = sec x, du = sec x tan x dx.
Z
Z
tan3 x
dx = tan2 x (sec x tan x) dx
cos x
Z
= (sec2 x − 1)(sec x tan x) dx
Z
=
(u 2 −1) du
Z
Problem 6 Evaluate:
tan3 x
dx
cos x
Solution Simplify using tan2 x = sec2 x − 1 and
0
sec x = sec x tan x and make the substitution
u = sec x, du = sec x tan x dx.
Z
Z
tan3 x
dx = tan2 x (sec x tan x) dx
cos x
Z
= (sec2 x − 1)(sec x tan x) dx
Z
=
1
(u 2 −1) du = u 3 −u+C
3
Z
Problem 6 Evaluate:
tan3 x
dx
cos x
Solution Simplify using tan2 x = sec2 x − 1 and
0
sec x = sec x tan x and make the substitution
u = sec x, du = sec x tan x dx.
Z
Z
tan3 x
dx = tan2 x (sec x tan x) dx
cos x
Z
= (sec2 x − 1)(sec x tan x) dx
Z
=
1
1
(u 2 −1) du = u 3 −u+C = sec3 x −sec x +C .
3
3
Z
Problem 7 Evaluate:
x −1
dx
x2 − x − 2
x −1
dx
x2 − x − 2
Solution Use partial fractions.
Z
Problem 7 Evaluate:
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
x2 − x − 2
Z
Problem 7 Evaluate:
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
x −1
=
x 2 − x − 2 (x − 2)(x + 1)
Z
Problem 7 Evaluate:
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
B
x −1
A
+
.
=
=
x 2 − x − 2 (x − 2)(x + 1) x − 2 x + 1
Z
Problem 7 Evaluate:
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
B
x −1
A
+
.
=
=
x 2 − x − 2 (x − 2)(x + 1) x − 2 x + 1
Z
Problem 7 Evaluate:
Clearing denominators, yields
x − 1 = (x + 1)A + (x − 2)B.
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
B
x −1
A
+
.
=
=
x 2 − x − 2 (x − 2)(x + 1) x − 2 x + 1
Z
Problem 7 Evaluate:
Clearing denominators, yields
x − 1 = (x + 1)A + (x − 2)B. For x = 2, 1 = 3A;
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
B
x −1
A
+
.
=
=
x 2 − x − 2 (x − 2)(x + 1) x − 2 x + 1
Z
Problem 7 Evaluate:
Clearing denominators, yields
x − 1 = (x + 1)A + (x − 2)B. For x = 2, 1 = 3A;
for x = −1, −2 = −3B.
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
B
x −1
A
+
.
=
=
x 2 − x − 2 (x − 2)(x + 1) x − 2 x + 1
Z
Problem 7 Evaluate:
Clearing denominators, yields
x − 1 = (x + 1)A + (x − 2)B. For x = 2, 1 = 3A;
for x = −1, −2 = −3B. So, A = 31 , B = 23 .
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
B
x −1
A
+
.
=
=
x 2 − x − 2 (x − 2)(x + 1) x − 2 x + 1
Z
Problem 7 Evaluate:
Clearing denominators, yields
x − 1 = (x + 1)A + (x − 2)B. For x = 2, 1 = 3A;
for x = −1, −2 = −3B. So, A = 31 , B = 23 .
Z
x −1
dx
x2 − x − 2
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
B
x −1
A
+
.
=
=
x 2 − x − 2 (x − 2)(x + 1) x − 2 x + 1
Z
Problem 7 Evaluate:
Clearing denominators, yields
x − 1 = (x + 1)A + (x − 2)B. For x = 2, 1 = 3A;
for x = −1, −2 = −3B. So, A = 31 , B = 23 .
Z
x −1
dx =
x2 − x − 2
Z
1
1
2
1
+
dx
3 x −2
3 x +1
x −1
dx
x2 − x − 2
Solution Use partial fractions.
x −1
B
x −1
A
+
.
=
=
x 2 − x − 2 (x − 2)(x + 1) x − 2 x + 1
Z
Problem 7 Evaluate:
Clearing denominators, yields
x − 1 = (x + 1)A + (x − 2)B. For x = 2, 1 = 3A;
for x = −1, −2 = −3B. So, A = 31 , B = 23 .
Z
x −1
dx =
x2 − x − 2
=
Z
1
1
2
1
+
dx
3 x −2
3 x +1
1
2
ln |x − 2| + ln |x + 1| + C .
3
3
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral.
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral. Use
the substitution u = −x2 , du = −2x dx
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral. Use
the substitution u = −x2 , du = −2x dx
Z
2
x e −x dx
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral. Use
the substitution u = −x2 , du = −2x dx
Z
Z
1
−x 2
xe
dx = −
e u du
2
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral. Use
the substitution u = −x2 , du = −2x dx
Z
Z
1
1
2
−x 2
xe
dx = −
e u du = − e −x .
2
2
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral. Use
the substitution u = −x2 , du = −2x dx
Z
Z
1
1
2
−x 2
xe
dx = −
e u du = − e −x .
2
2
Z ∞
2
So,
x e −x dx
0
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral. Use
the substitution u = −x2 , du = −2x dx
Z
Z
1
1
2
−x 2
xe
dx = −
e u du = − e −x .
2
2
t
Z ∞
1
2
2
So,
x e −x dx = lı́m − e −x t→∞ 2
0
0
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral. Use
the substitution u = −x2 , du = −2x dx
Z
Z
1
1
2
−x 2
xe
dx = −
e u du = − e −x .
