The value of the integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\log \frac{\pi-x}{\pi+x}\right) \cos x d x$ is equal to

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Let $\mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left[x^2+\log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x \mathrm{~d} x$ $\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \mathrm{~d} x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \mathrm{~d} x$ Let $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2$ Where $\mathrm{I}_1=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \mathrm{~d} x$ and $\mathrm{I}_2=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \mathrm{~d} x$
Consider $\begin{aligned} & I_1=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \cdot d x \\ & =2 \int_0^{\frac{\pi}{2}} x^2 \cos x d x \end{aligned}$ $\ldots\left[x^2 \cos x\right.$ is an even function $]$ $=2\left[x^2 \cdot \int \cos x \mathrm{~d} x-\int \frac{\mathrm{d}}{\mathrm{d} x}\left(x^2\right)\left(\int \cos x \mathrm{~d} x\right) \mathrm{d} x\right]_0^{\frac{\pi}{2}}$ $\begin{aligned} & =2\left[x^2 \cdot \sin x-\int 2 x \cdot \sin x d x\right]_0^{\frac{\pi}{2}} \\ & =2\left[x^2 \sin x-2 \int x \cdot \sin x d x\right]_0^{\frac{\pi}{2}}\end{aligned}$ $=2\left[x^2 \sin x-2\left(x(-\cos x)-\int(-\cos x) \mathrm{d} x\right)\right]_0^{\frac{\pi}{2}}$ $\begin{aligned} & =2\left[x^2 \sin x+2 x \cos x-2 \sin x\right]_0^{\frac{\pi}{2}} \\ & =2\left[\frac{\pi^2}{4} \sin \frac{\pi}{2}+2 \times \frac{\pi}{2} \cos \frac{\pi}{2}-2 \sin \frac{\pi}{2}\right.\end{aligned}$ $\left.-0^2 \sin 0-2 \times 0 \times \cos 0+2 \sin 0\right]$ $\begin{aligned} & =2\left[\frac{\pi^2}{4}-2-0-0-0\right] \\ & =2\left[\frac{\pi^2}{4}-2\right]=\frac{\pi^2}{2}-4 \end{aligned}$ Consider $\begin{aligned} & \mathrm{I}_2=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \mathrm{~d} x \\ & \therefore \quad \mathrm{I}_2=0 \quad \ldots\left[\log \left(\frac{\pi-x}{\pi+x}\right) \cos x \text { is an odd function }\right] \\ & \therefore \quad \mathrm{I}=\mathrm{I}_1+\mathrm{I}_2 \\ & \Rightarrow I=\frac{\pi^2}{2}-4 \end{aligned}$

The value of the integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\log \frac{\pi-x}{\pi+x}\right)…

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