Discussion Forum

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

Answer:

Explanation:

Let   $I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2}\cos x}{1+e^{x}}dx$         ......(i)

$[\because \int_{a}^{b}  f(x) dx= \int_{a}^{b} f(a+b-x)dx]$

$\Rightarrow I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2}\cos (-x)}{1+e^{-x}}dx$  .................(ii)

On adding Eqs(i) and (ii) , we get

         $2 I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{2}\cos x[\frac{1}{1+e^{x}}+\frac{1}{1+e^{-x}}] dx$

          =$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{2}\cos x.(1)dx$

           [ $\therefore \int_{-a}^{a}f(x) dx$

                       $=2\int_{0}^{a} f(x) dx.whenf(-x)=f(x)]$

$\Rightarrow  2I=2\int_{0}^{\frac{\pi}{2}} x^{2}\cos x dx$

Using integration by parts , we get

                $2I= 2[x^{2}(\sin x)-(2x)(-\cos x)+(2)(-\sin x)]_{0}^{\frac{\pi}{2}}$

$\Rightarrow 2I=2[\frac{\pi^{2}}{4}-2]$

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