So as the question says I am dealing with $$\int_{0}^{\pi}4\sqrt{\sin(x)-\sin^3(x)}dx$$ So I'll show my steps and hopefully someone will point me in the right direction. $$4\int_{0}^{\pi}\sqrt{\sin(x)-\sin^3(x)}dx\rightarrow 4\int_{0}^{\pi}\sqrt{\sin(x)\cos^2(x)}dx\rightarrow 4\int_{0}^{\pi}\sqrt{\sin(x)}\cos(x)dx\rightarrow$$ $$u=\sin(x) \space du=\cos(x)\rightarrow 4\int_{0}^{\pi}u^{\frac{1}{2}}du\rightarrow \frac{8\sin(x)^\frac{3}{2}}{3}|_{0}^{\pi}$$ and after evaluating I got $4\int_{0}^{\pi}\sqrt{\sin(x)-\sin^3(x)}dx$$=0$ which isn't correct.

The correct solution with all the steps would be greatly appreciated, thanks for all the help in advance!


Notice that $\sqrt{\cos^{2}(x)}=\lvert \cos(x)\rvert$ and that $\cos(x)$ changes signs in the integral. Another problem is that $\sin(x)$ is not bijective over $[0,\pi]$ and so the substitution $u=\sin(x)$ is not appropriate over this interval. Another issue is that when you made your substitution you did not change the integration limits. Break the integral into two pieces, one over $[0,\frac{\pi}{2}]$ and another over $[\frac{\pi}{2},\pi]$, then make the substitution.

Alternatively notice that $\sin(x)=\sin(\pi-x)$ then:



so that by splitting the integral into two parts you get $2$ of the same integral. Now make the substitution you did before. Proceeding with the second idea we get:


I think you can take it from here.

  • $\begingroup$ Thanks a lot, really helped. $\endgroup$ – Kenshin Aug 8 '14 at 21:34
  • $\begingroup$ You're welcome. Glad it helped. $\endgroup$ – user71352 Aug 8 '14 at 21:42

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