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CBSE Class 12 Math 2008 Solved Paper

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Question : 17 of 29
Marks: +1, -0
Evaluate: ∫0πxsin⁡x1+cos⁡2x\int\limits_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x} dx
Solution:  
I = ∫0π\int\limits_{0}^{\pi} xisnx1+cos⁡2x\frac{xisnx}{1+\cos^2 x} dx ... (i)
I = ∫0π\int\limits_{0}^{\pi} π−xsin⁡π−x1+cos⁡2xπ−x\frac{\pi - x\sin\pi - x}{1+\cos^2 x\pi - x} dx
I = ∫0π\int\limits_{0}^{\pi} π−xsin⁡x1+cos⁡2x\frac{\pi - x\sin x}{1+\cos^2 x} dx
I = ∫0π\int\limits_{0}^{\pi} πsin⁡x1+cos⁡2x\frac{\pi \sin x}{1+\cos^2 x} dx - ∫0π\int\limits_{0}^{\pi} xsin⁡x1+cos⁡2x\frac{x\sin x}{1+\cos^2 x} dx ... (2)
Adding (1) and (2), we get:
2I = ∫1−1π−dt1+t2\int\limits_{1}^{-1} \frac{\pi - dt}{1+t^2}
2I = - π ∫1−1\int\limits_{1}^{-1} (11+t2)\left(\frac{1}{1+t^2}\right) dt
2I = - π ∣tan⁡−1t∣1−1\left| \tan^{-1} t \right|_{1}^{-1}
2I = π [tan⁡−11−tan⁡−11\tan^{-1}1 - \tan^{-1}1 - 1]
2I = π (π4−(−π4))\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right)
2I = π22\frac{\pi^2}{2}
∴ I = π24\frac{\pi^2}{4}
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