A definite integral

Evaluate the integral:

$$\mathcal{J}= \int_{-\pi/2}^{\pi/2} \frac{x^2\cos x}{1+x+\sqrt{x^2+1}}\, {\rm d}x$$

Solution

We have succesively:

\begin{align*}\mathcal{J}=\int_{-\pi/2}^{\pi/2} \frac{x^2 \cos x}{1+x+\sqrt{x^2+1}}\, {\rm d} x &\; \overset{u=-x}{=\! =\! =\! =\!} \int_{-\pi/2}^{\pi/2} \frac{x^2 \cos x}{1-x + \sqrt{x^2+1}}\, {\rm d} x \\ &= \int_{-\pi/2}^{\pi/2} \frac{x^2\cos x}{1+ \frac{1}{\sqrt{x^2+1}+x}}\, {\rm d} x\\ &= \int_{-\pi/2}^{\pi/2} \frac{\left ( \sqrt{x^2+1}+x \right )x^2 \cos x}{1+ x+\sqrt{x^2+1}}\, {\rm d}x =\mathcal{J}'\end{align*}

Adding the two integrals together we have that:

$$2\mathcal{J}= \int_{-\pi/2}^{\pi/2} x^2 \cos x \, {\rm d}x = \frac{\pi^2 -8}{2}$$

and the result follows.