 In this video, we're going to discuss the solution to question 13 from the practice midterm exam for calculus 2, math 12, 20. Number 13 here is going to represent our street fighting question for integration. That is, on question 13, everything goes. Anything goes. If you can do it, then you can do it. Those are the only rules. No stipulations on you have to do integration by parts or use substitution or whatever. So this is supposed to sort of test your judgment on what is the best technique to use right here. And so when I'm looking at this integral, if I were to see this for the first time, some things that I would notice would be the following. I see a rational function inside of a square root. I potentially might try to do a rationalizing substitution where I could set you equal to the square root 1 plus X over 1 minus X. That's kind of nice because we like to set square roots equal to you because you can do a rationalizing substitution here. But then you're going to have to do a DU, which by the chain rule, this is going to be kind of messy and see what we could do here. If you take the derivative here, you're going to get two times the square root of 1 plus X over 1 minus X here. You're going to get a DX. But then you have to take the inner derivative, which by the quotient rule, you're going to get something that's like 1 minus low D high minus high D low. That's going to give you a plus X right there. Square the bottom. Here we go. Like so. And so you could simplify this thing on the top. You're going to get a 2 over 2 times 1 minus X squared, the square root of 1 plus X over 1 minus X DX. The 2s cancel. And then you can get UDU equals DX over 1 minus X squared. It has me scratch my head a little bit what to do that 1 minus X squared. So I'm thinking a rationalizing substitution might not be the best approach. Whoops, that's the wrong problem. Go back to where we were. So instead we want to try something a little different here. And actually what I'm going to actually do in this situation, believe it or not, is I'm going to try to rationalize the numerator. That's another technique you want to be prepared to do. This is an algebraic technique. We're going to rationalize the numerator. You'd be surprised how useful that can be at times. So what that means is we're going to times. We're going to times the top and bottom by, in this case, 1 plus X. Why would I do that? Well, when you multiply the top and bottom by 1 plus X, you're going to see the following. You're going to get a 1 plus X squared on top. And you're going to get, when you foil it out on the bottom, you're going to get a square root of 1 minus X squared on the bottom, like so. Notice that if you take the square root of 1 plus X squared on top, the square root disappears. And so then we have a 1 plus X on the bottom and you get a square root of 1 minus X squared. Sorry, 1 plus X on the top, square root of 1 minus X squared on the bottom. And so now this algebraic idea of rationalization numerator helped us in so much that in this situation, it then starts to look like, well, maybe I should do some type of trigometric substitution. And this is something we can see now that we have this difference of squares. And so that's really what I was trying to create in this rationalizing the numerator. I was trying to create this difference of squares so I could do a trigonometric substitution. The trig sub I would want to do, not you, X would equal sine theta. So then DX equals cosine theta, d theta, and the square root of 1 minus X squared equals cosine theta. So with those substitutions, we're going to end up with 1 plus sine theta over cosine theta. And then we're going to times this by cosine theta, d theta. For which case the cosines cancel out and we're left to integrate 1 plus sine theta, d theta. Your antiderivative would look like theta plus, sorry, theta minus cosine theta plus a constant. Don't forget the plus C there. And then unraveling this theta based upon this identity right here is equal to sine inverse of X. So we get sine inverse of X. Antiderivative of cosine using this identity right here, you're going to get negative the square root of 1 minus X squared plus a constant like so. And so this one right here, this one right here, this gives us our final answer. And this one might be a little bit mysterious to us because when we first started it, we thought maybe a rationalization substitution. We thought maybe some type of partial fraction decomposition. I mean, that's another idea one could use here, but none of those would have been too fruitful here. Turns out trig sub is going to be the best approach. And I threw this one on the practice exam as a reminder of this technique of rationalizing the numerator or the denominator. You want to get a difference of squares going on here. And I believe you could rationalize the denominator as well. Had you done so, if you rationalize the denominator, you actually would have been times being by X minus one and X minus one right here. In the end, you would have instead had the integral of, in this case, you're going to get the square root of one minus X squared. And then on the bottom, you get an X minus one DX. You could then try doing a trig sub right there. That one's not going to be as clean as the approach we went. And the main reason why is that the square root is kind of stuck as a single object, but the one plus X or one minus X, they can be separated. Now, if something's inside the denominator, it's going to be stuck together as well. But things in the numerator could be separated, kind of like we handled over here. So since things in the, in a square root are stuck together and things that a denominator stuck together, it kind of makes sense to put all of those stuck things into one category. And so those things in the numerator, we can move around here. So rationalize the numerator does lead to a cleaner approach on this one here.