Math 166
September 25, 28, 2015
Trigonometric Substitution
In this section we learn how to handle various integrals by using substitutions
that involve trigonometric functions.
0.1
The technique
integral involves
a2 − x2
x2 + a2
2
x − a2 , x > a
0.2
substitution
x = a sin θ
x = a tan θ
x = a sec θ
θ range
−π/2 ≤ θ ≤ π/2
−π/2 < θ < π/2
0 ≤ θ < π/2
key identity
sin2 θ + cos2 θ = 1
tan2 θ + 1 = sec2 θ
tan2 θ + 1 = sec2 θ
Examples
Example 1
Z
√
dx
4 + x2
Let x = 2 tan θ where −π/2 < θ < π/2. Then dx = 2 sec2 θ dθ. Observe that
Z
Z
dx
2 sec2 θ
√
√
=
dθ
2
4+x
4 + 4 tan2 θ
Z
sec2 θ
√
=
dθ
1 + tan2 θ
Z
sec2 θ
=
dθ.
| sec θ|
1
Note that for the range of θ chosen above, | sec θ| = sec θ. Therefore,
Z
Z
dx
sec2 θ
√
=
dθ
| sec θ|
4 + x2
Z
= sec θ dθ
= ln | sec θ + tan θ| + C.
√
Note that (using triangle technique) sec θ = 21 x2 + 4. Thus,
p
Z
1
dx
1 √
= ln | sec θ + tan θ| + C = ln x2 + 4 + x + C.
2
2
4 + x2
Example 2
Z
x2 dx
√
9 − x2
Let x = 3 sin θ where −π/2 ≤ θ ≤ π/2. Then dx = 3 cos θ dθ. Observe that
Z
Z
x2 dx
9 sin2 θ cos θ
√
=3 p
dθ
9 − x2
9 − 9 sin2 θ
Z
9 sin2 θ cos θ
p
dθ
=
1 − sin2 θ
Z
9 sin2 θ cos θ
=
dθ
| cos θ|
Z
9 sin2 θ cos θ
=
dθ
cos θ
Z
= 9 sin2 θ dθ
Z
9
=
1 − cos 2θ dθ
2
9
1
=
θ − sin 2θ
2
2
9
= [θ − sin θ cos θ]
2
√
Note that (using triangle technique) cos θ = 31 9 − x2 . Thus
"
#
√
Z
x2 dx
9
9
x x 9 − x2
√
= [θ − sin θ cos θ] =
arcsin −
2
2
3
9
9 − x2
Example 3
Z
√
dx
,
25x2 − 4
2
x > 2/5
Notice that the integrand here doesn’t quite fit any of our three standard
forms. Therefore we start the problem with some algebra, observing that
Z
Z
Z
dx
1
dx
dx
√
q
q
=
=
.
2
5
25x − 4
25 x2 − 4
x2 − 4
25
25
Now we make the substitution x = 2/5 sec θ for 0 ≤ θ < π/2 . Then dx =
2/5 sec θ tan θ dθ and
Z
Z
dx
sec θ tan θ dθ
1
q
√
=
5
4
sec2 θ − 1
x2 − 25
Z
= sec θ dθ
= ln | sec θ + tan θ| + C.
Next note√that since sec θ = 5x/2, it follows using our reference triangle that
tan θ = 21 25x2 − 4. Therefore
Z
5
dx
1p
2
√
25x − 4 + C.
= ln | sec θ + tan θ| + C = ln x +
2
2
2
25x − 4
0.3
Problems
Problem 1
Z
√
dx
9 + x2
We use the substitution x = 3 tan θ for −π/2 < θ < π/2. Then dx =
3 sec2 θ dθ and
Z
Z
dx
3 sec2 θ dθ
√
√
=
9 + x2
9 + 9 tan2 θ
Z
= sec θ dθ
= ln | sec θ + tan θ| + C.
√
Next we note that since tan θ = x/3 it follows that sec θ = x2 + 9/3 and we
get that
p
Z
1
1 dx
2
√
= ln | sec θ + tan θ| + C = ln x + 9 + x + C.
3
3
9 + x2
3
Problem 2
√
Z
0
3/2
4x2 dx
(1 − x2 )3/2
We make the substitution x = sin θ for −π/2 ≤ θ ≤ π/2. Then dx = cos θ dθ
and
Z √3/2
Z π/3
4 sin2 θ cos θ dθ
4x2 dx
=
2
3/2
(1 − x )
(1 − sin2 θ)3/2
0
0
Z π/3
=4
tan2 θ dθ
0
Z
π/3
(sec2 θ − 1) dθ
0
π/3 = 4 tan θ − θ
0
√
= 4( 3 − π/3).
=4
Problem 3
Z √
x2 − 25
dx,
x3
x>5
We use the substitution x = 5 sec θ for 0 ≤ θ < π/2. Then dx = 5 sec θ tan θ dθ.
