JEE Advanced · Mathematics · 27. Definite Integration
The value of the integral \(\int_{-\pi / 2}^{\pi / 2}\left(x^{2}+\ln \frac{\pi+x}{\pi-x}\right) \cos x d x\) is
- A 0
- B \(\frac{\pi^{2}}{2}-4\)
- C \(\frac{\pi^{2}}{2}+4\)
- D \(\frac{\pi^{2}}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi^{2}}{2}-4\)
Step-by-step Solution
Detailed explanation
\(\int_{-\pi / 2}^{\pi / 2}\left[x^{2}+\ln \left(\frac{\pi+x}{\pi-x}\right)\right] \cos x d x\) \(=\int_{-\pi / 2}^{\pi / 2} x^{2} \cos x d x+\int_{-\pi / 2}^{\pi / 2} \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x d x\) \(=2 \int_{0}^{\pi / 2} x^{2} \cos x d x+0 \quad\left[\because x^{2} \cos x\right.\) is an even function and \(\ln \left(\frac{\pi+x}{\pi-x}\right) \cos x\) is an odd function] \(=2\left[x^{2} \sin x+2 x \cos x-2 \sin x\right]_{0}^{\pi / 2}=\frac{\pi^{2}}{2}-4\)
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