Question Video: Integration by Trigonometric Substitution Mathematics

Video Transcript

Use an appropriate substitution then a trigonometric one to evaluate the indefinite integral of the square root of 𝑥 timesed by the square root of one minus 𝑥 with respect to 𝑥.

First, we noticed that we have two square roots here. So in order to get rid of one of them, we can substitute in 𝑥 is equal to 𝑢 squared. And now in order to find 𝑑𝑢 in terms of 𝑑𝑥, we can use the fact that 𝑑𝑥 is equal to 𝑑𝑥 by 𝑑𝑢 times 𝑑𝑢. And now to find 𝑑𝑥 𝑑𝑢, we simply differentiate 𝑥 is equal to 𝑢 squared. We get that 𝑑𝑥 is equal to two 𝑢 times 𝑑𝑢.

We’re ready to make our substitution now. For every 𝑥, we substitute in 𝑢 squared and for 𝑑𝑥, we substitute in two 𝑢 𝑑𝑢. This gives us the indefinite integral of the square root of 𝑢 squared times the square root of one minus 𝑢 squared times two 𝑢 with respect to 𝑢. In this integral, we have the square root of 𝑢 squared. So we can simplify this to just 𝑢.

Now, we have a factor two in our integral, which we can factor out of the integral leaving us with two times the indefinite integral of 𝑢 squared times the square root of one minus 𝑢 squared 𝑑𝑢. We are still not able to integrate at this stage. So we need to use another substitution.

We noticed that we have the square root of one minus 𝑢 squared. And when we see this, we should immediately think of the trigonometric identity of cos squared 𝜃 plus sin squared 𝜃 equals one since we can rearrange this to cos 𝜃 is equal to the square root of one minus sin squared 𝜃. And so this is how we spot on 𝑥 substitution. Comparing this with the square root in our integral, we spot that we should let 𝑢 equal sin 𝜃.

Here, we need to find 𝑑𝑢 in terms of 𝑑𝜃. So similarly to before, we get that 𝑑𝑢 is equal to 𝑑𝑢 by 𝑑𝜃 times 𝑑𝜃. So differentiating 𝑢 with respect to 𝜃, we get 𝑑𝑢 is equal to cos 𝜃 𝑑𝜃. And now, we’re ready to make the substitution.

So substituting in sin 𝜃 for every 𝑢 and cos 𝜃 𝑑𝜃 for 𝑑𝑢, we get that two times the indefinite integral of 𝑢 squared times the square root of one minus 𝑢 squared 𝑑𝑢 is equal to two times the indefinite integral of sin squared 𝜃 times the square root of one minus sin squared 𝜃 times cos 𝜃 𝑑𝜃.

And now, we use the fact that cos 𝜃 is equal to the square root of one minus sin squared 𝜃. And we get two times the indefinite integral of sin squared 𝜃 times cos 𝜃 times cos 𝜃 𝑑𝜃, which is equivalent to two times the indefinite integral of sin squared 𝜃 cos squared 𝜃 𝑑𝜃.

At this stage, we’re still unable to integrate directly. So we can use our trigonometric identities in order to break down the cos squared 𝜃 and the sin squared 𝜃. First, let’s use the fact that cos squared 𝜃 is equal to one minus sin squared 𝜃. And we can substitute this in for the cos squared 𝜃. So we get two times the indefinite integral of sin squared 𝜃 times one minus sin squared 𝜃 𝑑𝜃, which we can expand to give us two times the indefinite integral of sin squared 𝜃 minus sin to the four 𝜃 𝑑𝜃.

Now, we can use the double angle formula for cos in order to remove some of the powers here. We have that cos of two 𝜃 is equal to one minus two sin squared 𝜃, which we can rearrange to sin squared 𝜃 is equal to one minus cos of two 𝜃 all over two.

Now, we can use the fact that sin to the four 𝜃 is equal to sin squared 𝜃 all squared to get that our integral is equal to two times the indefinite integral of one minus cos of two 𝜃 over two minus one minus cos of two 𝜃 over two squared 𝑑𝜃. Now, we can expand the square, leaving us with two times the indefinite integral of a half minus cos of two 𝜃 over two minus one minus cos of two 𝜃 plus cos squared of two 𝜃 all over four 𝑑𝜃.

Now, let’s multiply through by the two in front of the integral, which leaves us with the indefinite integral of one minus cos of two 𝜃 minus a half plus two timesed by cos of two 𝜃 all over two which simplifies just to cos of two 𝜃 minus cos squared of two 𝜃 over two 𝑑𝜃.

