The value of the integral $\int_{-\pi/2}^{\pi/2} \left( x^{2} + \ln\frac{\pi+x}{\pi-x} \right)\cos x\,dx$ is

The value of the integral ∫_(-π/2)^(π/2) ( x² + ln(π+x)/(π-x) )cos x dx is

$I = \int_{-\pi/2}^{\pi/2}\left(x^2+\ln\dfrac{\pi+x}{\pi-x}\right)\cos x\,dx$

$\text{symmetric interval } \left[-\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right]$

$\text{To find: the value of } I \text{ (among the options)}$

$1.\ \text{split } I$ $2.\ \text{odd part } = 0$ $3.\ \text{integrate the even part}$ $I = \underbrace{\int_{-\pi/2}^{\pi/2} x^2\cos x\,dx}_{I_1} + \underbrace{\int_{-\pi/2}^{\pi/2}\ln\left(\frac{\pi+x}{\pi-x}\right)\cos x\,dx}_{I_2}$ $f(x)=\ln\dfrac{\pi+x}{\pi-x}$ $f(-x)=\ln\dfrac{\pi-x}{\pi+x}=-\ln\dfrac{\pi+x}{\pi-x}=-f(x)$ $\Rightarrow f \text{ odd},\ \cos x \text{ even} \Rightarrow f(x)\cos x \text{ odd}$ $I_2 = \int_{-\pi/2}^{\pi/2}\underbrace{f(x)\cos x}_{\text{odd}}\,dx = 0$ $I_1 = \int_{-\pi/2}^{\pi/2}x^2\cos x\,dx = 2\int_{0}^{\pi/2}x^2\cos x\,dx$ $\int x^2\cos x\,dx = x^2\sin x+2x\cos x-2\sin x$ $\left[x^2\sin x+2x\cos x-2\sin x\right]_0^{\pi/2} = \left(\tfrac{\pi^2}{4}-2\right)-0 = \tfrac{\pi^2}{4}-2$ $I = 2\left(\tfrac{\pi^2}{4}-2\right) = \tfrac{\pi^2}{2}-4$ $\boxed{\frac{\pi^2}{2}-4} \quad \Rightarrow \text{ option (2)}$