Subject: Calculus

# Trigonometric Substitution

We have discussed many integration techniques and so far, the most noticeable and useful one is the integration by parts. Now, in this article let me introduce to you a new concept in integration techniques that uses trigonometry to solve not-so-easy and seemingly difficult integrals. This is called trigonometric substitution.

As its name suggests, this integration transforms integral into trigonometric functions and hola!, problem solved!. But not all integrals transformed into trigonometric functions via trigonometric substitution makes the integration easy. There are only a few integral forms where trigonometric substitution works best.

## Integral forms for trigonometric substitution

When your integral looks like the ones below, then integration by part cannot do it alone. You have to use trigonometric substitution which will be shown later how to perform.

### Form 1

When your integrand contains the form

b^2-x^2

### Form 2

When the integrand has the form

b^2+x^2

### Form 3

When the integrand has the form

x^2-b^2

If you have Form 1, all you have to do is let x = b\sin{\theta} and play with the trigonometric identities below.

If you have Form 2, all you have to do is let x = b\tan{\theta} and play with the trigonometric identities below.

If you have Form 3, all you have to do is let x = b\sec{\theta} and play with the trigonometric identities below.

### Useful Identities

\sin^2{\theta} + \cos^{2}{\theta} =1
\sec^2{\theta} - \tan^{2}{\theta} =1

## Illustration

Suppose we are to calculate the following integral;

1. \int \frac{dx}{16-x^2}

### Solution by Trigonometric Substitution

1. The first integral looks like the first form mentioned above where b = 4. So as suggested, we let x = 4\sin{\theta}. Then dx = 4\cos{\theta}d\theta. Plugging back into the integral, we get
\int \frac{dx}{16-x^2} = \int \frac{4\cos{\theta}d\theta}{16-\left (4\sin{\theta} \right )^2}
=\int \frac{4\cos{\theta}d\theta}{16-16\sin^2{\theta}}
=\frac{1}{16}\int \frac{4\cos{\theta}d\theta}{1-\sin^2{\theta}}
=\frac{1}{4} \int \frac{\cos{\theta}d\theta}{1-\sin^2{\theta}}

But according to the identity above, 1-\sin^2{\theta}=\cos^2{\theta}, thus

=\frac{1}{4} \int \frac{\cos{\theta}d\theta}{\cos^{2}{\theta}}
=\frac{1}{4} \int \sec{\theta}d\theta

Now this integral looks very simple compared to the original form. In fact, we can integrate directly such that

\frac{1}{4} \int \sec{\theta}d\theta= \frac{1}{4}\left [\ln{\sec{\theta}+\tan{\theta}}\right ]

You can do the other forms mentioned above by just doing almost similarly the methods above. Hope this helps you a lot. Happy INTEGRATING!

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