Download Calculus II The freecalc project team

Calculus II
Lecture 7
The freecalc project team
Reference Lectures
https://sourceforge.net/projects/freecalculus/
2016
The freecalc team
Lecture 7
2016
Outline
√
1
Integrals of form R(x, ax 2√+ bx + c)dx,
function
√ R - rational
√
Transforming to the forms x 2 + 1, −x 2 + 1, x 2 − 1
Table of Euler
√ and trig substitutions
The case √x 2 + 1
The case √−x 2 + 1
The case x 2 − 1
2
Rationalizing Substitutions
R
The freecalc team
Lecture 7
2016
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The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Integrals of form
function
R
√
R(x, ax 2 + bx + c)dx, R - rational
Let R(x, y ) be an arbitrary rational expression in two variables
(quotient of polynomials in two variables).
Question
Can we integrate
Z
p
R x, ax 2 + bx + c dx?
Yes. We will learn how in what follows.
The algorithm for integration is roughly:
to one of
√ substitution
√
RUse linear
R to transform
R three√integrals:
R(x, x 2 + 1)dx, R(x, −x 2 + 1)dx, R(x, x 2 − 1)dx.
Use trigonometric substitution or Euler substitution to transform to
trigonometric or rational function integral (no radicals).
Solve as previously studied.
We motivate why we need such integrals by examples such as
computing the area of an ellipse.
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Trigonometric Substitution
To
R √find the area of a circle or ellipse, one needs to compute
a2 − x 2 dx.
R √
For x a2 − x 2 dx, the substitution u = a2 − x 2 would work.
R√
For
a2 − x 2 dx, we need a more elaborate substitution.
Instead, substitute x = a sin θ.
q
q
p
p
a2 − x 2 = a2 − a2 sin2 θ = a2 (1 − sin2 θ) = a2 cos2 θ = a| cos θ|.
With u = a2 − x 2 , the new variable is a function of the old one.
With x = a sin θ, the old variable is a function of the new one.
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Transforming to the forms
p
x 2 + 1,
p
−x 2 + 1,
p
x2 − 1
p
Linear substitutions to simplify radicals ay 2 + by + c
p
Using linear substitutions, radicals of form ay 2 + by + c, a 6= 0,
b2 −√
4ac 6= 0 can be transformed to (multiple of):
2
√x + 1
2
√−x + 1
x 2 − 1.
We already studied how to do that using completing the square
when dealing with rational functions.
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Transforming to the forms
p
x 2 + 1,
p
−x 2 + 1,
Recall: linear substitution is subst. of the form u = px + q.
Example
√
2 + x + 1 to multiple of
Use linear substitution
to
transform
x
r
√
1
1
1
2
2
x +x +1 =
x +2· x + − +1
2
4
4
s
2
1
3
=
+
x+
2
4
v
!
u
2
u3 4
1
t
=
x+
+1
4 3
2
√ s
2
3
2
1
√
=
x+
+1
2
3
√2
3p 2
=
u + 1,
√
√
2 2
1
2 3
3
√
where u =
x+
=
x+
.
2
3
3
3
The freecalc team
Lecture 7
√
2016
p
x2 − 1
u 2 + 1.
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Transforming to the forms
p
x 2 + 1,
p
−x 2 + 1,
p
x2 − 1
Recall: linear substitution is subst. of the form u = px + q.
Example
√
2 + x + 1 to multiple of
Use linear subst. to transform
−2x
q
√
1
1
2
2
− 2 x − 2x − 2
− 2x + x + 1 =
q
1
1
1
1
2
=
−2 x − 2 4 x + 16 − 16 − 2
r
2
9
=
− 2 x − 14 − 16
r 2
9
1
16
=
−
x
−
+1
8
9
4
q
2
3
4
1
= √8 − 3 x − 4
+1
√
3
= √8 −u 2 + 1,
4
1
4
where u = 3 x − 4 = 3 x − 31 .
The freecalc team
Lecture 7
√
−u 2 + 1.
