# Question Video: Integration by Trigonometric Substitution Mathematics

Use an appropriate substitution and then a trigonometric one to evaluate β«βπ₯β(1 β π₯)ππ₯.

11:07

### Video Transcript

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

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

Weβre ready to make our substitution now. For every π₯, we substitute in π’ squared and for ππ₯, we substitute in two π’ ππ’. This gives us the indefinite integral of the square root of π’ squared times the square root of one minus π’ squared times two π’ with respect to π’. In this integral, we have the square root of π’ squared. So we can simplify this to just π’.

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 π’ squared times the square root of one minus π’ squared ππ’. 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 π’ squared. And when we see this, we should immediately think of the trigonometric identity of cos squared π plus sin squared π equals one since we can rearrange this to cos π is equal to the square root of one minus sin squared π. And so this is how we spot on π₯ substitution. Comparing this with the square root in our integral, we spot that we should let π’ equal sin π.

Here, we need to find ππ’ in terms of ππ. So similarly to before, we get that ππ’ is equal to ππ’ by ππ times ππ. So differentiating π’ with respect to π, we get ππ’ is equal to cos π ππ. And now, weβre ready to make the substitution.

So substituting in sin π for every π’ and cos π ππ for ππ’, we get that two times the indefinite integral of π’ squared times the square root of one minus π’ squared ππ’ is equal to two times the indefinite integral of sin squared π times the square root of one minus sin squared π times cos π ππ.

And now, we use the fact that cos π is equal to the square root of one minus sin squared π. And we get two times the indefinite integral of sin squared π times cos π times cos π ππ, which is equivalent to two times the indefinite integral of sin squared π cos squared π ππ.

At this stage, weβre still unable to integrate directly. So we can use our trigonometric identities in order to break down the cos squared π and the sin squared π. First, letβs use the fact that cos squared π is equal to one minus sin squared π. And we can substitute this in for the cos squared π. So we get two times the indefinite integral of sin squared π times one minus sin squared π ππ, which we can expand to give us two times the indefinite integral of sin squared π minus sin to the four π ππ.

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 π is equal to one minus two sin squared π, which we can rearrange to sin squared π is equal to one minus cos of two π all over two.

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

Now, letβs multiply through by the two in front of the integral, which leaves us with the indefinite integral of one minus cos of two π minus a half plus two timesed by cos of two π all over two which simplifies just to cos of two π minus cos squared of two π over two ππ.

And now, we have a plus cos of two π minus cos of two π. 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 π. And here, we can factor out the half.

And we can use the fact that sin squared of π₯ is equal to one minus cos squared of π₯. So this implies that sin squared of two π is equal to one minus cos squared of two π. And we get that our integral is equal to a half times the indefinite integral of sin squared of two π ππ.

Now, we can use the fact that sin squared of π is equal to one minus cos of two π over two and double each angle to get that sin squared of two π is equal to one minus cos of four π 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 π over two ππ. 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 π ππ, which now weβre able to integrate.

Letβ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 π to the zero since π to the zero is equal to one. So we increase the power by one giving us one timesed by π 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 π. And in order to integrate this, we can use the fact that the integral of cos π ππ is equal to sin π. So for this term, we get minus sin of four π. However, we mustnβ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 thatβs the derivative of four π with respect to π. This derivative is just four. So we need to multiply our minus sin of four π term by one over four.

And so we have calculated our integral. However, this was an indefinite integral. So we mustnβt forget to add on our constant of integration π. And now, what we need to do is use backward substitution to get our equation in terms of π₯. However, at the moment, our equation is in terms of sin of four π and we need it in terms of sin of π and cos of π so we can substitute. In order to do this, we can use the fact that sin of four π is also equal to two times cos of two π timesed by sin of two π.

So letβs substitute this into our equation. Now, we have a cos of two π term and a sin of two π term, which we need to get in terms of cos of π and sin of π. So we can use the fact that sin of two π is equal to two cos of π times sine of π and cos of two π is equal to one minus two sin squared π. Substituting this into our equation gives one-quarter times π minus a quarter of two timesed by one minus two sin squared π timesed by two cos π sin π plus π.

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 weβre left with one-quarter timesed by π minus sin π cos π timesed by one minus two sin squared π plus π.

And now weβre ready to backward substitute. We use our equations which we made before. So thatβs π₯ is equal to π’ squared and π’ is equal to sin π. First, weβll use the fact that π’ equals sin π. So for the first term, we have π. And if π’ equals sin π, that means that π equals inverse sine of π’. The next term is a sin π. So this is just π’. And now, we have a cos π.

So we can use the fact that cos π is equal to the square root of one minus sin squared π. And since sin π equals π’, we get that cos π equals the square root of one minus π’ squared. And then for our last term, we have a sin squared π. And since sin π equals π’, sin squared π equals π’ squared.

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

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