Trignometric Substitutions
I) Evaluate the following indefinite integrals:
Z
3
1) (36 − 9x2 )− 2 dx
√
Sol. Let x = 2 sin θ, so that dx = 2 cos θdθ, and 4 − x2 = 2 cos θ. Then
Z
Z
1
1
2 − 32
(36 − 9x ) dx =
3 dx
27
(4 − x2 ) 2
Z
2 cos θ
1
dθ
=
27
8 cos3 θ
Z
1
=
sec2 θdθ
108
1
=
tan θ + C
108
x
√
+C
=
108 4 − x2
Z
2)
√
dx
3 − 2x − x2
Z
dx
√
3 − 2x − x2
Z
dx
p
4 − (x + 1)2
Z
1
√
du,
=
4 − u2
=
u = x + 1.
Then let u = 2 sin θ; So that du = 2 cos θdθ. We have:
Z
2 cos θ
dθ = θ + C
2 cos θ
x+1
= sin−1 (
)+C
2
Z √
3)
9x2 − 25
5
dx, x >
3
x
3
5
Let x = sec θ, where θ ∈ [0, π2 ).
3
dx =
5
sec θ tan θdθ
3
and
p
9x2 − 25 = 5 tan θ
1
Thus,
Z √
9x2 − 25
dx =
x3
=
=
=
=
=
=
=
=
5 tan θ 35 sec θ tan θ
dθ
125
3
27 sec θ
Z
9
tan2 θ
dθ
5
sec2 θ
Z
9
sec2 θ − 1
dθ
5
sec2 θ
Z
9
(1 − cos2 θ)dθ
5
Z
9
sin2 θdθ
5
Z
9
(1 − cos 2θ)dθ
10
9θ 9 sin 2θ
−
+C
10
20
9θ 9 sin θ cos θ
−
+C
10
10 √
5
9 cos−1 ( 3x
)
9x2 − 25
−
+ C.
10
2x2
Z
x2
dx.
(25 + x2 )2
Let x = 5 tan θ, so that dx = 5 sec2 θdθ. Note that 25 + x2 = 25 sec2 θ. Thus,
Z
Z
25 tan2 θ × 5 sec2 θ
x2
dx
=
dθ
(25 + x2 )2
252 sec4 θ
Z
1
tan2 θ
=
dθ
5
sec2 θ
Z
sec2 θ − 1
1
dθ
=
5
sec2 θ
Z
1
=
(1 − cos2 θ)dθ
5
Z
1
=
sin2 θdθ
5
Z
1
(1 − cos 2θ)dθ
=
10
1
sin 2θ
=
(θ −
)+C
10
2
1
=
(θ − sin θ cos θ) + C
10
1
x
5x
=
(tan−1 ( ) −
)+C
10
5
25 + x2
Z
4)
II) Evaluate the following definite integrals:
2
Z
1)
1
dx
dx
x2 + 16
0
√
Let x = 4 tan θ, so that dx = 4 sec2 θdθ. Note that x2 + 16 = 4 sec θ.Thus,
√
Z
0
1
tan−1 ( 14 )
Z
dx
√
dx =
2
x + 16
0
tan−1 ( 14 )
Z
=
4 sec2 θ
dθ
4 sec θ
sec θdθ
0
tan−1 ( 14 )
= ln | sec θ + tan θ||0
√
17 + 1
= ln(
)
4
Z
4
2)
√4
3
dx
dx
− 4)
x2 (x2
Let x = 2 sec θ, so that dx = 2 sec θ tan θdθ and x2 − 4 = 4 tan2 θ. Thus
Z
4
√4
3
dx
dx =
2
x (x2 − 4)
=
=
=
=
=
π
3
Z
2 sec θ tan θ
dθ
4 sec2 θ × 4 tan2 θ
π
6
1
8
Z
1
8
Z
1
8
Z
π
3
cos2 θ
dθ
sin θ
π
3
1 − sin2 θ
dθ
sin θ
π
6
π
6
π
3
csc θ − sin θdθ
π
6
π
1
(− ln | csc θ + cot θ| + cos θ)| π3
8
√6
√
√
1
1− 3
(− ln( 3(2 − 3)) +
).
8
2
3