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Table of Euler and trig substitutions
Let R be a rational function in two variables.
So far,
transformations we converted all integrals of the
Z with linear
p
form R(x, ax 2 + bx + c)dx to one of the three forms:
√
√
√
R
R
R
R(x, x 2 + 1)dx, R(x, −x 2 + 1)dx , R(x, x 2 − 1)dx.
Each of the above integrals can be transformed to a rational
trigonometric integral using 3 pairs of substitutions:
x = tan θ, x = cot θ; x = sin θ, x = cos θ; x = csc θ, x = sec θ.
We studied that trigonometric integrals are converted to rational
function integrals via θ = 2 arctan t.
The resulting 3 pairs of substitutions are called Euler substitutions:
x = tan(2 arctan t), x = cot(2 arctan t); x = sin(2 arctan t),
x = cos(2 arctan t); x = csc(2 arctan t), x = sec(2 arctan t).
The Euler substitutions directly transform the integral to a rational
function integral.
We will demonstrate that the Euler substitutions are rational.
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Table of Euler and trig substitutions
Trigonometric substitution and Euler substitution
Expression Substitution Variable range
Relevant identity
π π
2
2θ
√
x
=
tan
θ
θ
∈
−
1
+
tan
θ
=
sec
,
2 2
x2 + 1
x = cot θ
θ ∈ (0, π)
1 + cot2 θ = csc2 θ
π π
2
2θ
√
,
x
=
sin
θ
θ
∈
−
1
−
sin
θ
=
cos
2 2
−x 2 + 1
x = cos θ
θ ∈ (0, π)
1 − cos2 θ = cos2 θ
π 3π 2 θ − 1 = cot2 θ
√
x
=
csc
θ
θ
∈
0,
∪
π,
csc
2 π2 3π
x2 − 1
x = sec θ
θ ∈ 0, 2 ∪ π, 2
sec2 θ − 1 = tan2 θ
Euler substitution by applying in addition θ = 2 arctan t
2t
√
x
=
−1 < t < 1
(?)
2
2
1−t
x +1
x = 12 1t − t 0 < t
(?)
2t
√
x
=
−1 ≤ t ≤ 1
(?)
2
2
1+t
−x + 1
1−t 2
x = 1+t 2
0<t
(?)
1
1
√
x = 2 t + t t ∈ (−∞, −1) ∪ [0, 1)
(?)
2
x −1
1+t 2
x = 1−t 2
t ∈ (−∞, −1) ∪ [0, 1)
(?)
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 + 1
Trigonometric substitution x = cot θ for
√
x2 + 1
√
The trigonometric substitution x = cot θ, θ ∈ (0, π) for x 2 + 1:
p
p
cot 2 θ + 1
x2 + 1 =
s
cos2 θ
=
+1
2
s sin θ
cos2 θ + sin2 θ
=
sin2 θ
s
when θ ∈ (0, π) we have
1
1
sin θ ≥ 0 and so
p
=
=
p
sin2 θ
sin2 θ
sin2 θ = sin θ
1
=
= csc θ .
sin θ
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 + 1
Trigonometric substitution x = cot θ for
√
x2 + 1
√
The trigonometric substitution x = cot θ, θ ∈ (0, π) for x 2 + 1:
p
1
2
= csc θ .
x +1 =
sin θ
The differential dx can be expressed via dθ from x = cot θ. To
summarize:
Definition
√
The trigonometric substitution x = cot θ, θ ∈ (0, π) for x 2 + 1 is given
by:
x = cot θ
√
1
2
x +1 =
= csc θ
sin θ
dθ
2
dx = −
=
−
csc
θ?dθ
2
sin θ
θ = arccotx .
