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Evaluate the integral.
$ \displaystyle \int_0^1 x \sqrt{2 - \sqrt{1 - x^2}}\ dx $
$\frac{2^{9 / 2}-14}{15}$
Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 5
Strategy for Integration
Integration Techniques
Missouri State University
Harvey Mudd College
University of Nottingham
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let's start off by doing the use substitution here. So here I'm basically taking you to be this radical, which is inside of the larger rentable. And then here use the chain rule to take the derivative. So here we get negative X over square root, one minus x square and then here we should go ahead and rewrite this. So this is negative X over you the eggs and let's go ahead and write. Multiply the negative. Let's take this equation here and just push this negative in the D you to the other side by multiplying So we get negative. You do You equals X T X and receiving the X here with the DX term. So also for our integral Remember, we have a definite numeral here, so we will have to change the limits of integration. So plug in X equals zero and X equals one into this X over here. And after doing that, we have our new limits for one and zero. And then for the inner girl. Well, this big right uncle right here that's just radical to minus you and what's left over. And blue. It's just this x, The X, Sir and we already have that over here. That's just negative. You and then I also need my to you. So this is our new integral. And then here we can go ahead and pull out the minus sign looks one zero. And then the next thing to do would be we could do another use up here. Let's take Vita. Be two minus you. Oops. Two minus you. That's just based on the radical here was inside for the negative DVD equals do you? So here I have a minus from here. But then we'll also get another minus from over here. It is a definite integral still. So I will have to change the limits. So plug in U equals one and zero into this equation. Oh, so plugging in one first we get to minus one is one and then plugging in zero for you We get to So then hear this radical. This is just square root of thie and then we're left over with you. So here you is just equal to two minus V. So we just have square ruin and hear that schist two minus v. So before we start integrating, let's first of all cancelled, this double minus that's not becomes a plus and louse. Just rewrite this integral so that it will be easier to use the power rule. So I'm doing here is rewriting the radical as Vito the one half. And then I just multiplied this radical turbo terms. So going on to the next page, I used the power rule. Now that's to be Now we have three halves and then we have to divide by three house. And then the next one becomes me too. The five half divided by five house plugging your end points wants it too. So here first, let's just multiply the constant out before we plug into too. So it's for over three and then we have two to the three house. And then here we have two or five times two of the five over two. So this is from plugging into now plug in one. Now we're basically finished here. The last thing we could do is just go ahead and trying to clean up our answer a little bit to make it look a little nicer. So here you can write. This is to square, or if you'd like to the four over, too. And then go ahead and combine these two exponents. That's two to the seven over too. And we still have the three in the bottom. Similarly, over here we have to to the one or flee want two to the two divided by two. So combining the two over two and five over two and then here combining these fractions, we'LL end up with fourteen over fifteen. Such is combining these here get a common denominator. Also here we should get a come denominator as well. So let's multiply top and bottom by five top and bottom here by three. So then we'LL have five of these up here, but then we're subtracting three of them. So that will leave us with two times through the seven halfs. Once again, I'll simplify the exponents so that it's two to the nine halfs And then we have minus fourteen from this and all divided by fifteen. What? And that's your final answer
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