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JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}\,x}}{{1 + {2^x}}}dx} \) is
- A \( \pi \)
- B \(\frac{\pi}{2}\)
- C \(4 \pi \)
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
(d) \(I = \int\limits_{\pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \) \(......(1)\) \( \Rightarrow I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 - {2^{ - x}}}}dx} \) by replacing \(x\) by \(\left(\frac{\pi}{2}-\frac{\pi}{2}-x\right)\)…
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