$\int_{-a}^{a} \sin x \, f(\cos x) \, dx = $

If $f(t) = \int_0^t \tan^{(2n-1)} x \, dx$,$n \in N$,then $f(t+\pi) =$

$\int \limits_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x$ is equal to:

$\int_0^{\frac{\pi}{2}} \sqrt{\tan x} \, dx =$

Let $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$ for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the correct expression$(s)$ is(are):
$(A) \int_0^{\pi/4} x f(x) dx = \frac{1}{12}$
$(B) \int_0^{\pi/4} f(x) dx = 0$
$(C) \int_0^{\pi/4} x f(x) dx = \frac{1}{6}$
$(D) \int_0^{\pi/4} f(x) dx = 1$

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