Question:
The value of \(\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2 x}{1+2^x}\,dx\) is :
The value of \(\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2 x}{1+2^x}\,dx\) is :
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Use \(\int_{-a}^{a}f(x)\,dx=\int_{-a}^{a}f(-x)\,dx\) and add the two forms.
Updated On: Oct 31, 2026
- \(\pi\)
- \(\frac{\pi}{2}\)
- \(4\pi\)
- \(\frac{\pi}{4}\)
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The Correct Option is D
Solution and Explanation
Let \(I=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2x}{1+2^x}dx\). Replacing \(x\to -x\): \(I=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2x\cdot 2^x}{1+2^x}dx\).
Adding: \(2I=\displaystyle\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=2\int_0^{\pi/2}\sin^2x\,dx=2\cdot\dfrac{\pi}{4}=\dfrac{\pi}{2}\).
Therefore \(I=\dfrac{\pi}{4}\).
Answer: (4) \(\frac{\pi}{4}\)
Adding: \(2I=\displaystyle\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=2\int_0^{\pi/2}\sin^2x\,dx=2\cdot\dfrac{\pi}{4}=\dfrac{\pi}{2}\).
Therefore \(I=\dfrac{\pi}{4}\).
Answer: (4) \(\frac{\pi}{4}\)
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