$\int_{-1}^{1} x|x| \, dx = $

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx = $ . . . . . . .

Let $f$ be a non-constant continuous function for all $x \geq 0$. Let $f$ satisfy the relation $f(x) f(a-x)=1$ for some $a \in R^{+}$. Then, $I=\int_{0}^{a} \frac{d x}{1+f(x)}$ is equal to

$\int_{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x=$

$\int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}\,dx = } $

$\int_{0}^{1} x(1 - x)^{5} dx = . . . . . .$