Evaluate $\int_{-\pi}^{\pi} \frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$ I came across this integral:$$\int_{-\pi}^{\pi} \frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$$
I tried $u=x+\pi$
$$\int_{-\pi}^{\pi} \frac{(x+\pi)^2}{1-\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$$
but had no success.
I also tried $u=-x$:
$$\int_{-\pi}^{\pi} \frac{x^2}{1-\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$$
Does this help?  Any suggestions.
Answer is $\dfrac{\pi^3}{3}$ by the way.

Rationalizing the denominator makes it much easier to proceed as follows
$$\frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}}=\frac{(1+\sin x-\sqrt{1+\sin^2x})x^2}{(1+\sin{x}+\sqrt{1+\sin^2{x}})(1+\sin{x}-\sqrt{1+\sin^2{x}})}$$
$$=\frac{(1+\sin{x}-\sqrt{1+\sin^2{x}})x^2}{2\sin x}$$
$$\therefore I=\int_{-\pi}^{\pi}\frac{(1+\sin{x}-\sqrt{1+\sin^2{x}})x^2}{2\sin x}\ dx\tag 1$$
Now, substituting $x=-x$, we get
$$I=-\int_{-\pi}^{\pi}\frac{(1-\sin{x}-\sqrt{1+\sin^2{x}})x^2}{2\sin x}\ dx\tag 2$$
Adding (1) & (2), we get
$$2I=\int_{-\pi}^{\pi} \frac{2x^2\sin x}{2\sin x} \mathop{dx}$$
$$I=\int_{0}^{\pi} x^2dx$$$$=\color{blue}{\frac{\pi^3}{3}}$$
