The value of _-/2^/21+^2x1+^ x,dx is

Question

The value of $\displaystyle\int_{-\pi/2}^{\pi/2}\frac{1+\sin^2x}{1+\pi^{\sin x}}\,dx$ is

Accepted Answer

Let $I=\displaystyle\int_{-\pi/2}^{\pi/2}\frac{1+\sin^2x}{1+\pi^{\sin x}}dx\ \ldots(i)$

Using $x\to-x$: $I=\displaystyle\int_{-\pi/2}^{\pi/2}\frac{1+\sin^2x}{1+\pi^{-\sin x}}dx=\int_{-\pi/2}^{\pi/2}\frac{\pi^{\sin x}(1+\sin^2x)}{1+\pi^{\sin x}}dx\ \ldots(ii)$

Adding (i) and (ii): $2I=\displaystyle\int_{-\pi/2}^{\pi/2}(1+\sin^2x)\,dx=2\int_0^{\pi/2}(1+\sin^2x)\,dx$ (even function)
$$I=\int_0^{\pi/2}\left(1+\frac{1-\cos2x}{2}\right)dx=\left[\frac{3x}{2}-\frac{\sin2x}{4}\right]_0^{\pi/2}=\frac{3\pi}{4}$$

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