Question

$\int \frac{1}{\sin x\sqrt{\sin x\cos x}}dx=$

• A. $\sqrt{\tan x}+c$
• B. $-2\sqrt{\tan x}+c$
• C. $-2\sqrt{\cot x}+c$
• D. $2\sqrt{\cot x}+c$

Answer 1

Answer: (C)

$-2\sqrt{\cot x}+c$

$\int \frac{1dx}{\sin x\sqrt{\sin x.\cos x}}$

$=\int \frac{1}{{\sin }^{2}x}\sqrt{\frac{\sin x}{\cos x}}dx$ (multiply and divide by $\sin x$)

$=\int \frac{{\text{cosec}}^{2}x}{\sqrt{\cot x}}dx$

Let $\cot x=t$

$-{\text{cosec}}^{2}x.dx=dt$

$\int \frac{-dt}{\sqrt{t}}=-2\sqrt{t}+c$

$=-2\sqrt{\cot }x+c$

$\Rightarrow \int \frac{1}{\sin x\sqrt{\sin x\cos x}}dx=-2\sqrt{\cot x}+c$