The value of the integral int_{0}^{pi / 2} fr | vedclass

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(D) Let $I = \int_{0}^{\pi / 2} \frac{1}{1+(	an x)^{-101}} dx$.Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, we get:$I = \int_{

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$\frac{\pi}{4}$

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(D) Let $I = \int_{0}^{\pi / 2} \frac{1}{1+(	an x)^{-101}} dx$.Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, we get:$I = \int_{0}^{\pi / 2} \frac{1}{1+(	an(\frac{\pi}{2}-x))^{-101}} dx$$I = \int_{0}^{\pi / 2} \frac{1}{1+(\cot x)^{-101}} dx$$I = \int_{0}^{\pi / 2} \frac{1}{1+(\frac{1}{	an x})^{-101}} dx = \int_{0}^{\pi / 2} \frac{1}{1+(	an x)^{101}} dx$.Adding the two expressions for $I$:$2I = \int_{0}^{\pi / 2} (\frac{1}{1+(	an x)^{-101}} + \frac{1}{1+(	an x)^{101}}) dx$$2I = \int_{0}^{\pi / 2} (\frac{(	an x)^{101}}{(	an x)^{101}+1} + \frac{1}{1+(	an x)^{101}}) dx$$2I = \int_{0}^{\pi / 2} \frac{1+(	an x)^{101}}{1+(	an x)^{101}} dx = \int_{0}^{\pi / 2} 1 dx = \frac{\pi}{2}$.Therefore, $I = \frac{\pi}{4}$.

The value of the integral $\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x$ is equal to

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