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For $\int_0^{\pi}\sqrt{\sin^2x-\sin^5x}dx$, I've tried to several different ways, like $$\int_0^{\pi}\sqrt{\sin^2x-\sin^5x}dx=2\int_0^{\pi/2}\sqrt{\sin^2x-\sin^5x}dx,$$ or let $u=\sqrt{1-\sin^3x}$, but neither of them works. I'm doubting if there is a solution.

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    There must be a solution. And the answer is about 1.12982. – 2017-01-10
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    Was it an exam question? – 2017-01-10
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    If the $5$ would have been a $4$ or a $6$ one could easily compute the integral. – 2017-01-10
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    It's not an exam, but someone ask me for help. I sadly found I can't solve it analytically. – 2017-01-12

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$$I=2\int_{0}^{\pi/2}\sqrt{\sin^2 x-\sin^5 x}\,dx = \int_{0}^{1}2t\,\sqrt{\frac{1+t+t^2}{1+t}}\,dt $$ and by using the substitution $t=\frac{1}{2}\left(-1+u+\sqrt{u^2+2u-3}\right)$ we get:

$$ I = \frac{9\sqrt{6}-8}{20}+\color{blue}{\int_{1}^{3/2}\frac{(u+2)\sqrt{u(u-1)}}{\sqrt{u+3}}\,du} $$ where the blue integral is an elliptic integral, equal to: $$ 2\int_{2}^{3/\sqrt{2}}(u^2-1)\sqrt{(u^2-4)(u^2-3)}\,du.$$ Numerically, $I\approx 1.1298158$.

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    The explicit value of the integral is $$\frac{1}{5} \left(3 \sqrt{6}-2+12 i ~F\left(i~ \text{csch}^{-1}\left(\sqrt{2}\right)|\frac{1}{4}\right)-12 i ~E\left(i~ \text{csch}^{-1}\left(\sqrt{2}\right)|\frac{1}{4}\right)\right)=$$ $$=1.1298158061973205974...$$ – 2017-01-10
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    @YuriyS: thank you for the addendum! – 2017-01-10
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    @ Jack D'Aurizio, don't mention it. In the end I just used Mathematica on the blue integral – 2017-01-10