Question

The value of the definite integral $\int\limits_0^{2 \pi} x \ln \left(\frac{3+\cos x}{3-\cos x}\right) d x$ 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=\int\limits_0^{2 \pi} x \ln \left(\frac{3+\cos x}{3-\cos x}\right) d x$ .....(1)
Also, $ I=\int\limits_0^{2 \pi}(2 \pi-x) \ln \left(\frac{3+\cos x}{3-\cos x}\right) d x$ ...(2)(Using King Property)
$\therefore $ On adding (1) and (2), we get
$ I =\pi \int\limits_0^{2 \pi} \ln \left(\frac{3+\cos x}{3-\cos x}\right) dx $
$\Rightarrow I =2 \pi \int\limits_0^\pi \ln \left(\frac{3+\cos x}{3-\cos x}\right) dx \text { (Using Queen Property) }$ ....(3)
$ I =2 \pi \int\limits_0^\pi \ln \left(\frac{3-\cos x}{3+\cos x}\right) dx \text { (Using King Property) }$....(4)
$\therefore $ On adding (3) and (4), we get
$2 I =2 \pi \int\limits_0^\pi \ln 1 dx =0$
Hence $I=0$. Ans. $]$