Question:
The value of
\(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^2 x}{1+3^x}\,dx\)
is :
The value of
\(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^2 x}{1+3^x}\,dx\)
is :
\(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^2 x}{1+3^x}\,dx\)
is :
Show Hint
Use \(\int_{-a}^{a}f(x)\,dx=\int_0^a\left[f(x)+f(-x)\right]dx\); the \(3^x\) terms combine to 1.
Updated On: Nov 4, 2026
- \(2\pi\)
- \(\frac{\pi}{2}\)
- \(4\pi\)
- \(\frac{\pi}{4}\)
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The Correct Option is D
Solution and Explanation
Let \(I=\int_{-\pi/2}^{\pi/2}\frac{\cos^2x}{1+3^x}dx\). Using \(\int_{-a}^{a}f(x)dx=\int_0^a[f(x)+f(-x)]dx\):
\(I=\int_0^{\pi/2}\cos^2x\left[\frac{1}{1+3^x}+\frac{1}{1+3^{-x}}\right]dx\).
Since \(\frac{1}{1+3^{-x}}=\frac{3^x}{1+3^x}\), the bracket equals 1.
\(I=\int_0^{\pi/2}\cos^2x\,dx=\int_0^{\pi/2}\frac{1+\cos2x}{2}dx=\frac{\pi}{4}\).
Answer: (4) \(\frac{\pi}{4}\)
\(I=\int_0^{\pi/2}\cos^2x\left[\frac{1}{1+3^x}+\frac{1}{1+3^{-x}}\right]dx\).
Since \(\frac{1}{1+3^{-x}}=\frac{3^x}{1+3^x}\), the bracket equals 1.
\(I=\int_0^{\pi/2}\cos^2x\,dx=\int_0^{\pi/2}\frac{1+\cos2x}{2}dx=\frac{\pi}{4}\).
Answer: (4) \(\frac{\pi}{4}\)
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