2
2
t
Z ∞
1
2
2
So,
x e −x dx = lı́m − e −x t→∞ 2
0
0
1 2 1
= lı́m − e −t +
t→∞ 2
2
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z ∞
2
x e −x dx
0
Solution First evaluate the indefinite integral. Use
the substitution u = −x2 , du = −2x dx
Z
Z
1
1
2
−x 2
xe
dx = −
e u du = − e −x .
2
2
t
Z ∞
1
2
2
So,
x e −x dx = lı́m − e −x t→∞ 2
0
0
1 2 1 1
= lı́m − e −t + =
t→∞ 2
2 2
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Solution First evaluate the indefinite integral.
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Solution First evaluate the indefinite integral.
Z
1
dx
(x − 1)3
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Solution First evaluate the indefinite integral.
Z
1
1
(x − 1)−2
=
−
dx
3
2
(x − 1)
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Solution First evaluate the indefinite integral.
Z
1
1
1
1
−2
(x
−
1)
·
=
−
=
−
dx
2
2 (x − 1)2
(x − 1)3
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Solution First evaluate the indefinite integral.
Z
1
1
1
1
−2
(x
−
1)
·
=
−
=
−
dx
2
2 (x − 1)2
(x − 1)3
Z
1
So,
−2
1
dx
(x − 1)3
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Solution First evaluate the indefinite integral.
Z
1
1
1
1
−2
(x
−
1)
·
=
−
=
−
dx
2
2 (x − 1)2
(x − 1)3
t
Z 1
1
1
1
So,
dx
=
l
ı́
m
−
·
3
2
+
t→1
2 (x − 1) −2
−2 (x − 1)
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Solution First evaluate the indefinite integral.
Z
1
1
1
1
−2
(x
−
1)
·
=
−
=
−
dx
2
2 (x − 1)2
(x − 1)3
t
Z 1
1
1
1
So,
dx
=
l
ı́
m
−
·
3
2
+
t→1
2 (x − 1) −2
−2 (x − 1)
1
1
1 1
= lı́m+ − ·
+
·
t→1
2 (t − 1)2 2 9
Problem 8 Determine whether each improper
integral converges and, if it does, find its value.
Z 1
1
dx
3
−2 (x − 1)
Solution First evaluate the indefinite integral.
Z
1
1
1
1
−2
(x
−
1)
·
=
−
=
−
dx
2
2 (x − 1)2
(x − 1)3
t
Z 1
1
1
1
So,
dx
=
l
ı́
m
−
·
3
2
+
t→1
2 (x − 1) −2
−2 (x − 1)
1
1
1 1
= lı́m+ − ·
+
· = −∞.
t→1
2 (t − 1)2 2 9
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry.
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry. For example;
1
f (x) = cos3 x
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry. For example;
1
2
f (x) = cos3 x
f (x) = sin5 x cos2 x
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry. For example;
1
2
3
f (x) = cos3 x
f (x) = sin5 x cos2 x
f (x) = sin2 x.
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry. For example;
1
2
3
f (x) = cos3 x
f (x) = sin5 x cos2 x
f (x) = sin2 x.
Integrating such functions involve several techniques
and strategies which we will describe today.
Strategy for Evaluating
R
sinm x cosn x dx
(a) If the power of cosine is odd (n = 2k + 1), save
one cosine factor and use cos2 x = 1 − sin2 x to
express the remaining factors in terms of sine:
Strategy for Evaluating
R
sinm x cosn x dx
(a) If the power of cosine is odd (n = 2k + 1), save
one cosine factor and use cos2 x = 1 − sin2 x to
express the remaining factors in terms of sine:
Z
Z
m
2k+1
sin x cos
x dx = sinm x(cos2 x)k cos x dx
Strategy for Evaluating
R
sinm x cosn x dx
(a) If the power of cosine is odd (n = 2k + 1), save
one cosine factor and use cos2 x = 1 − sin2 x to
express the remaining factors in terms of sine:
Z
Z
m
2k+1
sin x cos
x dx = sinm x(cos2 x)k cos x dx
Z
=
sinm x(1 − sin2 x)k cos x dx
Strategy for Evaluating
R
sinm x cosn x dx
(a) If the power of cosine is odd (n = 2k + 1), save
one cosine factor and use cos2 x = 1 − sin2 x to
express the remaining factors in terms of sine:
Z
Z
m
2k+1
sin x cos
x dx = sinm x(cos2 x)k cos x dx
Z
=
sinm x(1 − sin2 x)k cos x dx
Then substitute u = sin x.
Strategy for Evaluating
R
sinm x cosn x dx
(b) If the power of sine is odd (m = 2k + 1), save
one sine factor and use sin2 x = 1 − cos2 x to
express the remaining factors in terms of cosine:
Strategy for Evaluating
R
sinm x cosn x dx
(b) If the power of sine is odd (m = 2k + 1), save
one sine factor and use sin2 x = 1 − cos2 x to
express the remaining factors in terms of cosine:
Z
sin2k+1 x cosn x dx =
Z
(sin2 x)k cosn x sin x dx
Strategy for Evaluating
R
sinm x cosn x dx
(b) If the power of sine is odd (m = 2k + 1), save
one sine factor and use sin2 x = 1 − cos2 x to
express the remaining factors in terms of cosine:
Z
sin2k+1 x cosn x dx =
Z
=
Z
(sin2 x)k cosn x sin x dx
(1 − cos2 x)k cosn x sin x dx.
Strategy for Evaluating
R
sinm x cosn x dx
(b) If the power of sine is odd (m = 2k + 1), save
one sine factor and use sin2 x = 1 − cos2 x to
express the remaining factors in terms of cosine:
Z
sin2k+1 x cosn x dx =
Z
=
Z
(sin2 x)k cosn x sin x dx
(1 − cos2 x)k cosn x sin x dx.
Then substitute u = cos x.
Strategy for Evaluating
R
sinm x cosn x dx
(c) If the powers of both sine and cosine are even,
use the half-angle identities
Strategy for Evaluating
R
sinm x cosn x dx
(c) If the powers of both sine and cosine are even,
use the half-angle identities
sin2 x = 12 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
Strategy for Evaluating
R
sinm x cosn x dx
(c) If the powers of both sine and cosine are even,
use the half-angle identities
sin2 x = 12 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
It is sometimes helpful to use the identity
Strategy for Evaluating
R
sinm x cosn x dx
(c) If the powers of both sine and cosine are even,
use the half-angle identities
sin2 x = 12 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
It is sometimes helpful to use the identity
sin x cos x = 12 sin 2x
Strategy for Evaluating
R
tanm x secn x dx
(a) If the power of secant is even (n = 2k, k ≥ 2),
save a factor of sec2 x and use sec2 = 1 + tan2 x
to express the remaining factors in terms of tan x:
Strategy for Evaluating
R
tanm x secn x dx
(a) If the power of secant is even (n = 2k, k ≥ 2),
save a factor of sec2 x and use sec2 = 1 + tan2 x
to express the remaining factors in terms of tan x:
Z
tanm x sec2k x dx =
Z
tanm x(sec2 x)k−1 sec2 x dx
Strategy for Evaluating
R
tanm x secn x dx
(a) If the power of secant is even (n = 2k, k ≥ 2),
save a factor of sec2 x and use sec2 = 1 + tan2 x
to express the remaining factors in terms of tan x:
Z
tanm x sec2k x dx =
Z
=
Z
tanm x(sec2 x)k−1 sec2 x dx
tanm x(1 + tan2 x)k−1 sec2 x dx
Strategy for Evaluating
R
tanm x secn x dx
(a) If the power of secant is even (n = 2k, k ≥ 2),
save a factor of sec2 x and use sec2 = 1 + tan2 x
to express the remaining factors in terms of tan x:
Z
tanm x sec2k x dx =
Z
=
Z
tanm x(sec2 x)k−1 sec2 x dx
tanm x(1 + tan2 x)k−1 sec2 x dx
Then substitute u = tan x.
Strategy for Evaluating
R
tanm x secn x dx
(b) If the power of tangent is odd (m = 2k + 1),
save a factor of sec x tan x and use
tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Strategy for Evaluating
R
tanm x secn x dx
(b) If the power of tangent is odd (m = 2k + 1),
save a factor of sec x tan x and use
tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Z
tan
2k+1
n
x sec x dx =
Z
(tan2 x)k secn−1 x sec x tan x dx
Strategy for Evaluating
R
tanm x secn x dx
(b) If the power of tangent is odd (m = 2k + 1),
save a factor of sec x tan x and use
tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Z
tan
2k+1
Z
=
n
x sec x dx =
Z
(tan2 x)k secn−1 x sec x tan x dx
(sec2 x − 1)k secn−1 x sec x tan x dx
Strategy for Evaluating
R
tanm x secn x dx
(b) If the power of tangent is odd (m = 2k + 1),
save a factor of sec x tan x and use
tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Z
tan
2k+1
Z
=
n
x sec x dx =
Z
(tan2 x)k secn−1 x sec x tan x dx
(sec2 x − 1)k secn−1 x sec x tan x dx
Then substitute u = sec x.
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Z
tan x dx = ln | sec x| + C .
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Z
tan x dx = ln | sec x| + C .
The next formula can be checked by differentiating
the right hand side.
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Z
tan x dx = ln | sec x| + C .
The next formula can be checked by differentiating
the right hand side.
Z
sec x dx = ln | sec x + tan x| + C .
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Z
tan x dx = ln | sec x| + C .
The next formula can be checked by differentiating
the right hand side.
Z
sec x dx = ln | sec x + tan x| + C .
Also, don’t forget that
d
dx sec x = sec x tan x.
d
dx
tan x = sec2 x and
Table of Trigonometric Substitution
Expression
Substitution
√
2
2
x = a sin θ, − π2 ≤ θ ≤ π2
√a − x
2
2
x = a tan θ, − π2 < θ < π2
√a + x
x 2 − a2
x = a sec θ, 0 ≤ θ < π2
or π ≤ θ < 3π
2
Identity
1 − sin2 θ = cos2 θ
1 + tan2 θ = sec2 θ
sec2 θ − 1 = tan2 θ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
= 1.
Find the area enclosed by the ellipse
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
= 1.
Find the area enclosed by the ellipse
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
a2 − x 2 = a cos θ.
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
a2 − x 2 dx = a cos θ · a cos θ dθ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos θ · a cos θ dθ
= a2 cos2 θ dθ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ = 21 a2 (θ + 21 sin 2θ).
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ = 21 a2 (θ + 21 sin 2θ).
Ra√
A = 4b
a2 − x 2
a 0
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ = 21 a2 (θ + 21 sin 2θ).
π
Ra√
A = 4b
a2 − x 2 = 2ab θ + 12 sin 2θ 0
a 0
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ = 21 a2 (θ + 21 sin 2θ).
π
Ra√
A = 4b
a2 − x 2 = 2ab θ + 12 sin 2θ 0 = πab.
a 0
Integration by partial fractions
A function f (x) is called a rational function if it
can be expressed as the ratio of two polynomials.
Integration by partial fractions
A function f (x) is called a rational function if it
can be expressed as the ratio of two polynomials.
For example,
f (x) =
2
x−1
−
1
x+2
=
2(x+2)−(x−1)
(x−1)(x+2)
is a rational function.
