int_{-pi / 2}^{pi / 2} frac{1}{1+ e^{sin x}} | vedclass

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(D) ધારો કે $I = \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} dx$  $(1)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા:$I

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) ધારો કે $I = \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} dx$  $(1)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા:$I = \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin(-\pi / 2 + \pi / 2 - x)}} dx = \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{-\sin x}} dx$$I = \int_{-\pi / 2}^{\pi / 2} \frac{e^{\sin x}}{e^{\sin x} + 1} dx$  $(2)$સમીકરણ $(1)$ અને $(2)$ નો સરવાળો કરતા:$2I = \int_{-\pi / 2}^{\pi / 2} \left( \frac{1}{1+e^{\sin x}} + \frac{e^{\sin x}}{1+e^{\sin x}} \right) dx$$2I = \int_{-\pi / 2}^{\pi / 2} \frac{1+e^{\sin x}}{1+e^{\sin x}} dx = \int_{-\pi / 2}^{\pi / 2} 1 dx$$2I = [x]_{-\pi / 2}^{\pi / 2} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$$I = \frac{\pi}{2}$

$\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+ e^{\sin x}} dx$ ની કિંમત શોધો.

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