Question Video: Finding the Integration of a Function Involving a Root in Its Denominator by Completing the Square Mathematics • Higher Education

Video Transcript

Evaluate the integral of one divided by the square root of five plus four 𝑥 minus 𝑥 squared with respect to 𝑥.

In this question, we’re asked to evaluate a definite integral. And we can see this is in a form which we don’t know how to integrate. This means we’re going to need to manipulate our integral into a form which we can integrate. And there’s a few different things we might want to try. For example, we could try factoring our quadratic. And if we were to do this, we would see it factors fully. It’s equal to negative one multiplied by 𝑥 plus one times 𝑥 minus five. However, writing this quadratic in this way doesn’t help us. For example, we can’t simplify the square root, and we still don’t know how to evaluate this integral.

But this is not the only thing we could’ve done to manipulate our quadratic. We also could’ve tried completing the square, so we’ll try completing the square. First, we see that our coefficient of 𝑥 squared is negative one. So remember this means we need to multiply by negative one. The next step is usually to divide our coefficient of 𝑥 by two. But remember, we’re multiplying through by negative one. So in fact, we need to divide by negative two. This gives us negative one multiplied by 𝑥 minus two all squared. Now we can ask the question, what happens if we were to distribute the exponent over our parentheses?

By either using the FOIL method or binomial expansion, we get negative one multiplied by 𝑥 squared minus four 𝑥 plus four, which if we distribute negative one over our parentheses, we get negative 𝑥 squared plus four 𝑥 minus four. And this is almost exactly what we have in our denominator. However, we can see the constant in the quadratic in our denominator is positive five. And to make our constant five, we need to add nine to negative four. Therefore, if we add nine to negative one times 𝑥 minus two all squared plus nine, we get the quadratic in our denominator, five plus four 𝑥 minus 𝑥 squared. And we can then use this to rewrite our integral.

Doing this, we get the integral given to us in the question is equal to the integral of one divided by the square root of negative one times 𝑥 minus two all squared plus nine with respect to 𝑥. And at first, it might be hard to see how this helps us evaluate this integral. However, this is actually very similar to an integral we know how to evaluate. We need to recall the following integral result. For any real constant 𝑎, the integral of one over the square root of 𝑎 squared minus 𝑥 squared with respect to 𝑥 is equal to the inverse sin of 𝑥 over 𝑎 plus the constant of integration 𝐶.

And we can see that our integrand is almost in this form. First, we need to rearrange the two terms inside of the square root in our denominator. Doing this, we get the following expression. Next, we can see instead of subtracting 𝑥 squared in our denominator, we’re instead subtracting 𝑥 minus two all squared. So to use this result, we instead need our variable squared in this position. We can do this by using a substitution. We want to use the substitution 𝑢 is equal to 𝑥 minus two. If 𝑢 is equal to 𝑥 minus two, we’ll need to differentiate both sides of this expression with respect to 𝑥. We get d𝑢 by d𝑥 is equal to one.

And of course we know d𝑢 by d𝑥 is not a fraction. However, when we’re using integration by substitution, it can help to treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials, d𝑢 is equal to d𝑥. And we can now evaluate this integral by using our substitution. We replace 𝑥 minus two in our denominator with 𝑢, and because d𝑢 is equal to d𝑥, we’re now integrating with respect 𝑢. This gives us the integral of one over the square root of nine minus 𝑢 squared with respect to 𝑢. And this is now almost exactly in the form we need for our integral rule.

All we need to do now is find the value of the constant 𝑎. And to find this, we just rewrite nine as three squared. Now we can see the value of our constant 𝑎 is equal to three. This means we can apply our integral rule with the variable 𝑥 equal to 𝑢. So by setting our value of 𝑎 equal to three and applying our integral rule, we get the inverse sin of 𝑢 over three plus our constant of integration 𝐶. However, remember, our original integral is in terms of 𝑥, so we should give our answer in terms of 𝑥. We can do this by using our substitution 𝑢 is equal to 𝑥 minus two. And by doing this, we get the inverse sin of 𝑥 minus two all over three plus the constant of integration 𝐶, which is our final answer.

Therefore, by completing the square in our denominator, we were able to show the integral of one over the square root of five plus four 𝑥 minus 𝑥 squared with respect to 𝑥 is equal to the inverse sin of 𝑥 minus two all over three plus the constant of integration 𝐶.