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How do you go about this question?

Calculate $ \int \cos (x) (1− \sin x)^2 dx$ .

Can you integrate the different products separately?

Does it have something to do with integration by parts?

I have tried letting $u=(1− \sin x)^2$ but I don't think I'm heading in the right direction! Can anyone help?

Thanks

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    $\begingroup$ You have asked several questions and you have only accepted one of them. Maybe you should start accepting answers so that way people will be willing to help you in the future. By the way, you get $2$ points when you accept an answer. $\endgroup$ – user139708 Oct 14 '14 at 5:48
  • $\begingroup$ i did not know that. @HoracioOliveira. thank you $\endgroup$ – ojando Oct 14 '14 at 6:09
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Put $u = 1 - \sin x \implies du = -\cos x dx $. Hence

$$ \int \cos x (1 - \sin x)^2 dx = - \int u^2 du = -\frac{u^3}{3} + C$$

Therefore, you dont need to use integration by parts.

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  • $\begingroup$ why is it negative cosdx? $\endgroup$ – ojando Oct 15 '14 at 1:00
  • $\begingroup$ how do you find c? @HoraciaOliveira $\endgroup$ – ojando Oct 15 '14 at 1:02
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Try substituting $u = \sin(x)$ and $du = \cos(x)$ $$∫ \cos (x) (1−\sin x)^2 dx = ∫ (1-u)^2 du$$

You should be able to integrate that.

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Hint: $\cos x$ is the derivative of which function ?

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Let $\cos x = {1 \over 2} (e^{ix}+e^{-ix})$, $\sin x = {1 \over 2i} (e^{ix}-e^{-ix})$. Then we obtain (by expanding and then simplifying using the same rules): $\int \cos x (1 - \sin x)^2 dx = {1 \over 4} \int (5 \cos x - 4 \sin (2x) -\cos(3x) )dx$.

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  • $\begingroup$ Why the downvote? $\endgroup$ – copper.hat Oct 14 '14 at 14:30
  • $\begingroup$ what down vote? $\endgroup$ – ojando Oct 15 '14 at 1:00
  • $\begingroup$ I have two upvotes and one downvote. I was just wondering why the downvote? $\endgroup$ – copper.hat Oct 15 '14 at 2:16

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