Question:
The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1+\sin ^{2} x}{1+\pi^{\sin x}}\right) d x$ is
Correct Option: , 3
Solution:
$I=\int_{0}^{\pi / 2} \frac{\left(1+\sin ^{2} x\right)}{\left(1+\pi^{\sin x}\right)}+\frac{\pi^{\sin x}\left(1+\sin ^{2} x\right)}{\left(1+\pi^{\sin x}\right)} d x$
$I=\int_{0}^{\pi / 2}\left(1+\sin ^{2} x\right) d x$
$I=\frac{\pi}{2}+\frac{\pi}{2} \cdot \frac{1}{2}=\frac{3 \pi}{4}$
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