Question

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

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

Answer 1

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

Given, $\int \frac{dx}{\sqrt{{\sin }^{3}x\cos x}}$

Multiply numerator and denominator by ${\sec }^{2}x$

$=\int \frac{se{c}^{2}dx}{\sqrt{{\sin }^{2}x{\sec }^{3}x}}$

$=\int \frac{{\sec }^{2}xdx}{\sqrt{{\tan }^{3}x}}$

$=\int \frac{{\sec }^{2}xdx}{(\tan x{)}^{3/2}}$

Let $z=\tan x$

$dz={\sec }^{2}xdx$

$\int \frac{dz}{z^{3/2}}=\int z^{-3/2}dz$

$=\frac{z^{-3/2+1}}{\frac{-3}{2}+1}$

$=\frac{Z^{-1/2}}{\frac{-1}{2}}=\frac{-2}{\sqrt{z}}$

$=\frac{-2}{\sqrt{\tan x}}$