Question

The value of ${\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx$ is

• A. $2$
• B. $\frac{3}{4}$
• C. $0$
• D. $-2$

Answer 1

Answer: (C)

$0$

Let $I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx$ .............. (1)
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}log\left[\frac{4+3\sin (\frac{\pi }{2}-x)}{4+3\cos (\frac{\pi }{2}-x)}\right]dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\cos x}{4+3\sin x}\right)dx$ ........... (2)
Adding (1) and (2), we obtain
$2I={\int }_0^{\frac{\pi }{2}}\{\log \left(\frac{4+3\sin x}{4+3\cos x}\right)+\log \left(\frac{4+3\cos x}{4+3\sin x}\right)\}dx$
$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log (\frac{4+3\sin x}{4+3\cos x}\times \frac{4+3\cos x}{4+3\sin x})dx$
$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log 1dx={\int }_0^{\frac{\pi }{2}}0dx$
$\Rightarrow I=0$
Hence, the correct Answer is C.