int_0^{pi / 2} frac{1}{1+sqrt{tan x}} d x= | vedclass

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(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{	an x}} dx$ $\qquad ....(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:$I = \int

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$\frac{\pi}{4}$

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(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{	an x}} dx$ $\qquad ....(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:$I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{	an(\frac{\pi}{2}-x)}} dx$$I = \int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\cot x}} dx = \int_0^{\frac{\pi}{2}} \frac{\sqrt{	an x}}{1+\sqrt{	an x}} dx$ $\qquad ....(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^{\frac{\pi}{2}} \frac{1+\sqrt{	an x}}{1+\sqrt{	an x}} dx$$2I = \int_0^{\frac{\pi}{2}} 1 dx = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$$I = \frac{\pi}{4}$

$\int_0^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x=$

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