# How to solve $\int_{0}^{\pi}4\sqrt{\sin(x)-\sin^3(x)}dx$?

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$$=0 which isn't correct. The correct solution with all the steps would be greatly appreciated, thanks for all the help in advance! ## 1 Answer Notice that \sqrt{\cos^{2}(x)}=\lvert \cos(x)\rvert and that \cos(x) changes signs in the integral. Another problem is that \sin(x) is not bijective over [0,\pi] and so the substitution u=\sin(x) is not appropriate over this interval. Another issue is that when you made your substitution you did not change the integration limits. Break the integral into two pieces, one over [0,\frac{\pi}{2}] and another over [\frac{\pi}{2},\pi], then make the substitution. Alternatively notice that \sin(x)=\sin(\pi-x) then:$$\int_{\frac{\pi}{2}}^{\pi}\sqrt{\sin(x)-(\sin(x))^{3}}dx=\int_{\frac{\pi}{2}}^{\pi}\sqrt{\sin(\pi-x)-(\sin(\pi-x))^{3}}dx=\int_{\frac{\pi}{2}}^{0}\sqrt{\sin(u)-(\sin(u))^{3}}(-1)du=\int_{0}^{\frac{\pi}{2}}\sqrt{\sin(u)-(\sin(u))^{3}}du$$so that by splitting the integral into two parts you get 2 of the same integral. Now make the substitution you did before. Proceeding with the second idea we get:$$\int_{0}^{\pi}\sqrt{\sin(x)-(\sin(x))^{3}}dx=2\int_{0}^{\frac{\pi}{2}}\sqrt{\sin(x)}\cos(x)dx=2\int_{0}^{1}\sqrt{u}du$\$

I think you can take it from here.

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