JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1+\sin ^{2} \mathrm{x}}{1+\pi^{\sin \mathrm{x}}}\right)\, \mathrm{dx}\) is
- A \(\frac{\pi}{2}\)
- B \(\frac{5 \pi}{4}\)
- C \(\frac{3 \pi}{4}\)
- D \(\frac{3 \pi}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \pi}{4}\)
Step-by-step Solution
Detailed explanation
\(I=\int_{0}^{\pi / 2} \frac{\left(1+\sin ^{2} x\right)}{\left(1+\pi^{\sin x}\right)}+\frac{\pi^{\sin x}\left(1+\sin ^{2} x\right)}{\left(1+\pi^{\sin x}\right)} \,d x\) \(I=\int_{0}^{\pi / 2}\left(1+\sin ^{2} x\right) \,d x\)…
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