The value of int_0^{pi / 2} frac{d x}{1+(tan | vedclass

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(D) Let $I = \int_0^{\pi / 2} \frac{1}{1+(	an x)^{\sqrt{2018}}} dx$  $(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \in

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$\frac{\pi}{4}$

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(D) Let $I = \int_0^{\pi / 2} \frac{1}{1+(	an x)^{\sqrt{2018}}} dx$  $(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \int_0^{\pi / 2} \frac{1}{1+(	an(\pi / 2 - x))^{\sqrt{2018}}} dx$$I = \int_0^{\pi / 2} \frac{1}{1+(\cot x)^{\sqrt{2018}}} dx$$I = \int_0^{\pi / 2} \frac{1}{1+\frac{1}{(	an x)^{\sqrt{2018}}}} dx$$I = \int_0^{\pi / 2} \frac{(	an x)^{\sqrt{2018}}}{1+(	an x)^{\sqrt{2018}}} dx$  $(ii)$Adding equations $(i)$ and $(ii)$:$2I = \int_0^{\pi / 2} \left( \frac{1}{1+(	an x)^{\sqrt{2018}}} + \frac{(	an x)^{\sqrt{2018}}}{1+(	an x)^{\sqrt{2018}}} ight) dx$$2I = \int_0^{\pi / 2} \frac{1+(	an x)^{\sqrt{2018}}}{1+(	an x)^{\sqrt{2018}}} dx$$2I = \int_0^{\pi / 2} 1 dx$$2I = [x]_0^{\pi / 2} = \frac{\pi}{2}$$I = \frac{\pi}{4}$

The value of $\int_0^{\pi / 2} \frac{d x}{1+(\tan x)^{\sqrt{2018}}}$ is equal to

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