The value of int_{-frac{pi}{2}}^{frac{pi}{2}} | vedclass

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(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^{2} x}{1+\pi^{\sin x}} \, dx$. Using the property $\int_{-a}^{a} f(x) \, dx = \int_{0}

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^{2} x}{1+\pi^{\sin x}} \, dx$. Using the property $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx$,we have: $I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+\sin^{2} x}{1+\pi^{\sin x}} + \frac{1+\sin^{2}(-x)}{1+\pi^{\sin(-x)}} \right) \, dx$ Since $\sin^{2}(-x) = \sin^{2} x$ and $\sin(-x) = -\sin x$: $I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+\sin^{2} x}{1+\pi^{\sin x}} + \frac{1+\sin^{2} x}{1+\pi^{-\sin x}} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \left( \frac{1}{1+\pi^{\sin x}} + \frac{\pi^{\sin x}}{\pi^{\sin x}+1} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \left( \frac{1+\pi^{\sin x}}{1+\pi^{\sin x}} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \, dx = \int_{0}^{\frac{\pi}{2}} (1 + \frac{1-\cos 2x}{2}) \, dx = \int_{0}^{\frac{\pi}{2}} (\frac{3}{2} - \frac{1}{2}\cos 2x) \, dx$ $I = [\frac{3}{2}x - \frac{1}{4}\sin 2x]_{0}^{\frac{\pi}{2}} = \frac{3\pi}{4} - 0 = \frac{3\pi}{4}$.

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1+\sin^{2} x}{1+\pi^{\sin x}}\right) \, dx$ is

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