And now, we have a plus cos of two 𝜃 minus cos of two 𝜃. So these two terms cancel out with one another. So we can simplify our integral to leave us with the indefinite integral of a half minus a half times cos squared of two 𝜃. And here, we can factor out the half.

And we can use the fact that sin squared of 𝑥 is equal to one minus cos squared of 𝑥. So this implies that sin squared of two 𝜃 is equal to one minus cos squared of two 𝜃. And we get that our integral is equal to a half times the indefinite integral of sin squared of two 𝜃 𝑑𝜃.

Now, we can use the fact that sin squared of 𝜃 is equal to one minus cos of two 𝜃 over two and double each angle to get that sin squared of two 𝜃 is equal to one minus cos of four 𝜃 over two. And we can substitute this into our integral. And so we get a half times the indefinite integral of one minus cos of four 𝜃 over two 𝑑𝜃. And again, we can factor the half out of the integral, leaving us with one-quarter times the indefinite integral of one minus cos of four 𝜃 𝑑𝜃, which now we’re able to integrate.

Let’s do this term by term. So the constant in front of the integral will remain the same. So we just get a quarter. The first term in our integral is one. In order to integrate this, we can use the fact that one is equivalent to one timesed by 𝜃 to the zero since 𝜃 to the zero is equal to one. So we increase the power by one giving us one timesed by 𝜃 to the one and divide by the new power. So we divide by one.

The second term in our integral is minus cos of four 𝜃. And in order to integrate this, we can use the fact that the integral of cos 𝜃 𝑑𝜃 is equal to sin 𝜃. So for this term, we get minus sin of four 𝜃. However, we mustn’t forget to do the end of the chain rule here. And so we need to divide by the derivative of the inside of our function. So that’s the derivative of four 𝜃 with respect to 𝜃. This derivative is just four. So we need to multiply our minus sin of four 𝜃 term by one over four.

And so we have calculated our integral. However, this was an indefinite integral. So we mustn’t forget to add on our constant of integration 𝑐. And now, what we need to do is use backward substitution to get our equation in terms of 𝑥. However, at the moment, our equation is in terms of sin of four 𝜃 and we need it in terms of sin of 𝜃 and cos of 𝜃 so we can substitute. In order to do this, we can use the fact that sin of four 𝜃 is also equal to two times cos of two 𝜃 timesed by sin of two 𝜃.

So let’s substitute this into our equation. Now, we have a cos of two 𝜃 term and a sin of two 𝜃 term, which we need to get in terms of cos of 𝜃 and sin of 𝜃. So we can use the fact that sin of two 𝜃 is equal to two cos of 𝜃 times sine of 𝜃 and cos of two 𝜃 is equal to one minus two sin squared 𝜃. Substituting this into our equation gives one-quarter times 𝜃 minus a quarter of two timesed by one minus two sin squared 𝜃 timesed by two cos 𝜃 sin 𝜃 plus 𝑐.

And now on the second term here, we notice that we have a quarter outside the brackets and inside the brackets we have a two here and a two here. So this two times two will cancel with the quarter. So we’re left with one-quarter timesed by 𝜃 minus sin 𝜃 cos 𝜃 timesed by one minus two sin squared 𝜃 plus 𝑐.

And now we’re ready to backward substitute. We use our equations which we made before. So that’s 𝑥 is equal to 𝑢 squared and 𝑢 is equal to sin 𝜃. First, we’ll use the fact that 𝑢 equals sin 𝜃. So for the first term, we have 𝜃. And if 𝑢 equals sin 𝜃, that means that 𝜃 equals inverse sine of 𝑢. The next term is a sin 𝜃. So this is just 𝑢. And now, we have a cos 𝜃.

So we can use the fact that cos 𝜃 is equal to the square root of one minus sin squared 𝜃. And since sin 𝜃 equals 𝑢, we get that cos 𝜃 equals the square root of one minus 𝑢 squared. And then for our last term, we have a sin squared 𝜃. And since sin 𝜃 equals 𝑢, sin squared 𝜃 equals 𝑢 squared.

And so now, we have our equation in terms of 𝑢. And we just need to make one more substitution using the fact that 𝑥 is equal to 𝑢 squared. And this means that 𝑢 is equal to the square root of 𝑥. So for every 𝑢, we simply substitute in root of 𝑥.

As we know, we have our equation back in terms of 𝑥. So we obtain the solution that the indefinite integral of the square root of 𝑥 times the square root of one minus 𝑥 with respect to 𝑥 is equal to sin inverse of root 𝑥 over four minus the square root of 𝑥 times the square root of one minus 𝑥 times one minus two 𝑥 all over four plus 𝑐.