JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of the integral \(\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x\) is :
- A \(\frac{\pi^2}{6}\)
- B \(\frac{\pi^2}{12 \sqrt{3}}\)
- C \(\frac{\pi^2}{3 \sqrt{3}}\)
- D \(\frac{\pi^2}{6 \sqrt{3}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi^2}{6 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\(I=\int \limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x\) \(x \rightarrow-x\) \(I=\int \limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{-x+\frac{\pi}{4}}{2-\cos 2 x} d x\) \((1)\) + \((2)\)…
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