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

The correct option is C $0$

As,

$\begin{array}{c}{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(x-a\right)dx\\ So,\\ I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)\\ ={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin \left(\frac{\pi }{2}-x\right)}{4+3\cos \left(\frac{\pi }{2}-x\right)}\right)\\ ={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\cos x}{4+3\sin x}\right)\end{array}$

Adding the above two integral,

$\begin{array}{c}I+I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx+{\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\cos x}{4+3\sin x}\right)dx\\ 2I={\int }_0^{\frac{\pi }{2}}\left[\log \left(\frac{4+3\sin x}{4+3\cos x}\right)+\log \left(\frac{4+3\cos x}{4+3\sin x}\right)\right]dx\\ As,\\ \log a+\log b=\log ab\\ So,\\ 2I={\int }_0^{\frac{\pi }{2}}\log \left[\left(\frac{4+3\sin x}{4+3\cos x}\right)\times \left(\frac{4+3\cos x}{4+3\sin x}\right)\right]dx\\ ={\int }_0^{\frac{\pi }{2}}\log \left(1\right)dx\\ ={\int }_0^{\frac{\pi }{2}}0dx\\ 2I=0\\ I=0\end{array}$

Thus Option C