Question

The value of $\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx$ is

(a) 2

(b) $\frac{3}{4}$

(c) 0

(d) −2

Answer 1

(c) 0

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left[\frac{4+3\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{4+3\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{4+3\cos x}{4+3\sin x}\right)dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\log \left(\frac{4+3\sin x}{4+3\cos x}\right)+l\mathrm{og}\left(\frac{4+3\cos x}{4+3\sin x}\right)\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\times \frac{4+3\cos x}{4+3\sin x}\right)dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log 1dx=0 \\ \mathrm{Hence}I=0 \end{gathered}$$