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
Example
Z
x2
√
p
ax 2 + bx + c)dx, R - rational function
1
x2
+9
dx =
The freecalc team
1
Z
q
x 23
x 2
The case
dx
p
x2 + 1
+1
Z
1
√
=
d(3 cot θ)
2
2
Z (3 cot θ) 3 cot θ+ 1
1
2
√
=
−
3
csc
θ dθ
2
2
Z27 cot θ 2 csc θ
1
− csc θ
=
dθ
2
9 Z cot θ csc θ
Z
1
− sin θ
1
1
=
dθ =
d( cos θ)
2
2
9
9
cos θ
cos θ
Z
1
du
1
sec θ
=
= −
+C =−
+C
2
9
9u
9
u
√
x2 + 9
=−
+C
9x
3
Lecture 7
Set
x
3 = cot θ
x = 3 cot θ
θ ∈ (0, π)
θ ∈ (0, π) ⇒
csc θ > 0
Set u = cos θ
p
x2 + 9
θ
x
2016
3
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
x2 + 1
x 2 + 1 corresponding to x = cot θ
√
x = cot θ transforms dx, x, x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
x
=
=
=
=
=
cot θ
cot (2 arctan t)
1 − tan2 (arctan t)
2 tan(arctan t)
1 − t2
2t
1 1
−t
.
2 t
The freecalc team
cos(2z)
1 − tan2 z
|Recall: cot(2z) =
=
sin(2z)
2 tan z
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
x2 + 1
x 2 + 1 corresponding to x = cot θ
√
x = cot θ transforms dx, x, x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
1 1
x =
−t
.
2 t
We can furthermore compute
s
2
p
1 1
2
x +1 =
−t
+1
4 t
s
2
1
1
=
−t
+4
2
t
The freecalc team
Lecture 7
1
t
−t
2
+4=
1
t
+t
2016
2
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
x2 + 1
x 2 + 1 corresponding to x = cot θ
√
x = cot θ transforms dx, x, x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
1 1
x =
−t
.
2 t
We can furthermore compute
s
2
p
1 1
2
x +1 =
−t
+1
4 t
s
2
1
1
=
+t
2
t
1 1
=
+t
.
2 t
The freecalc team
Lecture 7
q
1
t
2
+ t = 1t + t
because t > 0
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
x2 + 1
x 2 + 1 corresponding to x = cot θ
√
x = cot θ transforms dx, x, x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
1 1
x =
−t
.
2 t
We can furthermore compute
p
1 1
2
x +1 =
+t
.
2 t
Finally compute
dx
=
t
=
The freecalc team
1 1
1 1
d
−t
= −
+ 1 dt
2
2 t
2 t p
1 1
1 1
+t −
− t = x2 + 1 − x
2 t
2 t
Lecture 7
.
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
x2 + 1
x 2 + 1 corresponding to x = cot θ
√
x = cot θ transforms dx, x, x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
Definition
The Euler substitution for
by:
The freecalc team
√
x 2 + 1 corresponding to x = cot θ is given
1 1
−t ,
t >0
2 t
1 1
+t
2 t
1 1
−
+ 1 dt
2
√ 2 t
x2 + 1 − x .
x
=
p
x2 + 1
=
dx
=
t
=
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
1
2
Euler substitution: x =
√
t = x 2 + 1 − x, dx = − 21
Example
1
t
The case
p
x2 + 1
√
1 1
2
−t , x +1= 2 t +t ,
1
+ 1 dt. Recall t > 0.
t2
Z p
Z
1 1
1 1
2
+t
+ 1 dt
x + 1 dx= −
2
2 t
Z 2 t
1
1
1
=−
+ 2 + t dt
3
4 t
t
2
−2
t
1
t
+ 2 ln |t| +
= −
−
+C
2 4 2 1 −1
1
1 1 −1
t −t
t +t
− ln t + C
=
2 2
2
2
p
1 p 2
1
= x x + 1 − ln
x2 + 1 − x + C
2
2
√
p
2+1+x
x
1
1
√
+C
= x x 2 + 1 + ln √
2
2
x2 + 1 − x
x2 + 1 + x
1 p 2
1 p 2
= x x + 1 + ln
x +1+x +C
2
2
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
Example
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 + 1
√
√
2
Find the area locked b-n the hyperbolas y = ± x + 1 and x = ±2 2.
