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Problem 59 Medium Difficulty

Change of Variables In Exercises 53-60, find the indefinite integral by making a change of variables.
$$\int \cos ^{3} 2 x \sin 2 x d x$$

Answer

$-\frac{1}{8} \cos ^{4} 2 x+c$

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Adam87 K.

March 14, 2021

Hi

Video Transcript

in this video we'll look at how to solve an indefinite integral using the change of variables method. So here we have the indefinite integral of co sign to X cubed signed two x d. X. The first thing I'm gonna do is rewrite this so it's a little bit easier to look at. So I'm gonna write co sign two x quantity, cubed and then leave the rest the same. So with the change of variables method, we want to be able to use U substitution in order. Thio rearrange our indefinite integral into something that we know how to integrate. So here what I might do first is look for an inter function, uh, that I can express as you to make this simpler. So I'm gonna look at this component right here, co sign of two X and I'm going to let that be you. Now, if I take the derivative of both sides of this equation, I'm gonna end up with D U on the left and then negative two times a sign of two x d x on the right. Make sure you don't forget that DX that could be really easy to leave off So one of the things that will notice now is this a sign of two x appears or sign of two x d. X appears in our original interval. So what I want to be able to do is manipulate this equation. So that way I can directly substitute a component of it back into the integral. So what I'm gonna do is I'm gonna divine both sides by negative to sew on my left. I end up having negative 1/2 do you? And on the right I have sign of two x d. X. So what I'm gonna do then is plug both of these my you in my negative 1/2 to you into my original integral. So here this becomes you cubed times negative 1/2 d you. So now we have an indefinite integral that is entirely in terms of you, but it's something that we know how to integrate. So I'm gonna pull that constant term out in front who have the integral of you. Cute. Do you? If I take that integral, I'm going to be left with 1/4 times you to the fourth power. You don't forget my plus C on the end. So simplifying this a bit. We have negative 18 you to the fourth plus. See, Now I'm not quite finished. I need to substitute you back into the equation. So that way I have something that looks similar to what I was originally given the correct variable here. So I'm gonna have negative 1/8 co sign to the fourth power two X plus C. Now, the nice thing about using the change variables method and working with integration is that we can always check our work by going backwards and taking the derivative. Let's check our work down here. So here I want to take the derivative with respect to X negative 1/8 co sign to X to the fourth plus C and notice. I'm just writing that co sign to ex term same way that we re route the original integral just to make it a little bit easier to look at a little bit easier to differentiate. So because I'm taking the derivative of a constant that C term is going to go away and that applying the chain rule, we're going to have a negative 1/8 pull my constant out times four co signed two X cube times Negative sign of two x times two. And if you don't remember how to use the chain rule, you want to go back and refresh yourself on how to do that. This is a direct application of it here when we check our work. So now if I simplify that will notice that my negative 1/8 my four, my two and the negative sign in front of side of two ex those all multiply together to get one mark those out there and we're left with the co sign of two x cubed quantity, huge times the sign of two X, which is exactly what we integrated at the beginning. So this is our final answer.

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