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To take $$\int{\cot^5(x)}dx$$

I substitute $\csc(x)=u$, $du=-\csc(x)\cot(x)dx$

and then it seems like $$du = -u\cot(x)dx \quad\implies\quad \cot(x)dx=-\frac{du}{u} \tag1$$ is it not?

but then $$\begin{align} \int{\cot^5(x)}dx &= \int{[\cot^2(x)]^2}\,\cot(x)dx \tag2\\ &=\int{(u^2-1)^2\cdot\frac{-du}{u}} \tag3\\ &=-\int{\frac{u^4-2u^2+1}{u}du} \tag4\\ &=-\frac14\csc^4(x)+\csc^2(x)-\log(\csc(x)) +C \tag5 \end{align}$$

which seems wrong.

I feel like my substitution is a sham, but I don't understand why. Where is my mistake?

(note: I know that the correct solution is to substitute $\sin(x)$, don't bother. )

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  • $\begingroup$ The easiest way to check whether an integral result is correct is to take its derivative and see if you get back the original function. Give it a go, maybe? $\endgroup$ Commented Aug 14, 2023 at 17:32
  • $\begingroup$ The proof is in the pudding. Have you derived your answer? And does it match the function to be integrated? Can you find a few trig identities to reconcile the 2 answers. If so, then you don't need our help. $\endgroup$ Commented Aug 14, 2023 at 17:33
  • $\begingroup$ Heck, you're right. The last time I tried to take the derivative I made a mistake. Seems the solution is right, though it doesn't look like the one in textbook $\endgroup$ Commented Aug 14, 2023 at 17:35
  • $\begingroup$ Related (if not a duplicate): Getting different answers when integrating using different techniques $\endgroup$ Commented Aug 14, 2023 at 18:02

1 Answer 1

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This is perfectly fine. In fact, even WolframAlpha agrees (once one utilizes the identity $\log(\csc(x)) = -\log(\sin(x))$).

If you wish to check it for yourself, differentiate it and see if you can do the algebraic manipulations required to get $\cot^5(x)$ back out.

(note: I know that the correct solution is to substitute $sin(x)$, don't bother. )

Debatable at best. Integrals can have many distinct paths to a solution, and those paths may even result in different looking solutions. Bear in mind the end goal is just to find a function that differentiates to your integrand, but there are infinitely-many such functions that all differ by a constant.

It might be the simplest and easiest solution, but that doesn't mean your method is "incorrect," just different. And mathematicians value the ability to take multiple paths to a correct answer.

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