If $f'(x) = x^2 + 5$ and $f(0) = -1$,then $f(x) = $

Let $f(x) = \tan^{-1}\left(\frac{1+\cos x}{\sin x}\right)$ and $g(x) = \tan^{-1}\left(\frac{\sin x}{1-\cos x}\right)$. Then,evaluate $\int (f(x) + g(x)) \, dx$.

If $\int \frac{1}{1-\cos x} dx = \tan \left(\frac{x}{\alpha} + \beta\right) + c$,then one of the values of $\frac{\pi \alpha}{4} - \beta$ is

$\int {{{\tan }^{ - 1}}\sqrt {\frac{{1 - \cos 2x}}{{1 + \cos 2x}}} } \;dx = $

If $\int (\sin 2x + \cos 2x) dx = \frac{1}{\sqrt{2}} \sin (2x - c) + a$,then the value of $a$ and $c$ is:

$\int \sqrt{1 + \sin \left( \frac{x}{4} \right)} \, dx$