Answer:
Option A
Explanation:
$I. \int_{-1}^{1} x|x|dx=0$ $[\because x|x| $ is an odd function]
II.Let I= $ \int_{0}^{\pi/2} \left[1+\left( \log \frac{4+3 \sin x}{4+3 \cos x}\right)\right]dx$
$ I= \int_{0}^{\pi/2}\left(1+ \left[ \log \frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left( \frac{\pi}{2}-x\right)})\right]\right)dx$
$I= \int_{0}^{\pi/2}\left(1+ \left[ \log \frac{4+3 \cos x}{4+3 \sin x}\right]\right)dx$
$\Rightarrow$ $2 I= \int_{0}^{\pi/2}\left(2+ \log \frac{4+3 \sin x}{4+3 \cos x}+\log \frac{4+3 \cos x}{4+3 \sin x}\right)dx$
$\Rightarrow$ $2 I= \int_{0}^{\pi/2}\left(2+ \log \frac{(4+3 \sin x)(4+3 \cos x)}{(4+3 \cos x)(4+3 \sin x)}\right)$
$\Rightarrow$ $2 I= \int_{0}^{\pi/2}(2+ \log 1)dx=\int_{0}^{\pi/2} 2dx= \pi$
$\Rightarrow$ $I= \frac{\pi}{2}$
III. $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$
IV. $\int_{-a}^{a} f(x) dx= \int_{0}^{a} f(x) dx+\int_{0}^{a} f(-x) dx$
= $\int_{0}^{a} [f(x) +f(-x)]dx$