=
x+5
x 2 +x−2 ,
Integration by partial fractions
A function f (x) is called a rational function if it
can be expressed as the ratio of two polynomials.
For example,
f (x) =
2
x−1
−
1
x+2
=
2(x+2)−(x−1)
(x−1)(x+2)
is a rational function.
So,
R x+5
R 1
1
dx
=
2
x −x−2
x−1 − x+2 dx =
2 ln |x − 1| − ln |x + 2| + C .
=
x+5
x 2 +x−2 ,
Proper and improper rational functions
Recall that the degree polynomial
P(x) = an x n + an−1 x n−1 + . . . + a1 x + a0
is n if an 6= 0.
Proper and improper rational functions
Recall that the degree polynomial
P(x) = an x n + an−1 x n−1 + . . . + a1 x + a0
is n if an 6= 0. We write this as deg(P) = n.
Proper and improper rational functions
Recall that the degree polynomial
P(x) = an x n + an−1 x n−1 + . . . + a1 x + a0
is n if an 6= 0. We write this as deg(P) = n.
P(x)
A rational function f (x) = Q(x)
is called proper if
the degree of P(x) is less than the degree of Q(x);
Proper and improper rational functions
Recall that the degree polynomial
P(x) = an x n + an−1 x n−1 + . . . + a1 x + a0
is n if an 6= 0. We write this as deg(P) = n.
P(x)
A rational function f (x) = Q(x)
is called proper if
the degree of P(x) is less than the degree of Q(x);
otherwise it is called improper.
Proper and improper rational functions
Recall that the degree polynomial
P(x) = an x n + an−1 x n−1 + . . . + a1 x + a0
is n if an 6= 0. We write this as deg(P) = n.
P(x)
A rational function f (x) = Q(x)
is called proper if
the degree of P(x) is less than the degree of Q(x);
otherwise it is called improper.
If deg(P) ≥ deg(Q), then after long division,
Proper and improper rational functions
Recall that the degree polynomial
P(x) = an x n + an−1 x n−1 + . . . + a1 x + a0
is n if an 6= 0. We write this as deg(P) = n.
P(x)
A rational function f (x) = Q(x)
is called proper if
the degree of P(x) is less than the degree of Q(x);
otherwise it is called improper.
If deg(P) ≥ deg(Q), then after long division,
P(x)
R(x)
f (x) = Q(x)
= S(x) + Q(x)
,
Proper and improper rational functions
Recall that the degree polynomial
P(x) = an x n + an−1 x n−1 + . . . + a1 x + a0
is n if an 6= 0. We write this as deg(P) = n.
P(x)
A rational function f (x) = Q(x)
is called proper if
the degree of P(x) is less than the degree of Q(x);
otherwise it is called improper.
If deg(P) ≥ deg(Q), then after long division,
P(x)
R(x)
f (x) = Q(x)
= S(x) + Q(x)
, where R(x) is the
remainder and deg(R) < deg(Q).
Example Find
R
x 3 +x
x−1
dx.
Example Find
R
x 3 +x
x−1
dx.
Solution After long division, we find
x3 + x
2
= x2 + x + 2 +
.
x −1
x −1
Example Find
R
x 3 +x
x−1
dx.
Solution After long division, we find
x3 + x
2
= x2 + x + 2 +
.
x −1
x −1
So,
Z
x3 + x
dx =
x −1
Z
x2 + x + 2 +
2
dx
x −1
Example Find
R
x 3 +x
x−1
dx.
Solution After long division, we find
x3 + x
2
= x2 + x + 2 +
.
x −1
x −1
So,
Z
x3 + x
dx =
x −1
Z
x2 + x + 2 +
2
dx
x −1
x3 x2
=
+
+ 2x + 2 ln |x − 1| + C .
3
2
Next step, factor Q(x) and express using partial
fractions.
Next step, factor Q(x) and express using partial
fractions. Any polynomial Q(x) can be factored into
a product of linear factors
Next step, factor Q(x) and express using partial
fractions. Any polynomial Q(x) can be factored into
a product of linear factors (ax + b)
Next step, factor Q(x) and express using partial
fractions. Any polynomial Q(x) can be factored into
a product of linear factors (ax + b) and quadratic
factors
Next step, factor Q(x) and express using partial
fractions. Any polynomial Q(x) can be factored into
a product of linear factors (ax + b) and quadratic
factors (ax2 + bx + c,
Next step, factor Q(x) and express using partial
fractions. Any polynomial Q(x) can be factored into
a product of linear factors (ax + b) and quadratic
factors (ax2 + bx + c, where b 2 − 4ac < 0).
Next step, factor Q(x) and express using partial
fractions. Any polynomial Q(x) can be factored into
a product of linear factors (ax + b) and quadratic
factors (ax2 + bx + c, where b 2 − 4ac < 0).
For example,
Q(x) = (x 2 − 4)(x 2 + 4) = (x − 2)(x + 2)(x 2 + 4).
Next step, factor Q(x) and express using partial
fractions. Any polynomial Q(x) can be factored into
a product of linear factors (ax + b) and quadratic
factors (ax2 + bx + c, where b 2 − 4ac < 0).
For example,
Q(x) = (x 2 − 4)(x 2 + 4) = (x − 2)(x + 2)(x 2 + 4).
R(x)
Next express Q(x)
as a sum of partial fractions of
the form
Next step, factor Q(x) and express using partial
fractions. Any polynomial Q(x) can be factored into
a product of linear factors (ax + b) and quadratic
factors (ax2 + bx + c, where b 2 − 4ac < 0).
For example,
Q(x) = (x 2 − 4)(x 2 + 4) = (x − 2)(x + 2)(x 2 + 4).
