int_{-frac{pi}{2}}^{frac{pi}{2}} frac{dx}{e^ | vedclass

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

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$ \frac{\pi}{2} $

Vedclass · Answer

(D) Let $ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin x}+1} $. 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}{e^{\sin x}+1} + \frac{1}{e^{\sin(-x)}+1} \right) dx $ $ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x}+1} + \frac{1}{e^{-\sin x}+1} \right) dx $ $ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x}+1} + \frac{e^{\sin x}}{1+e^{\sin x}} \right) dx $ $ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+e^{\sin x}}{1+e^{\sin x}} \right) dx $ $ I = \int_{0}^{\frac{\pi}{2}} 1 dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} $.

$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin x}+1} $ is equal to

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