### Video Transcript

Evaluate the integral of one divided by the square root of two π₯ minus π₯ squared with respect to π₯.

The question is asking us to evaluate the integral. This is the integral of one divided by the square root of two π₯ minus π₯ squared with respect to π₯. And itβs hard to see how to evaluate this integral. For example, by inspection, our integrand is not in the form π prime of π₯ divided by π of π₯ or π prime of π₯ times π of π of π₯. Our integrand is not the product of two functions. And even if we did try integration by parts, this would not help us evaluate this integral. So none of our standard tools for evaluating integrals seem to help us in this case. This means we should try manipulating our integrand into a form which we can evaluate.

To do this, weβre going to complete the square on the quadratic inside of the square root of our denominator. When we do this, we divide our coefficient of π₯ by two. We get negative one times π₯ minus one squared plus one. And this is because negative one times negative one squared gives us an extra negative one. So we need to add one to get rid of this.

So by completing the square, we can now rewrite the inner function of the denominator of our integrand. This gives us the integral of one divided by the square root of negative one times π₯ minus one squared plus one with respect to π₯. And at first, it might not seem like this has helped us to evaluate our integral. However, we can see that this resembles one of our integral rules for the inverse trigonometric functions.

We know for a constant π not equal to zero, the integral of one divided by the square root of π squared minus π squared is equal to the inverse sin of π over π plus a constant of integration πΆ. And we can see that our integrand is very similar to this one. We have one divided by the square root of a constant minus some function of our variable squared.

So weβll try to write our integral in this form. Weβll do this by using a π’ substitution. Weβll use π’ is equal to π₯ minus one. So if we have π’ is equal to π₯ minus one, weβll differentiate both sides with respect to π₯. We get dπ’ by dπ₯ is equal to one. And we know dπ’ by dπ₯ is not a fraction. However, when weβre using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement dπ’ is equal to dπ₯.

Now, we can rewrite our integral by using our π’ substitution. Weβll replace π₯ minus one with π’ and dπ₯ with dπ’. This gives us the integral of one divided by the square root of one minus π’ squared with respect to π’. And now this is exactly in the form for our integral rule. So we can just apply this to find the integral. In this case, our constant π is equal to one. So this gives us the inverse sin of π’ plus our constant of integration πΆ.

But remember, our original integral was in terms of π₯. So we want to give our answer in terms of π₯. Weβll do this by using our substitution π’ is equal to π₯ minus one. And this gives us our final answer. The integral of one divided by the square root of two π₯ minus π₯ squared with respect to π₯ is equal to the inverse sin of π₯ minus one plus πΆ.