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Question Number 216162 by Spillover last updated on 28/Jan/25

Answered by MrGaster last updated on 02/Feb/25

$$\mathrm{Let}x=\frac{\pi }{3}-t\Rightarrow dx=-dt$$$${\int }_0^{\frac{\pi }{3}}\sqrt{{\sin }^{3}x\sin (\frac{\pi }{3}-x)}dx{\int }_{\frac{\pi }{3}}^0\sqrt{{\sin }^3(\frac{\pi }{3}-t)\sin t}(-dt)={\int }_0^{\frac{\pi }{3}}\sqrt{{\sin }^3(\frac{\pi }{3}-t)\sin tdt}$$$${\int }_0^{\frac{\pi }{3}}\sqrt{{\sin }^{3}x\sin (\frac{\pi }{3}-x)}dx={\int }_0^{\frac{\pi }{3}}\sqrt{{\sin }^3(\frac{\pi }{3}-x)\sin xdx}$$$${\int }_0^{\frac{\pi }{3}}\sqrt{{\sin }^{3}x\sin (\frac{\pi }{3}-x)}dx=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}(\sqrt{{\sin }^{3}x\sin (\frac{\pi }{3}-x)}+\sqrt{{\sin }^3(\frac{\pi }{3}-x)\sin x})dx$$$${\int }_0^{\frac{\pi }{3}}\sqrt{{\sin }^{3}x\sin (\frac{\pi }{3}-x)}dx=\frac{1}{2}{\int }_0^{\frac{\pi }{3}}\sqrt{{\sin }^3\left(\frac{\pi }{3}/2\right)}dx=\frac{\pi }{16}$$$$\mathrm{so}:$$$${\int }_0^{\frac{\pi }{3}}\sqrt{{\sin }^{3}x\sin (\frac{\pi }{3}-x)}dx=\frac{\pi }{16}$$

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