MCQ
$\int_0^{\pi /2} {\frac{{\sqrt {\cot x} }}{{\sqrt {\cot x} + \sqrt {\tan x} }}\,dx = } $
- A$\pi $
- B$\frac{\pi }{2}$
- ✓$\frac{\pi }{4}$
- D$\frac{\pi }{3}$
$ = \int_0^{\pi /2} {\frac{{\sqrt {\cot \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\cot \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\tan \left( {\frac{\pi }{2} - x} \right)} }}dx} $
$ = \int_0^{\pi /2} {\frac{{\sqrt {\tan x} }}{{\sqrt {\tan x} + \sqrt {\cot x} }}dx} $.....$(ii)$
Now adding $(i)$ and $(ii),$ we get
$2I = \int_0^{\pi /2} {\frac{{\sqrt {\cot x} + \sqrt {\tan x} }}{{\sqrt {\tan x} + \sqrt {\cot x} }}dx = [x]_0^{\pi /2} \Rightarrow I = \frac{\pi }{4}} $.
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