Question

Find $I={\int }_0^{\frac{\pi }{2}}log\left(\frac{4+3sinx}{4+3cosx}\right)dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}log\left(\frac{4+3sinx}{4+3cosx}\right)dx...\left(1\right)$
$\Rightarrow I{\int }_0^{\frac{\pi }{2}}log\left[\frac{4+3sin(\frac{\pi }{2}-x)}{4+3cos(\frac{\pi }{2}-x)}\right]dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}log\left(\frac{4+3cosx}{4+3sinx}\right)dx...\left(2\right)$
From (1) + (2), we get
$2I={\int }_0^{\frac{\pi }{2}}[log\left(\frac{4+3sinx}{4+3cosx}\right)dx+log\left(\frac{4+3cosx}{4+3sinx}\right)dx]$
$={\int }_0^{\frac{\pi }{2}}[log\left(\frac{4+3sinx}{4+3cosx}\right)\times log\left(\frac{4+3cosx}{4+3sinx}\right)]dx$
$={\int }_0^{\frac{\pi }{2}}\left(log1\right)dx$
$={\int }_0^{\frac{\pi }{2}}\left(0\right)dx$
$=0$.