$\int_{0}^{\pi /2} {\sin 2x \log \tan x \, dx}$ is equal to

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 the integral $\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is

$\int\limits_0^{\frac{\pi }{2}} {\sqrt {\sin 2\theta } } \sin \theta \,d\theta$ is equal to :

If $\int_{0}^{1} \tan ^{-1} x \, dx = p$,then the value of $\int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) \, dx$ is

$\int_{-1}^{1} x|x| \, dx = $