Prove that: _0^ / 2 d x (1+ x) = 4 - Vidyadip

Question

Prove that: $\int_0^{\pi / 2} \frac{d x}{(1+\sqrt{\tan x})}=\frac{\pi}{4}$

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Answer

$\text { Let } y=\int_0^{\frac{\pi}{2}} \frac{d x}{(1+\sqrt{\tan x})}$
$y=\int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\frac{\sin x}{\cos x}} d x}$
$y=\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x}+\sqrt{\cos x})} ....(i)$
Using theorem of definite integral
$\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$y=\int_0^{\pi / 2} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\left(\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}\right)} d x$
$y=\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{(\sqrt{\cos x}+\sqrt{\sin x})} d x .....(ii)$
Adding eq $(i)$ and eq $(ii),$ we get
$2 y=\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x}+\sqrt{\cos x})} d x+\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{(\sqrt{\cos x}+\sqrt{\sin x})} d x$
$2 y=\int_0^{\pi / 2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{(\sqrt{\sin x}+\sqrt{\cos x})} d x$
$2 y=\int_0^{\pi / 2} 1 d x$
$2 y=(x)_0^{\frac{\pi}{2}}$
$y=\frac{\pi}{4}$

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