y =
p
x2 + 1
1
2
We studied v = u is called a
u
why do we call
hyperbola:
y+x y+x
v =0
( 2 , 2 )
v
√
y +x =0
y = x 2 + 1 hyperbola? Compute:
(y, x)
√
x2 + 1 = y
x2 + 1 = y2
y −x =0
2 − x2
u=0
y
= 1
√
√
2
2
1
p
(y
−
x)
(y
+
x)
=
2
y =− x +1
2
2
2
1
uv
=
Signed distance b-n (x, y ) and line
2
1
u =r0 equals
v = u2 ,
2 2
(x+y)
(x+y)
√
x− 2
+ y− 2
±
2
u
=
(y − x)
q
2
√
√
where
. Consider
2
1
2
2
v = 2 (y + x)
= ± 2 (y − x) = ± 2 (y − x) =
an arbitrary point (x, y).
u.
The freecalc team
(x, y)
Lecture 7
2016
Integrals of form
R
R(x,
Example
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 + 1
√
√
2
Find the area locked b-n the hyperbolas y = ± x + 1 and x = ±2 2.
y =
p
1
2
x2 + 1
We studied v = u is called a
why do we call
hyperbola:
v =0
√
y +x =0
y = x 2 + 1 hyperbola? Compute:
√
x2 + 1 = y
x2 + 1 = y2
y −x =0
2 − x2
u=0
y
= 1
√
√
2
2
1
p
(y
−
x)
(y
+
x)
=
2
y =− x +1
2
2
2
1
uv
=
Signed distance b-n (x, y ) and line
2
1
u = 0 equals u. Similarly
v = u2 ,
compute that signed distance b-n
√
(x, y) and the line v = 0 equals v .
u = √22 (y − x)
where
. Consider
2
⇒ y 2 − x 2 = 1 is the hyperbola
v = 2 (y + x)
1/2
v = u in the (u, v )-plane.
an arbitrary point (x, y).
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
Example
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 + 1
√
√
2
Find the area locked b-n the hyperbolas y = ± x + 1 and x = ±2 2.
y =
p
x2 + 1
The √area in question is:
2 2
Z
p
2 x 2 + 1dx
√
−2 h2
y =−
p
x2 + 1
p
= 2 x x2 + 1
p
i2√2
+ ln
x2 + 1 + x
√
0−2 2
q
√
√
= 2 2 2 (2 2)2 + 1
q
√
√
+ ln
(2 2)2 + 1 + 2 2
√
√ = 12 2 + 2 ln 3 + 2 2
≈ 20.496
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 + 1
Example
Find
R
√x
x 2 +4
dx.
We could use the trig substitution x = 2 tan θ.
But there is an easier way:
u = x 2 + 4.
du =Z2xdx.
Z
p
√
1
du
x
√
√ = u + C = x2 + 4 + C
dx =
2
u
x2 + 4
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
−x 2 + 1
Trigonometric substitution x = cos θ for
√
−x 2 + 1
√
The trigonometric substitution x = cos θ, θ ∈ [0, π] for −x 2 + 1:
p
p
1 − cos2 θ
−x 2 + 1 =
p
when θ ∈ [0, π] we
have
2
p
=
sin θ
sin θ ≥ 0 and so sin2 θ = sin θ
= sin θ .
To summarize:
Definition
The trigonometric substitution x = cos θ, θ ∈ [0, π] for
given by:
x = cos θ
√
−x 2 + 1 = sin θ
dx = − sin θdθ
θ = arccos x .
The freecalc team
Lecture 7
√
−x 2 + 1 is
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
−x 2 + 1
Example
Evaluate
R √9−x 2
x2
dx.
Let x = 3 sin θ, where
−π/2 ≤ θ ≤ π/2.
