Download Section 8.3: Trigonometric Substitution

Section 8.3: Trigonometric Substitution
Integration by trigonometric substitution is used if the integrand involves a radical and
u-substitution fails. There are three cases to consider:
√
1. If the√integrand involves a2 − x2 , then substitute x = a sin θ so that dx = a cos θdθ
and a2 − x2 = a cos θ.
√
2. If the√integrand involves x2 − a2 , then substitute x = a sec θ so that dx = a sec θ tan θdθ
and x2 − a2 = a tan θ.
√
3. If the√integrand involves x2 + a2 , then substitute x = a tan θ so that dx = a sec2 θdθ
and x2 + a2 = a sec θ.
Z
Example: Evaluate
x2
dx
√
.
25 − x2
√
Let x = 5 sin θ, dx = 5 cos θdθ, 25 − x2 = 5 cos θ. Then
Z
Z
5 cos θ
dx
√
=
dθ
2
2
25 sin2 θ(5 cos θ)
x 25 − x
Z
1
1
=
2 dθ
25
Z sin θ
1
csc2 θdθ
=
25
1
= − cot θ + C.
25
Drawing a reference triangle, we find that
√
Z
1
dx
25 − x2
√
= − cot θ + C = −
+ C.
25
25x
x2 25 − x2
5
𝑥
𝜃
25 − 𝑥 2
Z
Example: Evaluate
x2
dx
√
.
x2 − 4
√
Let x = 2 sec θ, dx = 2 sec θ tan θdθ, x2 − 4 = 2 tan θ. Then
Z
Z
2 sec θ tan θ
dx
√
=
dθ
2
2
4 sec2 θ(2 tan θ)
x x −4
Z
1
1
=
dθ
4
sec θ
Z
1
=
cos θdθ
4
1
sin θ + C.
=
4
Drawing a reference triangle, we find that
√
Z
dx
1
x2 − 4
√
= sin θ + C =
+ C.
4
4x
x2 x2 − 4
𝑥
𝑥2 − 4
𝜃
2
Z
Example: Evaluate
0
3
dx
√
.
9 + x2
√
Let x = 3 tan θ, dx = 3 sec2 θdθ, 9 + x2 = 3 sec θ. Then
Z 3
Z π/4
dx
3 sec2 θ
√
=
dθ
3 sec θ
9 + x2
0
0
Z π/4
=
sec θdθ
0
= ln | sec θ + tan θ||π/4
0
√
= ln | 2 + 1|.
Z
Example: Evaluate
x2
dx.
(4 − x2 )3/2
Let x = 2 sin θ, dx = 2 cos θdθ, (4 − x2 )3/2 = (2 cos θ)3 = 8 cos3 θ. Then
Z
Z
x2
4 sin2 θ(2 cos θ)
dθ
dx
=
(4 − x2 )3/2
8 cos3 θ
Z
sin2 θ
=
dθ
cos2 θ
Z
=
tan2 θdθ
Z
=
(sec2 θ − 1)dθ
= tan θ − θ + C.
Drawing a reference triangle, we find that
Z
x
x2
−1 x
dx = tan θ − θ + C = √
+ C.
+ sin
(4 − x2 )3/2
2
4 − x2
4
𝑥
𝜃
4 − 𝑥2
√
Note: If the integrand involves ax2 + bx + c, then complete the square and use the appropriate trigonometric substitution.
Z
Example: Evaluate
√
x2
dx
.
+ 4x + 8
First, complete the square
x2 + 4x + 8 = (x2 + 4x) + 8 = (x2 + 4x + 4) + 4 = (x2 + 2)2 + 4.
Let u = x + 2, du = dx. Then
Z
Z
dx
du
√
√
=
.
2
x + 4x + 8
u2 + 4
√
Let u = 2 tan θ, du = 2 sec2 θdθ, u2 + 4 = 2 sec θ. Then
Z
Z
2 sec2 θdθ
du
√
=
2 sec θ
u2 + 4
Z
=
sec θdθ
= ln | sec θ + tan θ| + C
√
u2 + 4 + u = ln +C
2
√
x2 + 4x + 8 + x + 2 = ln +C
2
√
= ln | x2 + 4x + 8 + x + 2| + C.
Example: Evaluate
Z √
16 − x2 dx.
√
Let x = 4 sin θ, dx = 4 cos θdθ, 16 − x2 = 4 cos θ. Then
Z
Z √
2
16 − x dx =
(4 cos θ)(4 cos θ)dθ
Z
= 16 cos2 θdθ
Z
= 8 (1 + cos 2θ)dθ
1
= 8 θ + sin 2θ + C
2
= 8θ + 8 sin θ cos θ + C √
x 16 − x2 −1 x
= 8 sin
+8
+C
4
4
4
x√16 − x2
−1 x
= 8 sin
+
+ C.
4
2