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Question Number 118242 by mathdave last updated on 16/Oct/20

$$solve$$$${\int }_0^{\pi }\frac{x\cos x}{(1+{\sin }^{2}x)}dx$$

Answered by TANMAY PANACEA last updated on 16/Oct/20

$$★\int \frac{cosx}{1+{sin}^{2}x}dx=\int \frac{d\left(sinx\right)}{1+{sin}^{2}x}={tan}^{-1}\left(sinx\right)★$$$$now$$$${\int }_0^{\pi }\frac{xcosx}{1+{sin}^{2}x}dx$$$$\mid {xtan}^{-1}\left(sinx\right){\mid }_0^{\pi }-{\int }_0^{\pi }1\times {tan}^{-1}\left(sinx\right)dx$$$$0-{\int }_0^{\pi }{tan}^{-1}\left(sinx\right)dx$$$$f\left(x\right)={tan}^{-1}\left(sinx\right)$$$$f\left(0\right)=0$$$$f\left(\pi \right)=0$$$$so{tan}^{-1}\left(sinx\right)makealoopinx\in [0,\pi ]$$$$inbetween[0,\pi ]{tan}^{-1}\left(sinx\right)musthavemaximum$$$$value...atx=\frac{\pi }{2}{tan}^{-1}\left(sin\frac{\pi }{2}\right)=\frac{\pi }{4}$$$${\int }_0^{\pi }\frac{\pi }{4}dx>{\int }_0^{\pi }{tan}^{-1}\left(sinx\right)dx>0$$$$\frac{{\pi }^2}{4}>{\int }_0^{\pi }{tan}^{-1}\left(sinx\right)dx>0$$$$(-1)\frac{{\pi }^2}{4}<{\int }_0^{\pi }-{tan}^{-1}\left(sinx\right)dx<0$$$$soI={\int }_0^{\pi }\frac{xcosx}{(1+{sin}^{2}x)}={\int }_0^{\pi }-{tan}^{-1}\left(sinx\right)dx$$$$\frac{-{\pi }^2}{4}<I<0$$$$$$$$$$

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