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Question Number 151851 by Tawa11 last updated on 23/Aug/21

Answered by OlafThorendsen last updated on 23/Aug/21

$$\mathrm{I}={\int }_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}(x^2+\ln \left(\frac{\pi +x}{\pi -x}\right))\cos xdx\left(1\right)$$$$\mathrm{Let}u=-x:$$$$\mathrm{I}={\int }_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}(u^2+\ln \left(\frac{\pi -u}{\pi +u}\right))\cos udu$$$$\mathrm{I}={\int }_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}(u^2-\ln \left(\frac{\pi +u}{\pi -u}\right))\cos udu\left(2\right)$$$$\frac{\left(1\right)+\left(2\right)}{2}:\mathrm{I}={\int }_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}{u}^2\cos udu$$$$\mathrm{I}={[x^2\sin x-2\sin x+2x\cos x]}_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}$$$$\mathrm{I}=\frac{{\pi }^2}{2}-4$$

Commented by Tawa11 last updated on 23/Aug/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by peter frank last updated on 24/Aug/21

$$\mathrm{good}$$

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