R(x)
Next express Q(x)
as a sum of partial fractions of
the form
A
(ax + b)i
.
or
Ax + B
(ax2 + bx + c)j
Case Q(x) is a product of linear factors
R 2 +2x−1
Example Evaluate 2xx3+3x
2 −2x dx.
Case Q(x) is a product of linear factors
R 2 +2x−1
Example Evaluate 2xx3+3x
2 −2x dx.
Solution First factor Q(x).
Case Q(x) is a product of linear factors
R 2 +2x−1
Example Evaluate 2xx3+3x
2 −2x dx.
Solution First factor Q(x).
Q(x) = 2x 3 + 3x 2 − 2x = x(2x 2 + 3x − 2)
Case Q(x) is a product of linear factors
R 2 +2x−1
Example Evaluate 2xx3+3x
2 −2x dx.
Solution First factor Q(x).
Q(x) = 2x 3 + 3x 2 − 2x = x(2x 2 + 3x − 2)
= x(2x − 1)(x + 2).
Case Q(x) is a product of linear factors
R 2 +2x−1
Example Evaluate 2xx3+3x
2 −2x dx.
Solution First factor Q(x).
Q(x) = 2x 3 + 3x 2 − 2x = x(2x 2 + 3x − 2)
= x(2x − 1)(x + 2).
x 2 + 2x − 1
A
B
C
So,
= +
+
.
x(2x − 1)(x + 2)
x
2x − 1 x + 2
Case Q(x) is a product of linear factors
R 2 +2x−1
Example Evaluate 2xx3+3x
2 −2x dx.
Solution First factor Q(x).
Q(x) = 2x 3 + 3x 2 − 2x = x(2x 2 + 3x − 2)
= x(2x − 1)(x + 2).
x 2 + 2x − 1
A
B
C
So,
= +
+
.
x(2x − 1)(x + 2)
x
2x − 1 x + 2
Now multiply both sides by Q(x):
x 2 +2x−1 = A(2x−1)(x+2)+Bx(x+2)+Cx(2x−1).
Next simplify:
x 2 +2x −1 = (2A+B +2C )x 2 +(3A+2B +C )x −2A.
Next simplify:
x 2 +2x −1 = (2A+B +2C )x 2 +(3A+2B +C )x −2A.
Next solve the linear equations:
2A + B + 2C = 1
3A + 2B − C = 2
−2A = −1
Next simplify:
x 2 +2x −1 = (2A+B +2C )x 2 +(3A+2B +C )x −2A.
Next solve the linear equations:
2A + B + 2C = 1
3A + 2B − C = 2
−2A = −1
1
.
Solving we get A = 12 , B = 15 , C = − 10
Next simplify:
x 2 +2x −1 = (2A+B +2C )x 2 +(3A+2B +C )x −2A.
Next solve the linear equations:
2A + B + 2C = 1
3A + 2B − C = 2
−2A = −1
1
. Thus,
Solving we get A = 12 , B = 15 , C = − 10
Next simplify:
x 2 +2x −1 = (2A+B +2C )x 2 +(3A+2B +C )x −2A.
Next solve the linear equations:
2A + B + 2C = 1
3A + 2B − C = 2
−2A = −1
1
. Thus,
Solving we get A = 12 , B = 15 , C = − 10
Z
x 2 + 2x − 1
dx
2x 3 + 3x 2 − 2x
Next simplify:
x 2 +2x −1 = (2A+B +2C )x 2 +(3A+2B +C )x −2A.
Next solve the linear equations:
2A + B + 2C = 1
3A + 2B − C = 2
−2A = −1
1
. Thus,
Solving we get A = 12 , B = 15 , C = − 10
Z
x 2 + 2x − 1
dx
2x 3 + 3x 2 − 2x
Z 1 1 1
1
1
1
=
· + ·
−
·
dx
2 x 5 2x − 1 10 x + 2
Next simplify:
x 2 +2x −1 = (2A+B +2C )x 2 +(3A+2B +C )x −2A.
Next solve the linear equations:
2A + B + 2C = 1
3A + 2B − C = 2
−2A = −1
1
. Thus,
Solving we get A = 12 , B = 15 , C = − 10
Z
x 2 + 2x − 1
dx
2x 3 + 3x 2 − 2x
Z 1 1 1
1
1
1
=
· + ·
−
·
dx
2 x 5 2x − 1 10 x + 2
=
1
1
1
ln |x| +
ln |2x − 1| −
ln |x + 2| + C .
2
10
10
Q(x) is a product of repeated linear factors
Example Find
R
4x
x 3 −x 2 −x+1
dx.
Q(x) is a product of repeated linear factors
Example Find
R
4x
x 3 −x 2 −x+1
dx.
Solution The first step is to factor the denominator
Q(x) = x 3 − x 2 − x + 1.
Q(x) is a product of repeated linear factors
Example Find
R
4x
x 3 −x 2 −x+1
dx.
Solution The first step is to factor the denominator
Q(x) = x 3 − x 2 − x + 1. Since Q(1) = 0, we know
that x − 1 is a factor.
Q(x) is a product of repeated linear factors
Example Find
R
4x
x 3 −x 2 −x+1
dx.
Solution The first step is to factor the denominator
Q(x) = x 3 − x 2 − x + 1. Since Q(1) = 0, we know
that x − 1 is a factor.
So, x 3 − x 2 − x + 1 = (x − 1)(x 2 − 1) =
(x − 1)(x − 1)(x + 1)
Q(x) is a product of repeated linear factors
Example Find
R
4x
x 3 −x 2 −x+1
dx.
