If $\int_{-a}^a f(x) dx = \int_0^a f(x) dx + \int_0^a g(x) dx$, then $g(x) =$

Let $f:[-1, 2] \rightarrow [0, \infty)$ be a continuous function such that $f(x) = f(1-x)$ for all $x \in [-1, 2]$. Let $R_1 = \int_{-1}^2 x f(x) dx$ and $R_2$ be the area of the region bounded by $y = f(x)$,$x = -1$,$x = 2$,and the $X$-axis. Then $R_2$ 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.....................

$\int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ is equal to

$\int_0^{2 \pi} \sin ^6 x \cos ^5 x \, dx$ is equal to

$\int_0^{\frac{\pi}{2}} \frac{300 \sin x+100 \cos x}{\sin x+\cos x} \,dx = \ldots$ (in $\pi$)