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∫ x·ln((3+cosx)/(3−cosx)) dx, 0 to 2π | JEE

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Question
∫02πx ln⁡ ⁣(3+cos⁡x3−cos⁡x)dx\displaystyle\int_{0}^{2\pi}x\,\ln\!\left(\dfrac{3+\cos x}{3-\cos x}\right)dx equals
Aπ2ln⁡3\dfrac{\pi}{2}\ln 3
Bπ6ln⁡3\dfrac{\pi}{6}\ln 3
Cπ12ln⁡3\dfrac{\pi}{12}\ln 3
D00correct
Solution
Step 1: Let g(x)=ln⁡ ⁣(3+cos⁡x3−cos⁡x)g(x)=\ln\!\left(\dfrac{3+\cos x}{3-\cos x}\right) and apply ∫02πx g(x) dx=∫02π(2π−x) g(2π−x) dx\displaystyle\int_{0}^{2\pi}x\,g(x)\,dx=\int_{0}^{2\pi}(2\pi-x)\,g(2\pi-x)\,dx. Since cos⁡(2π−x)=cos⁡x\cos(2\pi-x)=\cos x, g(2π−x)=g(x)g(2\pi-x)=g(x), giving
I=∫02π(2π−x) g(x) dx ⇒ 2I=2π∫02πg(x) dx ⇒ I=π∫02πg(x) dx.I=\int_{0}^{2\pi}(2\pi-x)\,g(x)\,dx\ \Rightarrow\ 2I=2\pi\int_{0}^{2\pi}g(x)\,dx\ \Rightarrow\ I=\pi\int_{0}^{2\pi}g(x)\,dx.
 Step 2: Reduce the remaining integral over [0,2π][0,2\pi] to [0,π][0,\pi]. Because cos⁡(2π−x)=cos⁡x\cos(2\pi-x)=\cos x, the function gg is symmetric about x=πx=\pi, so ∫02πg(x) dx=2∫0πg(x) dx\displaystyle\int_{0}^{2\pi}g(x)\,dx=2\int_{0}^{\pi}g(x)\,dx, and
I=2π∫0πln⁡ ⁣(3+cos⁡x3−cos⁡x)dx.I=2\pi\int_{0}^{\pi}\ln\!\left(\dfrac{3+\cos x}{3-\cos x}\right)dx.
 Step 3: Apply x↦π−xx\mapsto\pi-x, under which cos⁡x→−cos⁡x\cos x\to-\cos x, so the ratio inverts:
∫0πln⁡ ⁣(3+cos⁡x3−cos⁡x)dx=∫0πln⁡ ⁣(3−cos⁡x3+cos⁡x)dx=−∫0πln⁡ ⁣(3+cos⁡x3−cos⁡x)dx.\int_{0}^{\pi}\ln\!\left(\dfrac{3+\cos x}{3-\cos x}\right)dx=\int_{0}^{\pi}\ln\!\left(\dfrac{3-\cos x}{3+\cos x}\right)dx=-\int_{0}^{\pi}\ln\!\left(\dfrac{3+\cos x}{3-\cos x}\right)dx.
 Step 4: The integral equals its own negative, hence ∫0πln⁡ ⁣(3+cos⁡x3−cos⁡x)dx=0\displaystyle\int_{0}^{\pi}\ln\!\left(\dfrac{3+\cos x}{3-\cos x}\right)dx=0, and therefore I=2π⋅0=0.I=2\pi\cdot 0=0. Correct answer: (4)
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