θ
Then dx = 3 cos θdθ.
q
p
p
2
9 − x 2 = 9 − 9 sin θ = 9 cos2 θ = 3| cos θ| = 3 cos θ
Z √
9−
x2
x2
dx
Z
=
=
=
The freecalc team
3 cos θ
3 cos θdθ =
Z
cot2 θdθ
2
9
sin
θ
Z
(csc2 θ − 1)dθ = − cot θ − θ + C
√
x 9 − x2
−
− arcsin
+C
x
3
Lecture 7
2016
Integrals of form
R
R(x,
Example
p
ax 2 + bx + c)dx, R - rational function
The case
Find the area enclosed by the ellipse
y =b
(−a, 0)
r
x2
a2
+
p
−x 2 + 1
y2
b2
= 1, a, b > 0.
2
1 − x2
a
(a, 0)
r
2
y = −b 1 − x 2
a
The area in question is
Za r
x2
2b 1 − 2 dx
a
−a
Za r
x2
= 4 b 1 − 2 dx.
a
Express y via x:
x2 y2
+ 2 = 1
2
a
b
y2
x2
= 1− 2
2
b
a
2
x
2
2
y = b 1− 2
r a
x2
y = ±b 1 − 2
a
0
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
Example
p
ax 2 + bx + c)dx, R - rational function
The case
Find the area enclosed by the ellipse
y =b
r
2
1 − x2
a
r
2
y = −b 1 − x 2
a
The area in question is
Za r
x2
2b 1 − 2 dx
a
−a
Za r
x2
= 4 b 1 − 2 dx
a
x2
a2
The freecalc team
−x 2 + 1
y2
b2
= 1, a, b > 0.
Trig subst.: set x = a sin θ, θ ∈ 0, .
q
q
x2
a2 sin2 θ
Compute: 1 − a2 = 1 − a2 =
p
1 − sin2 θ = cos θ. When x = 0, θ = 0 and
when x = a, θ = π2 .
Z ar
Z π
2
2
x
1 − 2 dx =
cos θ d (a sin θ)
a
0
0
Z π
π
2
=
=
=
=
=
0
= 4b aπ
4 = πab
+
p
.
Lecture 7
a
2
Z0 π
cos2 θdθ
cos(2θ) + 1
a
dθ
2
h0
iθ= π
2
sin(2θ)
θ
a
+2
4
θ=0 π
a 0 + 4 − (0 + 0)
2
aπ
4
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
−x 2 + 1
Example
Evaluate
R
√
x
dx.
2
3−2x−x
Complete the square under the root sign:
3 − 2x − x 2 = 3 + 1 − (x 2 + 2x + 1) = 4 − (x + 1)2
Substitute u = x + 1. Then du = dx and x = u − 1.
R
R
R u−1
x
x
√
dx = √
dx = √ 2 du
2
2
3−2x−x
4−(x+1)
4−u
Let u = 2 sin θ, where −π/2 ≤ θ ≤ π/2. Then du = 2 cos θdθ.
p
√
√
2
4 − u 2 = 4 − 4 sin θ = 4 cos2 θ = 2| cos θ| = 2 cos θ
R
R u−1
R 2 sin θ−1
x
√
√
dx
=
du =
2 cos θdθ
2
cos
θ
2
2
4−u
R 3−2x−x
= (2
− 2 cos θ − θ + C
√sin θ − 1)dθ =
−1 u
2
= − √4 − u − sin
2 +C = − 3 − 2x − x 2 − sin−1 x+1
+C
2
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
−x 2 + 1
−x 2 + 1 corresponding to x = cos θ
√
x = cos θ transforms dx, x, −x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
x
=
=
=
=
cos θ
cos(2 arctan t)
1 − tan2 ( arctan t)
1 + tan2 ( arctan t)
1 − t2
1 + t2
The freecalc team
1 − tan2 z
cos(2z) =
1 + tan2 z
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
−x 2 + 1
−x 2 + 1 corresponding to x = cos θ
√
x = cos θ transforms dx, x, −x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
x
√
−x 2
+1
=
=
=
=
=
The freecalc team
1 − t2
2
1
+
t
s
t2
1−
1 + t2
2
1−
s
(1 + t 2 )2 − (1 − t 2 )2
(1 + t 2 )2
s
4t 2
(1 + t 2 )2
(1 + t 2 )2 − (1 − t 2 )2 = 4t 2
√
4t 2 = 2t because t > 0
2t
1 + t2
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
−x 2 + 1
−x 2 + 1 corresponding to x = cos θ
√
x = cos θ transforms dx, x, −x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
√
x
=
−x 2 + 1
=
t
=
dx
=
=
The freecalc team
1 − t2
1 + t2
2t
1 + t2
√
√
√
1−x 1+x
−x 2 + 1
√
√
=
1+x 1+x x +1
2
2
1−t
2 − (1 + t )
d
=d
2
1 + t2
1 + t
2
4t
d
−1 =−
dt
2
2
2
1+t
(1 + t )
Lecture 7
we use t > 0
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
Euler subst. for
√
The case
p
−x 2 + 1
−x 2 + 1 corresponding to x = cos θ
√
x = cos θ transforms dx, x, −x 2 + 1 to trig form.
θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form.
What if we compose the above? We get the Euler substitution:
Definition
The Euler substitution for
given by:
√
−x 2 + 1 corresponding to x = cos θ is
x
√
The freecalc team
=
−x 2 + 1 =
dx
=
t
=
1 − t2
,
t >0
2
1+t
2t
1 + t2
4t
dt
− 2
2
√(t + 1)
−x 2 + 1
.
x +1
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 − 1
Trigonometric substitution x = sec θ for
√
x2 − 1
3π ∪ π, 2 :
The trigonometric substitution x = sec θ, θ ∈ 0,
p
p
x 2 − 1 = rsec2 θ − 1
1
=
−1
2
s cos θ
sin2 θ
=
cos2 θ
π 3π p
when θ ∈ θ ∈ 0, p
2 ∪ π, 2 we have
2
=
tan θ
tan θ ≥ 0 and so tan2 θ = tan θ
= tan θ .
The freecalc team
Lecture 7
π
2
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 − 1
Trigonometric substitution x = sec θ for
The trigonometric substitution x = sec θ, θ ∈ 0,
p
x 2 − 1 = tan θ .
Definition
π
2
√
x2 − 1
3π ∪ π, 2 :
√
The trigonometric substitution x = sec θ, θ ∈ (0, π) for x 2 + 1 is given
by:
1
π
3π
x = sec θ =
θ ∈ 0,
∪ π,
cos θ
2
2
p
x 2 − 1 = tan θ
sin θ
dθ = sec θ tan θdθ
dx =
2
cos θ
θ = arcsecx .
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 − 1
Example
Find
R
√ dx
x 2 −a2
, a > 0.
x = a sec θ,
0 < θ < π/2 or
π < θ < 3π/2.
θ
dx = a sec θ tan θdθ.
p
p
p
x 2 − a2 = a2 sec2 θ − a2 = a2 tan2 θ = a| tan θ| = a tan θ
Z
Z
Z
dx
a sec θ tan θdθ
√
=
= sec θdθ
a tan θ
x 2 − a2
√
x
x 2 − a2 = ln | sec θ + tan θ| + C = ln +
+C
a
a
p
= ln x + x 2 − a2 + C1
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 − 1
Euler substitution x = sec θ, θ = 2 arctan t
√
x = sec θ transforms dx, x, x 2 − 1 to trig form.
θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ.
What if we compose the above? We get the Euler substitution:
1
x = sec θ =
cos θ
1
1 − tan2 z
=
cos(2z) =
cos(2 arctan t)
1 + tan2 z
1 + tan2 ( arctan t)
=
1 − tan2 ( arctan t)
1 + t2
2 − (1 − t 2 )
=
=
2
1 − t2
1−t
2
= −1+
1 − t2
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 − 1
Euler substitution x = sec θ, θ = 2 arctan t
√
x = sec θ transforms dx, x, x 2 − 1 to trig form.
θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ.
