### Video Transcript

Use a trigonometric substitution to evaluate the integral of one over the square root of nine plus π₯ squared with respect to π₯.

Weβre told to use a trigonometric substitution to evaluate this integral. But how do we decide which one we need? Well, there are three main situations we need to think about.

If weβre looking to evaluate an integral that involves the square root of some number squared minus π₯ squared, we use the substitution π₯ equals that number multiplied by sin of π’. If weβre looking to integrate the square root of some number squared plus π₯ squared, we let our substitution be π₯ equals that number multiplied by tan π’. And finally, if weβre looking to integrate the square root of π₯ squared minus some number squared, we set our substitution as π₯ is equal to that number times sec π’.

Our integral involves the square root of some number squared plus π₯ squared. So, weβre going to use the second situation. Nine is three squared, so the substitution weβre going to use is let π₯ equal three tan π’. Our next step is to differentiate this π₯ with respect to π’. The derivative of tan π’ is sec squared π’. So, differentiating three tan π’, and we get three sec squared π’. And this means dπ₯ must be equal to three sec squared π’ dπ’.

Before we replace dπ₯ in our integral with three sec squared π’ dπ’, we need to decide how weβre going to express the square root of nine plus π₯ squared in terms of π’. Well, replacing π₯ with three tan π’, and this becomes the square root of nine plus three tan π’ squared. Three tan π’ all squared is the same as nine tan squared π’, so we can rewrite the square root of nine plus π₯ squared as shown. Now we can actually further simplify this by factoring nine in nine plus nine tan squared π’.

And then, we can separate our roots. And it becomes the square root of nine times the square root of one plus tan squared π’. The square root of nine is three. And then, we use the identity one plus tan squared π’ equals sec squared π’. And this means the square root of one plus tan squared π’ is just sec π’. So, we can rewrite the square root of nine plus π₯ squared in our integral as three sec π’. And now weβre ready to make the substitutions.

We need to integrate three sec squared π’ divided by three sec π’ with respect to π’. Notice how we can cancel these threes. We then see that sec squared π’ divided by sec π’ is simply sec π’. And our final step is to integrate sec π’ with respect to π’. And thereβs a general result for this. The integral of sec π’ with respect to π’ is ln of sec π’ plus tan π’ plus that constant of integration π.

However, we were looking to evaluate our integral in terms of π₯. So, weβre going to need to find a way to substitute π₯ back into our expression. We go back to the original substitution. And we said let π₯ be equal to three tan π’. If we divide through both sides of this equation by three, we see that tan π’ is equal to π₯ over three.

Similarly, we saw that we could replace the square root of nine plus π₯ squared with three sec π’. If we divide through by three here, we see that sec π’ is equal to the square root of nine plus π₯ squared all over three. And we can now replace tan π’ and sec π’ with expressions in terms of π₯. And weβve evaluated our integral. Itβs ln of the square root of nine plus π₯ squared over three plus π₯ over three plus that constant of integration π.