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

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

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$\frac{\pi }{4}$

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(B) Let $I = \int_0^{\pi /2} \frac{1}{1 + \sqrt{	an x}} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we have: $I = \int_0^{\pi /2} \frac{1}{1 + \sqrt{	an(\pi/2 - x)}} \,dx = \int_0^{\pi /2} \frac{1}{1 + \sqrt{\cot x}} \,dx$. $I = \int_0^{\pi /2} \frac{1}{1 + \frac{1}{\sqrt{	an x}}} \,dx = \int_0^{\pi /2} \frac{\sqrt{	an x}}{\sqrt{	an x} + 1} \,dx$. Adding the two expressions for $I$: $2I = \int_0^{\pi /2} \left( \frac{1}{1 + \sqrt{	an x}} + \frac{\sqrt{	an x}}{1 + \sqrt{	an x}} ight) \,dx$. $2I = \int_0^{\pi /2} \frac{1 + \sqrt{	an x}}{1 + \sqrt{	an x}} \,dx = \int_0^{\pi /2} 1 \,dx$. $2I = [x]_0^{\pi /2} = \frac{\pi}{2}$. Therefore,$I = \frac{\pi}{4}$.

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

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