Question Number 107051 by Ar Brandon last updated on 08/Aug/20
$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cosx}}{\mathrm{2}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 08/Aug/20
![∫_0 ^π (dt/(2+t^2 )) (sinx=t) (1/(√2))tan^(−1) ((t/(√2)))+C=[(1/(√2))tan^(−1) (((sinx)/(√2)))]_0 ^π =0](Q107054.png)
$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{dt}}{\mathrm{2}+\mathrm{t}^{\mathrm{2}} }\:\:\left(\mathrm{sinx}=\mathrm{t}\right) \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{t}}{\sqrt{\mathrm{2}}}\right)+\mathrm{C}=\left[\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sinx}}{\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\pi} =\mathrm{0} \\ $$
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
$$\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cosx}}{\mathrm{2}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cosx}}{\mathrm{2}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=\mathrm{0} \\ $$