Question

The value of the definite integral $\underset{0}{\overset{2\pi }{\int }}x\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$ equals

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

Answer 1

Answer: (A)

Let $I=\underset{0}{\overset{2\pi }{\int }}x\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$ .....(1)
Also, $I=\underset{0}{\overset{2\pi }{\int }}(2\pi -x)\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$ ...(2)(Using King Property)
$\therefore$ On adding (1) and (2), we get
$I=\pi \underset{0}{\overset{2\pi }{\int }}\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$
$\Rightarrow I=2\pi \underset{0}{\overset{\pi }{\int }}\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx\text{ (Using Queen Property) }$ ....(3)
$I=2\pi \underset{0}{\overset{\pi }{\int }}\ln \left(\frac{3-\cos x}{3+\cos x}\right)dx\text{ (Using King Property) }$....(4)
$\therefore$ On adding (3) and (4), we get
$2I=2\pi \underset{0}{\overset{\pi }{\int }}\ln 1dx=0$
Hence $I=0$. Ans. $]$