I have encountered the following integral:

$$\int_{0}^{\pi} \sin ^{n}(\eta) d \eta=\underbrace{\left[\left(\sin ^{n-1}(\eta)\right)(-\cos (\eta))\right]_{\eta=0}^{\pi}}_{=0} -\int_{0}^{\pi}\left((n-1) \sin ^{n-2}(\eta) \cos (\eta)\right)(-\cos (\eta)) d \eta$$

But I'm not getting how did the integral evaluate to become what is on the right side of the above equation. I am trying to do integration by parts but in vain. Could you please show me the missing steps?

The right hand side can then be simplified to look like: $$=(n-1) \int_{0}^{\pi} \cos ^{2}(\eta) \sin ^{n-2}(\eta) d \eta$$

  • $\begingroup$ No, it's integration by parts with $u=\sin^{n-1}(x),\,\mathrm{d}v=\sin(x)\mathrm{d}x,\,v=-\cos(x)$. Would you please provide your integration by parts so we see where did you stuck and can point in right direction? Thanks. $\endgroup$ Jul 13 '20 at 18:39
  • $\begingroup$ Related question. $\endgroup$
    – Axion004
    Jul 13 '20 at 18:47
  • $\begingroup$ Thanks, @AlexeyBurdin. With your suggested substitutions, I've have been able to reach the right side. I wasn't using the right substitutions earlier. $\endgroup$ Jul 13 '20 at 18:54
  • $\begingroup$ @Axion004, no I wasn't looking for the form of the right-hand side given in that question. My question was different. $\endgroup$ Jul 13 '20 at 18:56
  • $\begingroup$ math.stackexchange.com/questions/3600745/… $\endgroup$
    – Guy Fsone
    Jul 13 '20 at 18:58

The key observation here is that you can write $\sin^{n}(\eta) = \sin^{n-1}(\eta)\sin(\eta).$ Then, to integrate by parts, we let

\begin{alignat}{3} u &= \sin^{n-1}(\eta) &&\implies du &&= (n-1)\sin^{n-2}(\eta)\cos(\eta)\\ dv &= \sin(\eta) &&\implies v &&= -\cos(\eta). \end{alignat}

Then, we get

$$\int_{0}^{\pi}\sin^{n}(\eta)\,d\eta = \left[\sin^{n-1}(\eta)(-\cos(\eta))\right]\bigg|_{0}^{\pi} -\int_{0}^{\pi}\left((n-1) \sin ^{n-2}(\eta) \cos (\eta)\right)(-\cos (\eta))\, d\eta.$$


$$\int_{0}^{\pi} \sin ^{n}(\eta) d \eta=\int_{0}^{\pi/2} \sin ^{n}(\eta) d \eta + \int_{\pi/2}^{\pi} \sin ^{n}(\eta) d \eta$$ Put $\eta \rightarrow \eta + \pi/2$ in the second integral to get: $$\int_{0}^{\pi} \sin ^{n}(\eta) d \eta=\int_{0}^{\pi/2} \sin ^{n}(\eta) d \eta + \int_{0}^{\pi/2} \cos ^{n}(\eta) d \eta$$ $$=\beta((n+1)/2, 1/2)$$ $$=\frac{\Gamma((n+1)/2)\sqrt\pi}{\Gamma(n/2 + 1)}$$


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