Chapter 7 – Techniques of Integration
7.3 Trigonometric Substitution
1
7.3 Trigonometric Substitution
Erickson
When do we use it?

Trigonometric substitution is used when you have
problems involving square roots with two terms under the
radical.
You will make one of the substitutions below depending
on what is inside your radical.

Expression
𝑎2
−
𝑢2
𝑎2 + 𝑢2
𝑢2 − 𝑎2
2
Substitution
𝜋
𝜋
≤𝜃≤
2
2
𝜋
𝜋
𝑢 = 𝑎 tan 𝜃, − ≤ 𝜃 ≤
2
2
𝑢 = 𝑎 sec 𝜃,
𝜋
3𝜋
0 ≤ 𝜃 ≤ or 𝜋 ≤ 𝜃 ≤
2
2
𝑢 = 𝑎 sin 𝜃, −
7.3 Trigonometric Substitution
Identity
cos 2 𝜃 = 1 − sin2 𝜃
sec 2 𝜃 = 1 + tan2 𝜃
tan2 𝜃 = sec 2 𝜃 − 1
Erickson
General Process
1.
Decide which formulas you will work with.
2.
Draw your triangle.
3.
Solve the triangle.
4.
Make your trig substitutions.
5.
Integrate.
6.
Convert back to the original variables.
3
7.3 Trigonometric Substitution
Erickson
Example 1

9  x2
dx
2
x
1. Choose your formula. This
matches the first set of formulas,
so x = 3sin
2.
3
Draw your triangle
•
x
sin = x/3

3. Solve the triangle
9  x2
4. Make your trig substitutions.
x  3sin 
dx  3cos  d
Note : cos  
4

9 x
3
2
3cos   9  x 2
9  x2
3cos 
dx

 9sin 2   3cos d 
x2
  cot 2  d
   csc 2   1 d
7.3 Trigonometric Substitution
Erickson
Example 1 continued
2
csc
    1 d   cot    C
5. Integrate
6. Convert Back
3
9  x2
x
 cot     C  
 sin 1  C
x
3
x


9  x2
9  x2
1 x
dx  
 sin
C
2
x
x
3
9  x2
5
7.3 Trigonometric Substitution
Erickson
Example 2

x2  4
dx
x
1. Choose your formula. This
matches the third set of formulas,
so x = 2sec
2.
x
Draw your triangle
x2  4
sec = x/2

3. Solve the triangle
2
4. Make your trig substitutions.
x  2sec 
dx  2sec  tan  d
Note : tan  
6
x 4
2
2

2 tan   x 2  4
x2  4
2 tan 
dx  
 2sec tan  d 
x
2sec 
 2 tan 2  d
 2  sec 2   1 d
7.3 Trigonometric Substitution
Erickson
Example 2 continued
2  sec 2   1 d  2  tan      C
5. Integrate
6. Convert Back
 x2  4

1 x
2  tan      C  2 
 sec
C
 2
2 

x
x2  4


x2  4
2
1 x
dx  x  4  2sec
C
x
2
2
7
7.3 Trigonometric Substitution
Erickson
Example 3
x
1
4 x 2  16
dx
1. Choose your formula. This
matches the first set of formulas,
so 2x = 4tan
2.
Draw your triangle
•
4 x2  16
2x
tan = (2x)/4

3. Solve the triangle
4
4. Make your trig substitutions.
2 x  4 tan 
x  2 tan 
dx  2sec 2  d
Note : cos  
4
4 x 2  16
4 x 2  16 cos   4
4
4 x 2  16 
cos 
8
2sec 2  d
 x 4 x 2  16   2 tan  4sec
1 sec 
 
d
4 tan 
1
  csc  d
4
dx
4 x 2  16  4sec 
7.3 Trigonometric Substitution
Erickson
Example 3 continued
1
1
csc  d  ln csc   cot   C

4
4
5. Integrate
6. Convert Back
1
1
ln csc   cot   C  ln
4
4
1
 ln
4
dx
1

  ln
2
4
x 4 x  16
9
4 x 2  16 2
 C
2x
x
x2  4 2
 C
x
x
x2  4 2
 C
x
x
7.3 Trigonometric Substitution
Erickson
Examples

Evaluate the integral.
a.
b.
10


1  4 x dx
2

dx
2/3
2 /3
c.
1  x2
dx
x
x
5
9x 1
2
d.

7.3 Trigonometric Substitution

0
2
cos t
1  sin 2 t
Erickson
dt
Example

Find the volume of the solid obtained by rotating about
the x-axis the region enclosed by the curves
9
y 2
, y  0, x  0, and x  3
 x  9
11
7.3 Trigonometric Substitution
Erickson