Solution The first step is to factor the denominator
Q(x) = x 3 − x 2 − x + 1. Since Q(1) = 0, we know
that x − 1 is a factor.
So, x 3 − x 2 − x + 1 = (x − 1)(x 2 − 1) =
(x − 1)(x − 1)(x + 1) = (x − 1)2 (x + 1).
Q(x) is a product of repeated linear factors
Example Find
R
4x
x 3 −x 2 −x+1
dx.
Solution The first step is to factor the denominator
Q(x) = x 3 − x 2 − x + 1. Since Q(1) = 0, we know
that x − 1 is a factor.
So, x 3 − x 2 − x + 1 = (x − 1)(x 2 − 1) =
(x − 1)(x − 1)(x + 1) = (x − 1)2 (x + 1). Since the
linear factor x − 1 occurs twice, the partial fraction
decomposition is
Q(x) is a product of repeated linear factors
Example Find
R
4x
x 3 −x 2 −x+1
dx.
Solution The first step is to factor the denominator
Q(x) = x 3 − x 2 − x + 1. Since Q(1) = 0, we know
that x − 1 is a factor.
So, x 3 − x 2 − x + 1 = (x − 1)(x 2 − 1) =
(x − 1)(x − 1)(x + 1) = (x − 1)2 (x + 1). Since the
linear factor x − 1 occurs twice, the partial fraction
decomposition is
A
4x
B
C
=
+
+
.
(x − 1)2 (x + 1) x − 1 (x − 1)2 x + 1
Q(x) is a product of repeated linear factors
Example Find
R
4x
x 3 −x 2 −x+1
dx.
Solution The first step is to factor the denominator
Q(x) = x 3 − x 2 − x + 1. Since Q(1) = 0, we know
that x − 1 is a factor.
So, x 3 − x 2 − x + 1 = (x − 1)(x 2 − 1) =
(x − 1)(x − 1)(x + 1) = (x − 1)2 (x + 1). Since the
linear factor x − 1 occurs twice, the partial fraction
decomposition is
A
4x
B
C
=
+
+
.
(x − 1)2 (x + 1) x − 1 (x − 1)2 x + 1
Now solve for A, B, and C and integrate.
Q(x) contains irreducible quadratic factors
Example
Evaluate
R
2x 2 −x+4
; dx.
x 3 +4x
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
2x 2 − x + 4
A Bx + C
= + 2
.
2
x(x + 4)
x
x +4
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
2x 2 − x + 4
A Bx + C
= + 2
.
2
x(x + 4)
x
x +4
Multiplying by x(x 2 + 4), we have
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
2x 2 − x + 4
A Bx + C
= + 2
.
2
x(x + 4)
x
x +4
Multiplying by x(x 2 + 4), we have
2x 2 − x + 4 = A(x 2 + 4) + (Bx + C )x = (A + B)x 2 + Cx + 4A.
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
2x 2 − x + 4
A Bx + C
= + 2
.
2
x(x + 4)
x
x +4
Multiplying by x(x 2 + 4), we have
2x 2 − x + 4 = A(x 2 + 4) + (Bx + C )x = (A + B)x 2 + Cx + 4A.
Equating coefficients, we obtain
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
2x 2 − x + 4
A Bx + C
= + 2
.
2
x(x + 4)
x
x +4
Multiplying by x(x 2 + 4), we have
2x 2 − x + 4 = A(x 2 + 4) + (Bx + C )x = (A + B)x 2 + Cx + 4A.
Equating coefficients, we obtain
A+B =2
C = −1
4A = 4.
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
2x 2 − x + 4
A Bx + C
= + 2
.
2
x(x + 4)
x
x +4
Multiplying by x(x 2 + 4), we have
2x 2 − x + 4 = A(x 2 + 4) + (Bx + C )x = (A + B)x 2 + Cx + 4A.
Equating coefficients, we obtain
A+B =2
C = −1
4A = 4.
Thus A = 1, B = 1, and C = −1 and so
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
2x 2 − x + 4
A Bx + C
= + 2
.
2
x(x + 4)
x
x +4
Multiplying by x(x 2 + 4), we have
2x 2 − x + 4 = A(x 2 + 4) + (Bx + C )x = (A + B)x 2 + Cx + 4A.
Equating coefficients, we obtain
A+B =2
C = −1
4A = 4.
Thus A = 1, B = 1, and C = −1 and so
R 2x 2 −x+4
R 1
dx =
+ xx−1
dx.
2 +4
x 3 +4x
x
Q(x) contains irreducible quadratic factors
Example
R 2
Evaluate 2xx 3−x+4
; dx.
+4x
3
2
Solution Since x + 4x = x(x + 4),
2x 2 − x + 4
A Bx + C
= + 2
.
2
x(x + 4)
x
x +4
Multiplying by x(x 2 + 4), we have
2x 2 − x + 4 = A(x 2 + 4) + (Bx + C )x = (A + B)x 2 + Cx + 4A.
Equating coefficients, we obtain
A+B =2
C = −1
4A = 4.
Thus A = 1, B = 1, and C = −1 and so
R 2x 2 −x+4
R 1
dx =
+ xx−1
dx.
2 +4
x 3 +4x
x
In order to integrate the second term, split it into two parts.
