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(a) Prove the reduction formula$$ \int \cos^n x dx = \frac{1}{n} \cos^{n - 1} x \sin x + \frac{n - 1}{n} \int \cos^{n - 2} x dx $$(b) Use part (a) to evaluate $ \displaystyle \int \cos^2 x dx $.(c) Use parts (a) and (b) to evaluate $ \displaystyle \int \cos^4 x dx $.

(a) $\int \sin ^{2} x d x=-\frac{1}{2} \cos x \sin x+\frac{1}{2} x$$\int \sin ^{4} x d x=\frac{1}{4} \cos x \sin ^{3} x-\frac{3}{8} \cos x \sin x+\frac{3}{8} x$(b) $\int \cos ^{2} x d x=\frac{\sin (2 x)}{4}+\frac{x}{2}+C$(c) $\frac{1}{4} \cos ^{3}(x) \sin (x)+\frac{3}{16} \sin (2 x)+\frac{3}{8} x+C$

09:16

Wen Z.

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Missouri State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Lectures

01:11

In mathematics, integratio…

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(a) Prove the reduction fo…

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Use integration by parts t…

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so using integration by parts we are going thio improve a formula to be true. So the formula what we're gonna be starting off with is the indefinite integral of co sign to the annex DX. So when we perform in integration by parts we wanna pick are you D V D u and V. So let's see what that's gonna be. We'll have a you Equalling the co sign to the end minus one x that makes TV co sign X dx. We do that because if we combine them that it's just the same thing as this, we ultimately want to get it into the form of the integral of UDV. Then we'll have a d. U of negative and minus one co sign of and minus co sign to the n minus two x sine x dx. And this is by taking the derivative of this right here and using chain role and then we want RV, which is just gonna be by taking the anti derivative of this right here. And that's obviously just gonna be the sign of ax. So now with this in mind, we wanna look at our integration by parts formula which is U V minus the integral of V d u. So we combine all these components. What we're gonna have is u V is going to be side X Times Co sign to the n minus one X, that's UV and then we'll have minus the integral of V D u. So again we have signed X and then times d'you which is just this large expression right here. So we'll duplicate that and multiply it there. Now that we have this in place, we can replace thesis Ein of the sine squared of X, which will come about through multiplying knees with Theo the equivalent expression of one minus cosine squared X. So what we end up getting as a result is that this is equal tomb three exact same thing. This is gonna be the same. It's gonna be minus the integral will factor out the negative, so we'll make it. Plus the integral of and minus one co signed to the end minus two acts one minus coasting squared X instead of sine squared x, the X. Now that we have this, uh, we're gonna be able thio separate it into two different integral because we'll multiply these out and that's through the distributive property that we're able to do that. So now what we end up having is that it's equal to co sign to the n minus one x sine X plus and minus one factored out times the integral of cosign and minus two X DX minus and minus one times the integral of cosign of an goes under the end of X DX. So now that we have that in place, we see that, um, if we add the and minus one co sign portion, we can add it. It's right here if we added to both sides and divide both sides by n what we end up getting is what we wish to prove, which is that the cosine the anti vax D X is equal to one over and co sign to the n minus one x Synnex plus and minus one over and times thean a roll of co sign the end minus two x e x. And what we've shown is that that is all equal to this portion right here. So that will be our final proof of this formula. Then for part B, we want to see it in action More so So we're going to evaluate the cosine squared of X DX. So when we do that, we just substitute and equals two. So the integral of co sign squared X DX is not going to equal one half co sign of end minus ones to minus one X side X plus two minus 1/2. The inter role of co sign too. So I thought this in mind, we're gonna end up simplifying things further. We'll get one half co sign X Synnex plus one half thean of role of one DX. We know that one DX just gives us X plus c So our final answer is going to end up giving us, um, we'll get 1/4 to co sign ax sign ax plus X over two plus c. Obviously, the 2/4 would just give us one half, and that is the same thing. Um, that will end up getting as a result is going to be thesis ein of two X over four plus X over two plus c. Then lastly, we want to do part C. This is the answer. So for part C, we're gonna be using the exact same formula when we use this formula and plug in our values and simplify further what we end up getting is 1/4 co sign Cube X sine X plus 3/16 sign two X plus three x over eight plus k some constant or you could you see. And again. That's just by plugging in a four into theory journal equation that we solved in part a.

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