What if we compose the above? We get the Euler substitution:
2
x = −1+
1 − t2
s
2
2
√
1+t
2
x −1 =
−1
2
s 1−t
=
s
=
=
The freecalc team
(1 + t 2 )2 − (1 − t 2 )2
(1 − t 2 )2
(1 + t 2 )2 − (1 − t 2 )2 = 4t 2
4t 2
(1 − t 2 )2
t, 1 − t 2 have same sign
when t ∈ (−∞, −1) ∪ [0, 1)
2t
1 − t2
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 − 1
Euler substitution x = sec θ, θ = 2 arctan t
√
x = sec θ transforms dx, x, x 2 − 1 to trig form.
θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ.
What if we compose the above? We get the Euler substitution:
2
x = −1+
1 − t2
√
2t
2
x −1 =
1 − t2
1 + t2
x =
1 − t2
(1 − t 2 )x = 1 + t 2
(1 + x)t 2 = x − 1
x −1
2
t
=
x +r1
x −1
t = ±
x +1
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 − 1
Euler substitution x = sec θ, θ = 2 arctan t
√
x = sec θ transforms dx, x, x 2 − 1 to trig form.
θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ.
What if we compose the above? We get the Euler substitution:
2
x = −1+
1 − t2
√
2t
2
x −1 =
1 −rt 2
x −1
t = ±
x +1
2
dx = d − 1 +
1 − t2
4t
=
dt
2
2
(1 − t )
The freecalc team
Lecture 7
2016
Integrals of form
R
R(x,
p
ax 2 + bx + c)dx, R - rational function
The case
p
x2 − 1
Euler substitution x = sec θ, θ = 2 arctan t
√
x = sec θ transforms dx, x, x 2 − 1 to trig form.
θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ.
What if we compose the above? We get the Euler substitution:
Definition
The Euler substitution for
by:
√
x
=
x2 − 1
=
dx
=
t
=
The freecalc team
√
x 2 − 1 corresponding to x = sec θ is given
1 + t2
,
2
1−t
2t
1 − t2
4t
dt
2
2
(1 −
√t )
x2 − 1
±
x +1
Lecture 7
t ∈ (−∞, −1) ∪ [0, 1)
.
2016
Rationalizing Substitutions
Rationalizing Substitutions
Some nonrational fractions can be changed into rational fractions by
means of appropriate substitutions.p
In particular, when an integrand
p
n
contains an expression of the form g(x), the substitution u = n g(x)
may be effective.
The freecalc team
Lecture 7
2016
Rationalizing Substitutions
Example
√
Let u = x + 4. Then u 2 = x + 4, so x = u 2 − 4 and dx = 2udu.
Z √
Z
u
x +4
2udu
dx =
2
x
Z u −24
u
=2
du
2
u −4
Z 4
=2
du
long division
1+ 2
uZ − 4
Z
du
= 2 du + 8
u2 − 4
!
Z
Z
1
1
4
− 4
du partial fractions
= 2 du + 8
u−2 u+2
= 2u + 2(ln |u − 2|
− ln |u + 2|)
√
+C
√
x + 4 − 2
+C
= 2 x + 4 + 2 ln √
x + 4 + 2
The freecalc team
Lecture 7
2016
Rationalizing Substitutions
Example
Find the surfaceqarea of the ellipsoid obtained
y
by rotating y =
1−
x2
2
about the x axis.
x
z
The freecalc team
Lecture 7
2016
Rationalizing Substitutions
Example
Find the surfaceqarea of the ellipsoid obtained
y
by rotating y =
y0 =
x
z
(y 0 )2 =
Area =
=
=
=
=
The freecalc team
1−
x2
2
about the x axis.
1 q(−x)
2
2
1− x2
x2
4y 2
x
= − 2y
R2
q
x2
1 + 4y 2 dx
−2 2πy
q 2 2
R2
4y +x
2πy
dx
2
−2
4y
q 2 2
R2
4y +x
2πy
dx
−2 r
4y 2
R2
2
x
2 dx
π
4
1
−
+
x
2
−2
R2 √
2 dx
4
−
x
π
−2
Lecture 7
2016