 In this video we provide the solution to question number nine for the practice exam number two for math 1220 In which case we're asked to evaluate the indefinite integral x squared times sine Excuse me times the square root of four minus x squared dx sorry go ahead of myself there We look at the square root of four minus x squared that actually indicates to us We probably want to do a trigonometric substitution in particular. We would take a sine substitution x equals to sine theta Notice this tells us that dx is equal to two cosine theta d theta Don't forget that d theta there. This also tells us that the square root of four minus x squared is equal to two cosine For which if you want to consider the right triangle because this might be yet useful either now or later in the problem The sine ratio is that x over two is equal to sine theta So opposite over hypotenuse would be x over two then by the Pythagorean equation the other side is the square root of four minus x squared And so we could we could learn this identity from that, but this also might be useful later on So next what we want to do is we want to plug all these pieces in there The x squared becomes a two sine remember to square that the square root becomes a two cosine and Then the dx becomes a two cosine as well d theta Putting that together, of course, we're gonna have two squared times two times two so we get a coefficient of 16 I'm just gonna bring that out of the integral. We end up with a sine squared and Then we also have a cosine squared there d theta And so then we have to ask ourselves how we're gonna deal With these with these squares here. There's a couple ways here basically the half angle is gonna come into play here But what I'm actually gonna do is I'm gonna actually first use the identity that Sine of two theta is equal to two sine theta Cosine theta that is the double angle identity if I apply that here I can actually clean up my integral a little bit in which case we're gonna get four times the integral of Sine like I can write like this sine of two theta Squared d theta So I can have a single sine squared as opposed to a sine squared and a cosine squared, which is gonna be kind of nice here The next thing that I want to apply is gonna be the half angle identity because sine squared of two theta would equal one half one minus cosine of four theta Applying that identity in this situation We end up with two because one half times four is two just so you know the 16 I had to borrow two of the two of the four twos there in order to make this identity work That's why there's a four there now So we have two times the integral of one minus cosine of four theta d theta That's now a function that I'm ready to find the anti derivative of so taking the anti derivative We're going to get two times theta minus one fourth Sine of four theta plus a constant Okay, for which now I have to start translating these things back in terms of Theta somehow or another now with the sine for theta. We're gonna use this identity again, right? So what we get there I'll distribute that two through so you're gonna get two theta minus one half But then sine of four theta is the same thing as two times sine of two theta Cosine of two theta plus a constant this two and one half now cancel out here With regard to the theta you can take this sine ratio and solve for solve for theta there This is going to give you two times sine inverse of x over two Then for the next piece we have this Minus sine of two theta cosine two theta well again for sine of two theta You're gonna get two sine theta cosine theta We have to get it back into the language of theta and now for the cosines of two theta We use the double angle identity there cosine of two theta is equal to cosine squared theta minus sine squared theta and So that's what we're going to plug in for cosine of two theta We end up with this cosine squared theta minus sine squared theta plus our constant And so let's unravel them when we have here so the first part involving the sine inverse nothing's gonna happen there So we get a two sine inverse of x over two Then for the next part We have let's replace each of these pieces here the sine remembers the same thing as x over two the cosine If you solve for here you end up with the square root of four minus x squared over two Some things we can do is this two can cancel with that too Then we have a cosine squared Which is going to give you four minus x squared over four and then we're gonna get a sine squared Which is x squared over four plus a constant I do want to simplify this a little bit more because notice what happens you have this x squared over four That'll cancel with this x squared over four You end up with a four over four in that situation Actually, I take that back Official JK on that one. This is a minus and a minus is actually is going to double up So this will then become four minus two x squared over four Which is the same thing as two minus x squared over two And so therefore this denominator of two can combine with that one so we can get a coefficient of One fourth there. So let's write this again two sine inverse of x over two We're then going to get minus what we have here is a coefficient of one fourth We have an x. We have a square root of four minus x squared and then we also have a coefficient of Two minus x squared now if you want to you could flip the direction of this to get x squared minus two You fact out a negative one which makes this a positive and then lastly you have a plus C at the end Don't forget that plus C there and so this then gives us the anti derivative of the function here using trigonometric substitution