Question

$\int \frac{dx}{\sqrt{{\sin }^{3}x\cos x}}=g\left(x\right)+c\Rightarrow g\left(x\right)=$

• A. $\frac{-2}{\sqrt{\tan x}}$
• B. $\frac{2}{\sqrt{\cot x}}$
• C. $\frac{2}{\sqrt{\tan x}}$
• D. $\frac{1}{\sqrt{\cot x}}$

Answer 1

The correct option is A $\frac{-2}{\sqrt{\tan x}}$

Solution:

$I=\int \frac{1}{\sqrt{{\sin }^{3}x\cos x}}dx$

$=\int \frac{1}{\frac{\sqrt{{\sin }^{3}x}}{\sqrt{{\cos }^{3}x}}\times \sqrt{{\cos }^{3}x}\times \sqrt{\cos x}}dx$

$=\int \frac{{\sec }^{2}x.dx}{\sqrt{{\tan }^{3}x}}$ Let $\tan x=t$

$\Rightarrow {\sec }^{2}xdx=dt$

$=\int \frac{dt}{t^{3/2}}$

$=-2t^{-1/2}+C$

$=\frac{-2}{\sqrt{\tan x}}+C=g\left(x\right)+C$

$\Rightarrow g\left(x\right)=\frac{-2}{\sqrt{\tan x}}$