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

Let $f: [0, \frac{\pi}{2}] \rightarrow [0, 1]$ be the function defined by $f(x) = \sin^2 x$ and let $g: [0, \frac{\pi}{2}] \rightarrow [0, \infty)$ be the function defined by $g(x) = \sqrt{\frac{\pi x}{2} - x^2}$.
(There are two questions based on this paragraph. The questions given below are those two.)
$(1)$ The value of $2 \int_0^{\frac{\pi}{2}} f(x) g(x) dx - \int_0^{\frac{\pi}{2}} g(x) dx$ is
$(2)$ The value of $\frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} f(x) g(x) dx$ is

$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}\,d\theta = } $

$\int_0^{2n\pi } {\left( {|\sin x| - \left| {\frac{1}{2}\sin x} \right|} \right)} \;dx$ equals

The value of the definite integral $\int\limits_0^{\frac{1}{2}} \frac{\ln(1 + 2x)}{1 + 4x^2} \,dx$ is equal to:

$\int_0^{400 \pi} \sqrt{1-\cos 2 x} \, dx =$ (in $\sqrt{2}$)