Question

The value of ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{x^2\cos x}{1+e^x}dx$ is equal to

• A. $\frac{{\pi }^2}{4}-2$
• B. $\frac{{\pi }^2}{4}+2$
• C. ${\pi }^2{-e}^{\frac{\pi }{2}}$
• D. ${\pi }^2{+e}^{\frac{\pi }{2}}$

Answer 1

Answer: (A)

$\frac{{\pi }^2}{4}-2$

${\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x}{1+e^x}dx={\int }_0^{\pi /2}(\frac{x^2\cos x}{1+e^x}+\frac{x^2\cos x}{1+e^{-x}})dx$

(As numerator is even function but denominator is odd function)

$={\int }_0^{\pi /2}\frac{x^2\cos x+e^x\left(x^2\cos x\right)}{1+e^x}dx$

$={\int }_0^{\pi /2}{x}^2\cos xdx$

$=(x^2\sin x{)}_0^{\pi /2}-{\int }_0^{\pi /2}2x\sin xdx$

(uv rule of integration)

$=\frac{{\pi }^2}{4}-2\left[\right(-x\cos x{)}_0^{\pi /2}-{\int }_0^{\pi /2}-\cos xdx]$

(uv rule of integration)

$=\frac{{\pi }^2}{4}-2[0+(\sin xdx{)}_0^{\pi /2}]$

$=\frac{{\pi }^2}{4}-2$

This is the required solution.