Q(x) has irreducible repeated quadratic factors
Example Write out the form of the partial
fraction decomposition of the function
x 3 +x 2 +1
.
x(x−1)(x 2 +x+1)(x 2 +1)3
Q(x) has irreducible repeated quadratic factors
Example Write out the form of the partial
fraction decomposition of the function
x 3 +x 2 +1
.
x(x−1)(x 2 +x+1)(x 2 +1)3
Solution The form of the partial fraction
decomposition is
Q(x) has irreducible repeated quadratic factors
Example Write out the form of the partial
fraction decomposition of the function
x 3 +x 2 +1
.
x(x−1)(x 2 +x+1)(x 2 +1)3
Solution The form of the partial fraction
decomposition is
x 3 +x 2 +1
x(x−1)(x 2 +x+1)(x 2 +1)3
=
Q(x) has irreducible repeated quadratic factors
Example Write out the form of the partial
fraction decomposition of the function
x 3 +x 2 +1
.
x(x−1)(x 2 +x+1)(x 2 +1)3
Solution The form of the partial fraction
decomposition is
x 3 +x 2 +1
x(x−1)(x 2 +x+1)(x 2 +1)3
A
x
=
B
Ex+F
Gx+H
Ix+J
+ x−1
+ xCx+D
2 +x+1 + x 2 +1 + (x 2 +1)2 + (x 2 +1)3 .
Strategies: Simply the integrand if possible
Z
tan θ
dθ =
sec2 θ
Z
sin θ
cos2 θdθ =
cos θ
Z
1
=
sin 2θ dθ.
2
Z
sin θ cos θ dθ
Strategies: Simply the integrand if possible
Z
Z
tan θ
dθ =
sec2 θ
Z
sin θ
cos2 θdθ =
cos θ
Z
1
=
sin 2θ dθ.
2
(sin x+cos x)2 dx =
Z
Z
sin2 x+2 sin x cos x+cos2 x dx
Z
=
sin θ cos θ dθ
1 + 2 sin x cos x dx.
Strategies: Look for an obvious substitution
Z
1
x
dx
=
x2 − 1
2
Z
1
2x
dx
=
ln |x 2 − 1|.
2
x −1
2
Strategies: Look for an obvious substitution
Z
1
x
dx
=
x2 − 1
2
Z
1
2x
dx
=
ln |x 2 − 1|.
2
x −1
2
Here, u = x2 − 1, du = 2x dx.
Strategies: Classify integrand according to form
1
Trigonometric functions. Use recommended
substitutions.
Strategies: Classify integrand according to form
1
2
Trigonometric functions. Use recommended
substitutions.
Rational functions.
Strategies: Classify integrand according to form
1
2
Trigonometric functions. Use recommended
substitutions.
Rational functions.
Use partial fractions.
Strategies: Classify integrand according to form
1
Trigonometric functions. Use recommended
substitutions.
2
Rational functions.
3
Integration by parts.
Use partial fractions.
Strategies: Classify integrand according to form
1
2
3
Trigonometric functions. Use recommended
substitutions.
Rational functions.
Use partial fractions.
Integration by parts. Use if f (x) is a
product of x n and a transcendental function.
Strategies: Classify integrand according to form
1
2
3
4
Trigonometric functions. Use recommended
substitutions.
Rational functions.
Use partial fractions.
Integration by parts. Use if f (x) is a
product of x n and a transcendental function.
Radicals.
Strategies: Classify integrand according to form
1
2
3
4
Trigonometric functions. Use recommended
substitutions.
Rational functions.
Use partial fractions.
Integration by parts. Use if f (x) is a
product of x n and a transcendental function.
Radicals. Particular substitutions are
recommended.
Strategies: Classify integrand according to form
1
2
3
4
Trigonometric functions. Use recommended
substitutions.
Rational functions.
Use partial fractions.
Integration by parts. Use if f (x) is a
product of x n and a transcendental function.
Radicals. Particular substitutions are
recommended.
√
(a) If
±x 2 ± a2 occurs, use trig substitutions.
Strategies: Classify integrand according to form
1
2
3
4
Trigonometric functions. Use recommended
substitutions.
Rational functions.
Use partial fractions.
Integration by parts. Use if f (x) is a
product of x n and a transcendental function.
Radicals. Particular substitutions are
recommended.
√
(a) If √±x 2 ± a2 occurs, use trig substitutions.
(b) If ax
√ + b occurs, use the rationalizing substitution
u = ax + b.
Strategies: Try again
1
Try substitution.
Strategies: Try again
1
Try substitution.
2
Try parts.
Strategies: Try again
1
Try substitution.
2
Try parts.
3
Manipulate the integrand.
Strategies: Try again
1
Try substitution.
2
Try parts.
3
4
ForR example,
Manipulate
R dx
R the1 integrand.
1+cos2 x
1+cos x
dx
=
·
dx
=
1−cos x
1−cos x 1+cos x
sin2 x
Relate the problem to previous problems.
Strategies: Try again
1
Try substitution.
2
Try parts.
3
4
ForR example,
Manipulate
R dx
R the1 integrand.
1+cos2 x
1+cos x
dx
=
·
dx
=
1−cos x
1−cos x 1+cos x
sin2 x
Relate the problem to previous problems.
For
R example,
R
R
tan2 x sec x dx = sec3 x dx − sec x dx
Strategies: Try again
1
Try substitution.
2
Try parts.
3
4
ForR example,
Manipulate
R dx
R the1 integrand.
1+cos2 x
1+cos x
dx
=
·
dx
=
1−cos x
1−cos x 1+cos x
sin2 x
Relate the problem to previous problems.
For
R example,
R
R
3
tan2 x sec
x
dx
=
sec
x
dx
−
sec x dx and
R
3
you know sec x dx by previous work.
Strategies: Try again
1
Try substitution.
2
Try parts.
3
4
5
ForR example,
Manipulate
R dx
R the1 integrand.
1+cos2 x
1+cos x
dx
=
·
dx
=
1−cos x
1−cos x 1+cos x
sin2 x
Relate the problem to previous problems.
For
R example,
R
R
3
tan2 x sec
x
dx
=
sec
x
dx
−
sec x dx and
R
3
you know sec x dx by previous work.
