 In this video, we're going to discuss the solution to question 12 for the practice midterm exam for calculus to math 12 20 You can see that we have to evaluate the integral of x squared times the square root of 4 minus x squared Now question number 12 on this midterm exam. You can anticipate there to be a trig substitution But I hope that this square root of 4 minus x squared is really indicative to you that you want to do a trig substitution So notice here that we have a difference of squares a constant minus a variable So therefore we're going to do a sine substitution We want to take a sine sine theta taking the square root of the constant we get a 2 So that's the coefficient of sine taking the square root of the x square we get an x So this is the basis for the rest of the substitution Some important things we should mention is that if you solve for sine sine theta it's going to equal x over 2 We're going to do that one point We're also if you solve for theta you're going to theta equals sine inverse of x over 2 we're going to need that at one point Taking the derivative we get dx equals 2 cosine theta d theta. We'll need that We also need to set up for the square root the square root of 4 minus x squared That'll equal to cosine theta and then lastly notice that cosine theta Equals the square root of 4 minus x squared all over 2 and so this is all the Substitutions we're going to need in this situation here So our integral will be transformed x squared using this identity right here We're going to get a 4 sine squared theta and then the square root using this identity right here becomes 2 cosine and Then the dx using this identity becomes 2 cosine theta d theta So we end up with 16 times the integral of sine squared Theta cosine squared theta now in this question. There's a lot of tree identities We're going to need to use here the first of which we're going to want to use sine of 2 theta equals 2 sine theta Cosine theta this double angle identity for sine and if we do that notice we'll be get the following We're going to get 4 times the integral of 2 sine theta cosine theta And we're going to get another 2 sine theta cosine theta d theta the first one right here This is a sine 2 theta and then this next one right here is likewise a sine 2 theta Notice we borrowed some of the factors of 2 from the 16 from above it became a 4 and So we end up with 4 times the integral of sine squared of 2 theta d theta So that's a good start. So now we have a sine squared this now brings to our next identity sine squared theta is equal to one half one minus cosine of 2 theta So in particular the angle will double when you switch from sine to cosine So the consequence for us here is first of all this one half We can combine with the four that we already have to give us a 2 2 times the integral of 1 minus cosine This is the cosine of 4 theta because we went from 2 theta to 4 theta d theta Now we're in a pretty good position to find the anti-derivative Upon doing so we're going to get two times theta minus We're going to get 1 over 4 sine of 4 theta Plus a constant like so and so now we're in a position where we can start translating stuff back from theta into x using these Identities we did before but notice these identities all based upon theta theta Not a 4 theta here. So if we apply the double angle identity one more time We're going to end up with 2 theta minus 1 half. So if you distribute the 2 here 2 times 1 4th is 1 half. We're next going to get 2 sine of 2 theta cosine of 2 theta plus a constant This one half will cancel with the 2 that's right there. And so if we do the The double angle for sine one more time we can apply that here So we get 2 theta minus we're going to get 2 sine theta cosine theta that's going to be important. So now that's in terms of theta we can translate it back Now what do we do with this cosine of 2 theta? Well, there's one more identity We want to use here cosine of 2 theta equals cosine squared theta minus sine squared theta So if we make that substitution here, we're going to get cosine squared theta minus sine squared theta Plus a constant So now we're in position to switch everything back So we have to do a theta. So we're going to get a sine inverse of x over 2 Sine theta is equal to x over 2 and cosine theta is the square root over 2. So make those appropriate adjustments down here So like we said Theta becomes a sine inverse of x over 2 Sine is going to become x over 2 Cosine is going to become the square root of 4 minus x squared over 2. I should mention that this 2 cancels with that 2 and Then here cosine squared if you're going to square you're going to get 4 minus x squared over 4 And then square this sign you're going to get x squared over 4 plus a constant like so Combining like terms right here. You have a 4 you have an x squared and a negative x squared and a negative x squared You'll get 2 sine inverse of x over 2 we have a Minus so far one-half x times the square root of 4 minus x squared oops and Then you're going to get 4 minus 2x squared over 4 plus a constant Now what I want to do next is notice that here you have a 4 minus 2x which could be factored as if you take out a negative 4 sorry negative 2 you get negative 2 times x squared minus 2 like so this negative 2 can cancel with the Negative one-half we have right there and so now we've reached our final form as simplified as I can make it to sine inverse x over 2 We're going to get plus one-fourth Bringing out this one fourth right here one-fourth x times the square root of 4 minus x squared times Actually, I'm going to write this a little bit different I'm going to bring the I'm going to bring the x squared minus 2 in front and Then we get the square root. It's always polite to put the square roots in the back there and so then this gives us our final answer and So in terms of trig substitution the trig substitution once is so bad I think the hardest part of this question really comes down to all of these trig identities That we were using here the double angles here. And so I mean at one point we got up to a 4 theta And once we have to deal with that appropriately But if you're okay with those trig identities everything else is a fairly standard calculation on this one So it's going to be important that you include In your your notes as you're taking this test Lots of trig identities all the trig identities you would have seen in the lectures You're going to want to include those here as they're going to be very helpful for you