AP EAMCET · Maths · Definite Integration
If \(a>0\), then \(\int_{-\pi}^\pi \frac{\sin ^2 x}{1+a^x} d x\) is equal to
- A \(\frac{\pi}{2}\)
- B \(\pi\)
- C \(\frac{2 \pi}{2}\)
- D \(a \pi\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
Let \(l=\int_{-\pi}^\pi \frac{\sin ^2 x}{1+a^x} d x\) Put \(x=-x\), we get \(\begin{aligned} & I=-\int_\pi^{-\pi} \frac{\sin ^2(-x)}{1+a^{-x}} d x \\ \Rightarrow & I=\int_{-\pi}^\pi a^x \frac{\sin ^2 x}{1+a^x} d x\end{aligned}\) On adding Fqs. (i) and (ii), we get…
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