Suppose $f$ is such that $f(-x) = -f(x)$ for every real $x$ and $\int_{0}^{1} f(x) dx = 5$,then $\int_{-1}^{0} f(t) dt = $

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):
$(A) \int_0^{\pi/4} x f(x) dx = \frac{1}{12}$
$(B) \int_0^{\pi/4} f(x) dx = 0$
$(C) \int_0^{\pi/4} x f(x) dx = \frac{1}{6}$
$(D) \int_0^{\pi/4} f(x) dx = 1$

Let $g(x) = \int_{0}^{x} f(t) dt$,where $f$ is a continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

$\int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right] dx =$

If $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$,then: