Question:
The value of \(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{1}{1+e^{\sin x}}\,dx\) is:
The value of \(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{1}{1+e^{\sin x}}\,dx\) is:
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Use \(\int_{-a}^{a}f(x)dx=\int_{-a}^{a}f(-x)dx\) and add the two forms of the integral.
Updated On: Nov 5, 2026
- \(\pi\)
- \(\frac{3\pi}{2}\)
- \(\frac{\pi}{2}\)
- \(\frac{\pi}{4}\)
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The Correct Option is C
Solution and Explanation
Let \(I=\int_{-\pi/2}^{\pi/2}\frac{dx}{1+e^{\sin x}}\).
Replacing \(x\) by \(-x\): \(I=\int_{-\pi/2}^{\pi/2}\frac{dx}{1+e^{-\sin x}}=\int_{-\pi/2}^{\pi/2}\frac{e^{\sin x}}{1+e^{\sin x}}dx\).
Adding: \(2I=\int_{-\pi/2}^{\pi/2}\frac{1+e^{\sin x}}{1+e^{\sin x}}dx=\int_{-\pi/2}^{\pi/2}dx=\pi\).
So \(I=\frac{\pi}{2}\).
Answer: (3) \(\frac{\pi}{2}\)
Replacing \(x\) by \(-x\): \(I=\int_{-\pi/2}^{\pi/2}\frac{dx}{1+e^{-\sin x}}=\int_{-\pi/2}^{\pi/2}\frac{e^{\sin x}}{1+e^{\sin x}}dx\).
Adding: \(2I=\int_{-\pi/2}^{\pi/2}\frac{1+e^{\sin x}}{1+e^{\sin x}}dx=\int_{-\pi/2}^{\pi/2}dx=\pi\).
So \(I=\frac{\pi}{2}\).
Answer: (3) \(\frac{\pi}{2}\)
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