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Question Number 122979 by bemath last updated on 21/Nov/20

$${\int }_0^{\pi }\frac{x\sin x}{\sqrt{3+\sin {}^{2}x}}dx?$$

Answered by liberty last updated on 21/Nov/20

$$\psi \left(x\right)={\int }_0^{\pi }\frac{x\sin x}{\sqrt{4-\cos {}^{2}x}}dx$$$$replacingxby\pi -x,give$$$$\psi \left(x\right)={\int }_{\pi }^0\frac{(\pi -x)\sin (\pi -x)}{\sqrt{4-\cos {}^2(\pi -x)}}(-dx)$$$$\psi \left(x\right)={\int }_0^{\pi }\frac{(\pi -x)\sin x}{\sqrt{4-\cos {}^{2}x}}dx$$$$addingthebothintegral,weobtain$$$$2\psi \left(x\right)={\int }_0^{\pi }\frac{\pi \sin x}{\sqrt{{\left(2\right)}^2-\cos {}^2\left(x\right)}}dx$$$$\psi \left(x\right)=-\frac{\pi }{2}{\left[{\sin }^{-1}\left(\frac{\cos x}{2}\right)\right]}_0^{\pi }$$$$\psi \left(x\right)=-\frac{\pi }{2}(-\frac{\pi }{3})=\frac{\pi }{6}.▴$$$$$$

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