As you may know by now, indefinite integration is quite easy. Trigonometric substitution is one of the tricks you need to learn to solve integrals.

This is in fact a really clever trick. You may start watching this video:

Let's start with an example. Let's say we want to find the integral:

In fact, this integral is quite easy. There is more than one way to solve it. We'll try our trigonometric substitution with it.

Here we can't use simple integration by substitution. We don't have an x sitting in the numerator. The denominator looks like a trigonometric identity, though. Just imagine the x were a sin(x) or cos(x). It smells trigonometric, right?

Let's remember this from trigonometry:

If we subtract sin squared of θ from both sides:

Now, our denominator looks a lot like the right side of this identity. So, we'll try and use the substitution:

We have our x, equal to sin(θ). Now, to make this substitution we need to get dx. Deriving x with respect to θ:

So, our dx is:

Replacing this in our integral:

Now, the purpose of doing that substitution was to use the trigonometric identity I showed you at the beginning. Let's do that. We have the identity:

We can replace this in our denominator:

And our integral simplifies nicely:

And are we done? Not yet! Our original integral was in terms of x. So, our answer should also be a function of x. We made the substitution:

This means that:

This is just the definition of the inverse trigonometric functions. So, finally, our integral is:

Why Trigonometric Substitution Works

In the previous example we solved an integral that had a radical. Our goal with trigonometric substitution is always to simplify the radicals.

This is possible because of the trigonometric identities:

Generally, we can have three cases which we can solve using trigonometric substitution.

First Case

We have in the integral the expression:

Here a squared is a constant and x is the variable. In this case we can use the first identity involving sines and cosines, because we have a difference.

How do we do that? We factor the constant:

And we make the change of variable:

So, our radical becomes:

And this usually makes the integral easier to solve.

The second case is when we have an integral with the expression:

Here we'll use the second identity involving tangent and secant, because we have a sum.

Again, we follow the same steps. First, we factor the constant:

And we make the substitution:

So, we get:

Third Case

Here we use a variation of the second identity:

Here again we factor the constant:

And we use the substitution:

And we get:

And that's all trigonometric substitution is about. Let's do another example...

Watch this example solved on the video:

Let's find the integral:

As you may have seen, this integral falls into the third case. We will follow all the process. Let's factor the constant 16:

And now, let's use the substitution:

So, our x is:

Our x squared:

We also need to substitute dx. To get that, we derive x with respect to θ:

And you may find at a table of derivatives that:

This is easily proved using what we learned in derivatives of trigonometric functions.

So, the derivative of x is:

Substituting all this in our integral:

Here we use the trigonometric identity to substitute the denominator:

We simplify a bit and finally we have the integral:

Ok, I haven't proved this to you yet, but this integral is:

This is proved using integration by parts. This integral is quite tricky and I will pay special attention to it in a future page.

Let's accept this fact now, to get:

Now, we need to substitute back the secants and tangents. Let's remember what our substitution was:

So, putting secant in function of x:

So, we have:

Now, substituting back:

And finally, our integral is:

And that was a hairy problem. This page is long enough. In the links below you'll find more examples of trigonometric substitution.

- Trigonometric substitution is not hard. It is just a trick used to find primitives.
- It is usually used when we have radicals within the integral sign.
- There are three basic cases, and each follow the same process. The only difference between them is the trigonometric substitution we use.
- Remember to find dx as a function of θ before you actually make the substitution!

If you have just a general doubt about a concept, I'll try to help you. If you have a problem, or set of problems you can't solve, please send me your attempt of a solution along with your question. These will appear on a new page on the site, along with my answer, so everyone can benefit from it.

Click below to see contributions from other visitors to this page...

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