The value of the integral int_{-pi / 2}^{pi / | vedclass

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

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$\frac{\pi}{4}$

Vedclass · Answer

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

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

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