QJEEAdv12Ma6a07: JEE Advanced 2012 Mathematics - Class 12 JEEAdv12Ma6a07 Solution: The value of the integral is

We know the property

$\begin{aligned}& \int_{-a}^a f(x) d x \\& =\left\{\begin{array}{cc}2 \int_0^a f(x) d x, & \text { when } f(x) \text { is an even function } \\0, & \text { when } f(x) \text { is an odd function }\end{array}\right.\end{aligned}$

Let, $\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\ln \left(\frac{\pi+x}{\pi-x}\right)\right) \cos x \cdot d x$

$\Rightarrow \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot \cos x d x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x \cdot \ln \left(\frac{\pi+x}{\pi-x}\right) d x$

Here, $x^2 \cos x$ is an even function and $\cos x \cdot \ln \left(\frac{\pi+x}{\pi-x}\right)$ is an odd function.

$\begin{aligned}& \therefore \mathrm{I}=2 \int_0^{\frac{\pi}{2}} x^2 \cos x d x+0 \\& \Rightarrow \mathrm{I}=2\left[\left(\sin x \cdot x^2\right)_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 2 x \sin x \cdot d x\right] \\& \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4 \int_0^{\frac{\pi}{2}} x \sin x \cdot d x \\& \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4\left[(x(-\cos x))_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 1 \cdot(-\cos x) d x\right] \\& \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4 \int_0^{\frac{\pi}{2}} \cos x d x \\& \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4[\sin x]_0^{\frac{\pi}{2}} \\& \Rightarrow \mathrm{I}=\frac{\pi^2}{2}-4\end{aligned}$