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Question Number 107051 by Ar Brandon last updated on 08/Aug/20

$${\int }_0^{\pi }\frac{\mathrm{cosx}}{2+{\sin }^2\mathrm{x}}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 08/Aug/20

$${\int }_0^{\pi }\frac{\mathrm{dt}}{2+{\mathrm{t}}^2}(\mathrm{sinx}=\mathrm{t})$$$$\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\mathrm{t}}{\sqrt{2}}\right)+\mathrm{C}={\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\mathrm{sinx}}{\sqrt{2}}\right)\right]}_0^{\pi }=0$$

Commented by Ar Brandon last updated on 08/Aug/20

OK, I got it now��

Commented by Dwaipayan Shikari last updated on 08/Aug/20

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Answered by Ar Brandon last updated on 08/Aug/20

$$\mathcal{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cosx}}{2+{\sin }^2\mathrm{x}}\mathrm{dx}-{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cosx}}{2+{\sin }^2\mathrm{x}}\mathrm{dx}=0$$

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