Calculus:
Integral: Trigonometric Substitution

1. 
\(\int \frac{1}{16+{x}^{2}} \, dx\)
 Solution
2. 
\(\int \frac{3}{\sqrt{{x}^{2}-9}} \, dx\)
 Solution
3. 
\(\int \frac{2}{\sqrt{25-{x}^{2}}} \, dx\)
 Solution
4. 
\(\int \frac{\sqrt{{x}^{2}-16}}{x} \, dx\)
 Solution
5. 
\(\int \frac{1}{{(1+{x}^{2})}^{2}} \, dx\)
 Solution
6. 
\(\int \frac{1}{{x}^{4}\sqrt{4-{x}^{2}}} \, dx\)
 Solution
7. 
\(\int \frac{\sqrt{25{x}^{2}-4}}{x} \, dx\)
 Solution
8. 
\(\int \frac{1}{{x}^{2}\sqrt{{x}^{2}-9}} \, dx\)
 Solution

Trigonometric Substitution - Introduction

This tutorial assumes that you are familiar with trigonometric identities, derivatives, integration of trigonometric functions, and integration by substitution.
After simpler methods of integration failed, we should consider trigonometric substitution. The requirement is that the function contains the form
\({a}^{2}-{x}^{2}\)
, the form
\({a}^{2}+{x}^{2}\)
, or the form
\({x}^{2}-{a}^{2}\)
. Some examples are:
1.
\(\sqrt{{x}^{2}-25}\)

2.
\({(4-{y}^{2})}^{\frac{3}{2}}\)

3.
\(\sqrt{9-{h}^{2}}\)

4.
\(\frac{1}{16+{x}^{2}}\)
In fact, there are clear rules on what substitutions to make:
1. If the function contains
\({a}^{2}-{x}^{2}\)
, let
\(x=a\sin{u}\)
.
2. If the function contains
\({a}^{2}+{x}^{2}\)
, let
\(x=a\tan{u}\)
.
3. If the function contains
\({x}^{2}-{a}^{2}\)
, let
\(x=a\sec{u}\)
.

Using Trigonometric Substitution

Let's try an example. Solve this integral:
\(\int \frac{1}{16+{x}^{2}} \, dx\)
Since the denominator fits the form
\({a}^{2}+{x}^{2}\)
, let's use Rule #2 from the section above.
We let
\(x=a\tan{u}\)
, which is
\(4\tan{u}\)
.
Then, to get
\(dx\)
, we find the derivative of
\(4\tan{u}\)
, which is
\(4\sec^{2}u\)
.
Therefore, let
\(x=a\tan{u}\)
, and
\(dx=4\sec^{2}u\)
.
Substituting the variables from the above, we have:
\(\int \frac{1}{16+{(4\tan{u})}^{2}}\times 4\sec^{2}u \, du\)
After simplifying, we have:
\(\int \frac{1}{4} \, du\)
By the Power Rule, the integral is:
\(\frac{u}{4}\)
From the earlier steps, we know that
\(x=a\tan{u}\)
, and solving for
\(u\)
, we have:
\(u=\tan^{-1}{(\frac{1}{4}x)}\)
The only step remaining is to substitute the value of
\(u\)
back into the earlier integral, namely
\(\frac{u}{4}\)
. After the substitution, we have:
\(\frac{\tan^{-1}{(\frac{1}{4}x)}}{4}+C\)
We are done. We have successfully used trigonometric substitution to find the integral.

What's Next

Ready to dive deeper? You can try more practice problems at the top of this page to help you get more familiar with solving integral using trigonometric substitution. Want even more help? Sign up for Cymath Plus today. You can also download the Cymath app for iOS and Android to get step-by-step help on-the-go.