Question

Evaluate :
$I={\int }_0^{\frac{\pi }{2}}\log \left[\frac{4+3\sin x}{4+3\cos x}\right]dx$

Answer 1

$\begin{array}{c}LetI={\int }_0^{\frac{\pi }{2}}\log \left[\frac{4+3\sin x}{4+3\cos x}\right]dx---\left(1\right)\\ ={\int }_0^{\frac{\pi }{2}}\log \left[\frac{4+3\sin \left(\frac{\pi }{2}-x\right)}{4+3\cos \left(\frac{\pi }{2}-x\right)}\right]dx\\ ={\int }_0^{\frac{\pi }{2}}\log \left[\frac{4+3\cos x}{4+3\sin x}\right]dx-----\left(2\right)\\ now,adding\left(1\right)+\left(2\right)weget\\ I+I={\int }_0^{\frac{\pi }{2}}\left\{\log \left[\frac{4+3\sin x}{4+3\cos x}\right]+\log \left[\frac{4+3\cos x}{4+3\sin x}\right]\right\}dx\\ 2I={\int }_0^{\frac{\pi }{2}}\log \left\{\left[\frac{4+3\sin x}{4+3\cos x}\right]\times \left[\frac{4+3\cos x}{4+3\sin x}\right]\right\}dx\\ 2I={\int }_0^{\frac{\pi }{2}}\log \left(1\right)dx\\ 2I=0\\ \therefore I=0\\ Hence,\\ I={\int }_0^{\frac{\pi }{2}}\log \left[\frac{4+3\sin x}{4+3\cos x}\right]dx=1.\end{array}$