Use several methods.
Methods for approximate integration.
We have already
R b considered several methods for
estimating a f (x) dx.
Methods for approximate integration.
We have already
R b considered several methods for
estimating a f (x) dx.
1
left endpoint approximation
Methods for approximate integration.
We have already
R b considered several methods for
estimating a f (x) dx.
1
left endpoint approximation
2
right endpoint approximation
Methods for approximate integration.
We have already
R b considered several methods for
estimating a f (x) dx.
1
left endpoint approximation
2
right endpoint approximation
3
midpoint approximation
Trapezoidal Rule
Z b
f (x) dx ≈ Tn
a
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
where ∆x = (b−a)
and xi = a + i∆x.
n
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
where ∆x = (b−a)
and xi = a + i∆x.
n
Example Use the Trapezoidal
Rule to
R2 1
approximate the integral 1 x dx.
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
where ∆x = (b−a)
and xi = a + i∆x.
n
Example Use the Trapezoidal
Rule to
R2 1
approximate the integral 1 x dx.
Solution With n = 5, a = 1, and b = 2, we have
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
where ∆x = (b−a)
and xi = a + i∆x.
n
Example Use the Trapezoidal
Rule to
R2 1
approximate the integral 1 x dx.
Solution With n = 5, a = 1, and b = 2, we have
= 0.2,
∆x = (2−1)
5
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
where ∆x = (b−a)
and xi = a + i∆x.
n
Example Use the Trapezoidal
Rule to
R2 1
approximate the integral 1 x dx.
Solution With n = 5, a = 1, and b = 2, we have
= 0.2, and so the Trapezoidal Rule
∆x = (2−1)
5
gives
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
where ∆x = (b−a)
and xi = a + i∆x.
n
Example Use the Trapezoidal
Rule to
R2 1
approximate the integral 1 x dx.
Solution With n = 5, a = 1, and b = 2, we have
= 0.2, and so the Trapezoidal Rule
∆x = (2−1)
5
gives
R2 1
1 x dx ≈ T5
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
where ∆x = (b−a)
and xi = a + i∆x.
n
Example Use the Trapezoidal
Rule to
R2 1
approximate the integral 1 x dx.
Solution With n = 5, a = 1, and b = 2, we have
= 0.2, and so the Trapezoidal Rule
∆x = (2−1)
5
gives
R2 1
0.2
1 x dx ≈ T5 = 2 [f (1) + 2f (1.2) + 2f (1.4) +
2f (1.6) + 2f (1.8) + f (2)] =
Trapezoidal Rule
Z b
f (x) dx ≈ Tn =
a
∆x
2
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
where ∆x = (b−a)
and xi = a + i∆x.
n
Example Use the Trapezoidal
Rule to
R2 1
approximate the integral 1 x dx.
Solution With n = 5, a = 1, and b = 2, we have
= 0.2, and so the Trapezoidal Rule
∆x = (2−1)
5
gives
R2 1
0.2
1 x dx ≈ T5 = 2 [f (1) + 2f (1.2) + 2f (1.4) +
2f (1.6)
+ 2f (1.8) + f (2)] =
1
2
2
2
2
1
0.1 1 + 1.2 + 1.4 + 1.6 + 1.8 + 2 .
Example Use the Midpoint
Rule with n = 4 to
R
approximate the integral
8 x2
0 e
dx.
Example Use the Midpoint
Rule with n = 4 to
R
8
2
approximate the integral 0 e x dx.
Solution Since a = 0, b = 8, and n = 4, the
Midpoint Rule gives
Example Use the Midpoint
Rule with n = 4 to
R
8
2
approximate the integral 0 e x dx.
Solution Since a = 0, b = 8, and n = 4, the
Rule gives
RMidpoint
8 x2
dx ≈ ∆x[f (1) + f (3) + f (5) + f (7)]
0 e
Example Use the Midpoint
Rule with n = 4 to
R
8
2
approximate the integral 0 e x dx.
Solution Since a = 0, b = 8, and n = 4, the
Rule gives
RMidpoint
8 x2
0 e dx ≈ ∆x[f (1) + f(3) + f (5) + f (7)]
= 2 e 1 + e 9 + e 25 + e 49 .
Simpson’s Rule
Simpson’s Rule generalizes the method in the
Trapezoidal Rule by using parabola
approximations (see book).
Simpson’s Rule
Simpson’s Rule generalizes the method in the
Trapezoidal Rule by using parabola
approximations (see book). It gives a much better
approximation.
Simpson’s Rule
Simpson’s Rule generalizes the method in the
Trapezoidal Rule by using parabola
approximations (see book). It gives a much better
approximation.
Rb
Simpson’s Rule: a f (x) dx ≈ Sn
Simpson’s Rule
Simpson’s Rule generalizes the method in the
Trapezoidal Rule by using parabola
approximations (see book). It gives a much better
approximation.
Rb
Simpson’s Rule: a f (x) dx ≈ Sn
= ∆x
3 [f (x0 ) + 4f (x1 ) + 2(x2 ) + 4f (x3 ) + · · · +
2f (xn−2 ) + 4f (xn−1 ) + f (xn )],
Simpson’s Rule
Simpson’s Rule generalizes the method in the
Trapezoidal Rule by using parabola
approximations (see book). It gives a much better
approximation.
Rb
Simpson’s Rule: a f (x) dx ≈ Sn
= ∆x
3 [f (x0 ) + 4f (x1 ) + 2(x2 ) + 4f (x3 ) + · · · +
2f (xn−2 ) + 4f (xn−1 ) + f (xn )], where n is even and
∆x = (b−a)
n .