$\int_0^\pi x f(\sin x) dx = $

$\int\limits_{\frac{1}{2}}^2 {\frac{1}{x}{{\tan }^{2015}}\left( {x - \frac{1}{x}} \right)dx} $ is equal to

The value of the integral $\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ is :

The value of $\int_0^1 (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} dx$ is equal to:

Let $m, n, p, q$ be four positive integers. If $\int_0^{2 \pi} \sin^m x \cos^n x \, dx = 4 \int_0^{\pi/2} \sin^m x \cos^n x \, dx$, $\int_0^{2 \pi} \sin^p x \cos^n x \, dx = 0$, $\int_0^{\pi} \sin^p x \cos^q x \, dx = 0$, $a = m + n + p$ and $b = m + n + q$, then:

$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ is equal to