Therefore
Z √
Z √ 2
x − 25
25 sec2 θ − 25
dx =
5 sec θ tan θ dθ
3
x
125 sec3 θ
Z
1
tan2 θ
=
dθ
5
sec2 θ
Z
1
=
sin2 θ dθ
5
Z
1
=
(1 − cos 2θ) dθ
10
1
1
=
θ − sin 2θ + C
10
2
1
=
[θ − sin θ cos θ] + C.
10
Now note when sec θ = x/5
√ we have that cos θ = 5/x, and using a reference
triangle we find that sin θ = x2 − 25/x. Therefore
Z √ 2
x − 25
1
1
5p 2
−1 x
dx =
[θ − sin θ cos θ]+C =
sec
− 2 x − 25 +C.
x3
10
10
5
x
4
Problem 4
Z
100
dx
36 + 25x2
First we note that
100
100
=
=
2
36 + 25x2
25 36
25 + x
36
25
4
.
+ x2
Next, we use the trigonometric substitution x = 6/5 tan θ for −π/2 < θ < π/2.
Then dx = 6/5 sec2 θ dθ and we observe that
Z
Z
4
100
dx
=
dx
36
36 + 25x2
+
x2
25
Z
sec2 θ
24
dθ
=
36
5
+ 36 tan2 θ
Z 25 25
10
=
dθ
3
10
θ+C
=
3
10
5x
−1
=
tan
+ C.
3
6
Problem 5
Z
0
ln 4
√
et dt
e2t + 9
We start by making the substitution x = et . Then dx = et dt and
Z ln 4
Z 4
et dt
dx
√
√
=
.
2t
e +9
x2 + 9
0
1
Now we use the trigonometric substitution x = 3 tan θ for −π/2 < θ < π/2.
Therefore dx = 3 sec2 θ dθ and
Z 4
Z
dx
3 sec2 θ dθ
√
√
=
x2 + 9
9 tan2 θ + 9
1
Z
= sec θ dθ
x=4
= ln | sec θ + tan θ|
.
x=1
Next
note that since tan θ = x/3, using a reference triangle we find that sec θ =
√
x2 + 9/3 and we have that
p
Z 4
x=4
1
dx
1 x=4
8 1√
2
√
= ln | sec θ+tan θ|
= ln x + 9 + x = ln
−
10 .
3
3
3 3
x=1
x=1
x2 + 9
1
5
Problem 6
e
Z
dy
y
1
p
1 + (ln y)2
We start by making the substitution x = ln y. Then dx = dy/y. Therefore
Z
1
e
Z
dy
y
p
1+
(ln y)2
1
=
0
√
dx
.
1 + x2
Next we make the substitution x = tan θ for −π/2 < θ < π/2. Then dx =
sec2 θ dθ and
Z 1
Z π/4
dx
sec2 θ dθ
√
√
=
1 + x2
1 + tan2 θ
0
0
Z π/4
=
sec θ dθ
0
π/4
= ln | sec θ + tan θ|
√
1
= ln | 2| = ln 2.
2
Problem 7
Z
0
1
0
4y dy
p
1 − y4
We start by making the substitution u = y 2 . Then du = 2y dy and
Z 1
Z 1
4y dy
2 du
p
√
=
.
4
1
− u2
1−y
0
0
Next we make the substitution u = sin θ for −π/2 ≤ θ ≤ π/2. Then du =
cos θ dθ and
Z 1
Z π/2
2 du
2 cos θ dθ
√
p
=
= π.
2
1−u
0
0
1 − sin2 θ
Problem 8 Find the area enclosed by the ellipse
x2
y2
+ 2 = 1,
2
a
b
where a, b > 0 are constants.
6
Note that by solving the defining Cartesian equation for y, we find that the
ellipse is given by the following two functions:
r
bp 2
x2
y = ±b 1 − 2 = ±
a − x2 .
a
a
Moreover, note that by plugging in y = 0 it is easy to see that the two xintercepts of the ellipse are x = ±a. Therefore, by symmetry it follows that the
area enclosed by ellipse is given by the integral
Z a p
b
a2 − x2 dx.
4
a
0
Now, to compute this integral we’ll use the substitution x = a sin θ for
−π/2 ≤ θ ≤ π/2. Observe that dx = a cos θ dθ and
Z a p
Z p
b
4
a2 − x2 dx = 4b
a2 − a2 sin2 θ cos θ dθ
a
0
Z
= 4ab cos2 θ dθ
Z
= 2ab (1 + cos 2θ) dθ
x=a
1
= 2ab θ + sin 2θ 2
x=0
x=a
= 2ab (θ + sin θ cos θ) x=0
a
p
x
x
= 2ab sin−1 + 2 a2 − x2 a a
0
= πab.
7