Evaluate the integral: $\int_0^{50 \pi} \sqrt{1-\cos 2x} \, dx$ (in $\sqrt{2}$)

Let $f : R \rightarrow R$ be a continuous function satisfying $f(x) + f(x + k) = n$,for all $x \in R$,where $k > 0$ and $n$ is a positive integer. If $I_{1} = \int_{0}^{4nk} f(x) dx$ and $I_{2} = \int_{-k}^{3k} f(x) dx$,then:

The value of $I=\int_{\sqrt{\log _e 2}}^{\sqrt{\log _e 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log _e 6-x^2\right)} d x$ is

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$.

$\int_{ - 1}^1 {\log \left( \frac{2 - x}{2 + x} \right)\,dx} = $

If $f$ is a continuous function and $f(x+T)=f(x)$ for all $x \in R$,it is given that $\int_0^{NT} f(t) dt = N \int_0^T f(t) dt$ (where $N$ is a natural number). Then,evaluate $\int_0^{50\pi} \sqrt{1-\cos 2x} dx$.