Ive asked this question a few times but still don't understand how to go about and there have been a few answers that a different so can anyone clarify what the correct answer for this integral would be - including the $c$ value?

Calculate $∫\cos(x)(1−\sin x)^2 dx$ .

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?


  • 1
    $\begingroup$ There is no correct $c$ value. This is an indefinite integral, which means that its answers are the class of functions whose derivative is $\cos(x)(1 - \sin(x))^2$. This happens to be unique up to addition by a constant, and that constant can be anything. It's often called $c$. I encourage you to ask this question, exactly as you've asked us, to your teacher/instructor/professor, as there is a fundamental misunderstanding that should be cleared up. Finally, you don't want to integrate this by parts, but instead use a substitution. $\endgroup$ – davidlowryduda Oct 16 '14 at 2:07

One standard way to do the integral is to let $1-\sin x=u$. Then $du=\cos x\,dx$, and we end up with the integral $$\int -u^2\,du,$$ which is easy. Our integral turns out to be $$-\frac{1}{3}(1-\sin x)^3+C.\tag{1}$$

Another way of doing it is to expand the square. So we want to integrate $\cos x -2\cos x\sin x+\cos x\sin^2 x$. For the calculations of the integrals of the last two parts, $u=\sin x$ is again useful. We end up with $$\sin x-\sin^2 x+\frac{1}{3}\sin^3 x+C.\tag{2}$$

It is easy to verify that $-\frac{1}{3}(1-\sin x)^3$ is not the same function as $\sin x-\sin^2 x+\frac{1}{3}\sin^3 x$. However, they differ by a constant, so both (1) and (2) are valid answers to our integration problem. Another valid answer would be $x+\cos^2 x+\frac{1}{3}\sin^3 x+C$, because $x+\cos^2 x+\frac{1}{3}\sin^3 xx$ and $x-\sin^2 x+\frac{1}{3}\sin^3 x$ differ by a constant.

Remark: We could undoubtedly attack the problem using integration by parts. But it would be harder work than the simple substitution $u=1-\sin x$ that we used in the first calculation.


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