The value of $\int_0^{2\pi } {{\cos }^{99}x\,dx} $ is

If $\int_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \, dx}{(1+e^{\sin x})(1+\sin ^4 x)} = \alpha \pi + \beta \log _e(3+2 \sqrt{2})$,where $\alpha, \beta$ are integers,then $\alpha^2+\beta^2$ equals.....................

Evaluate the definite integral: $\int_{\pi / 4}^{\pi / 2} \frac{3 \, dx}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}}$

Let $\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{{\tan x}}{2}} \right)} dx = \alpha $. Then $\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{{\tan x - 2\cot x}}{3}} \right)} dx$ is equal to:

$\int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx = $

Let $(a, b)$ be the point of intersection of the curve $x^2=2y$ and the straight line $y-2x-6=0$ in the second quadrant. Then the integral $I=\int_a^b \frac{9x^2}{1+5^x} dx$ is equal to: