2.3 Trigonometric Substitutions
This method works on integrals involving a square root of a quadratic. The simplest cases are
a2 + x2 and
a2 - x2,
x2 - a2 where a is some number. In the first case we can substitute x = a sin  or
x = a tanh . In the second case substitute x = a tan  or x = a sinh . In the third substitute x = a sec  or
x = a cosh .. In each case this gets rid of the radical and converts the integral into a trigonometric or
hyperbolic-trigonometric integral of the form considered in section 2.2. Sometimes this new integral is
fairly easy to do while in other situations it can be difficult. Let's look at some examples.
Example 1. Find
2

 9 - x dx.
A typical situation where this type of integral arises is in finding areas of parts of circles. For example, the
2
definite integral 

9 - x2 dx is the area of part of the circle of radius 3 centered at the origin, i.e. the region
1
inside the circle, above the x axis and between the lines x = 1 and x = 2.
To do this integral we let x = 3 sin . Then dx = 3 cos  d and the integral becomes
2
2
2

 9 - x dx = 
 9 - 9 sin  (3 cos ) d = 3 
 9(1 - sin ) cos  d
= 3

= 9

2
9 cos2 cos  d = 3 
 3 (cos ) (cos ) d = 9 
 cos  d
1 + cos 2
9
9
d =
 + 4 sin 2
2
2
x
Now we have to convert from  back to x. One has  = sin-1 . For sin 2 we can use a trig identity. We
3
get
9 -1x 
9
9 -1x 
9 x 
2
2

 9 - x dx = 2 sin 3 + 4 (2 sin  cos ) = 2 sin 3 + 2 3 1 - sin 
=
9 -1x 
3
sin
2
3 + 2 x
Example 2. Find
x 2
9 -1x 
1
1-  =
sin
2
3
3 + 2 x
9 - x2
2

 9 + x dx.
To do this integral we let x = 3 sinh . Then dx = 3 cosh  d and the integral becomes
2
2
2

 9 + x dx = 
 9 - 9 sinh  (3 cosh ) d = 3 
 9(1 + sinh ) cosh  d
2.3 - 1
= 3

2
9 cosh2 cosh  d = 3 
 3 (cosh ) (cosh ) d = 9 
 cosh  d
= 9

1 + cosh 2
9
9
9
x
9
d =
 + 4 sinh 2 = 2 cosh-13 + 4 (2 sinh  cosh )
2
2
 
=
9
x
9 x
cosh-1  +  
2
3 2 3
1 + sinh2
=
9
x
3
sinh-1  + x
2
3 2
x 2
9
x
1
1+  =
sinh-1  + x
2
3
3 2
9 + x2
We can either leave it like this or use the formula for sinh-1 in terms of ln. If we do this we get
9 x
 9 + x2 dx = 2 ln3 +

=
However, since
9
ln(x +
2
1 + x2) -
x 2
1
1+  + x
3  2
9
1
ln 3 + x
2
2
9 + x2 =
9 x + 1 + x2
1
ln
2 
3
 + 2x
9 + x2
9
ln 3 is a constant we can equally well write
2
9
1
 9 + x2 dx = 2 ln(x + 1 + x2) + 2 x 9 + x2

Example 3. Find
 x2 - 9 dx.

To do this integral we let x = 3 cosh . Then dx = 3 sinh  d and the integral becomes
2
2
2

 x - 9 dx = 
 9 cosh  - 9 (3 sinh ) d = 3 
 9(cosh  - 1) cosh  d
= 3

9 sinh2 sinh  d = 3 
 3 (sinh ) (sinh ) d = 9 
 sinh2  d
= 9

1 - cosh 2
9
9
9
x
9
d =
 - 4 sinh 2 = 2 cosh-13 - 4 (2 sinh  cosh )
2
2
 
=
9
x
9 x
cosh-1  -  
2
3
  2 3
=
9
x
1
cosh-1  - x
2
3 2
cosh2 - 1 =
9
x
3
cosh-1  - x
2
3
  2
x  - 1
3
2
x2 - 9
We can either leave it like this or use the formula for cosh-1 in terms of ln. If we do this we get
2.3 - 2
9 + x2
9 x
2

 x - 9 dx = 2 ln3 +
=
However, since
9
ln(x +
2
x2 - 1) -
2
2
x  - 1 - 1 x x2 - 9 = 9 lnx + x - 1 - 1 x x2 - 9
2 
3
3  2
 2
9
1
ln 3 - x
2
2
x2 - 9
9
ln 3 is a constant we can equally well write
2
9
1
2
2
2

 9 + x dx = 2 ln(x + x - 9) - 2 x x - 9
2.3 - 3