
We integrate by substitution with the appropriate trigonometric function.

We have seen that the definite integral can be interpreted as giving us the signed area under the curve. Consider the integral

This integral can be interpreted as area of the following region:

Using the graph above and basic facts from geometry, compute:

We now know a number of integration techniques, yet we are still unable to evaluate the above integral without resorting to a geometric interpretation! This section introduces trigonometric substitution, a method of integration that will give us a new tool in our quest to compute more antiderivatives. This technique works on the same principle as substitution. Recall the substitution formula.

The idea behind substitution is that by changing the variable of integration from $x$ to $u$, then in many cases we can simplify the integrand to a form that we can integrate directly. In previous sections, we used this result to go from the left integral to the right integral. But what if we tried going from right to left?

Let’s rewrite the substitution formula in a more suggestive manner for our purposes:

This is just the substitution formula again but now we are thinking of transforming from left-to-right. Initially this may seem to complicate matters. We go from a simple integral to one that looks more complicated. However in certain cases we may not know how to integrate $f(x)$, but we may be able to integrate $f(x(\theta )) x'(\theta )$.

We start by applying this technique to compute the area that we previously used geometry to compute.

If then

The key idea with this example was that the expression $\sqrt {1-x^{2}}$ was difficult to work with directly and using the substitution $x=\sin (\theta )$ allowed us to simplify the expression and turn the integral into a trig integral that was easier to calculate.

There are other situations where this kind of strategy is useful. Let’s look at a few more examples.

So far we have seen that we can use trig substitutions to simplify expressions like $x^{2}-a^{2}$, $a^{2}-x^{2}$, and $x^{2}+a^{2}$. However, sometimes the integral may not be in a form where it is clear that a trig substitution will help. One very useful tool is to complete the square, and then choose a trigonometric function which allows the use of a Pythagorean identity to simplify the expression.

Let’s recall how to complete the square for a general quadratic $ax^2+bx+c$. First we divide through by $a$.

Now, consider the following example:

Thus using the technique of completing the square, we can always rewrite any quadratic in a form where a trig substitution can be used to simplify the expression.

### Summary

In general, when doing a trig substitution:

• If the quadratic has the form $a^2 + x^2$ after completing the square, make the substitution $u = a\tan (\theta )$.
• If the quadratic has the form $a^2 - x^2$ after completing the square, make the substitution $u = a\sin (\theta )$
• If the quadratic has the form $x^2 - a^2$ after completing the square, make the substitution $u = a\sec (\theta )$

You should always remember the other integration techniques you have learned previously; they may allow you to compute the integral more easily than trig substitutions in some cases. In particular, go back and look at example 2 in this section and see if you can compute this integral using a more efficient integration technique! In some cases, though, trig substitution will be your only option.