Question

If $\int \frac{dx}{\sqrt{{\sin }^{3}x\cos x}}=g\left(x\right)+C$, then $g\left(x\right)=$

• 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

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

Explanation for the correct option.

Step 1. Simplify the given integral.

The integral $\int \frac{dx}{\sqrt{{\sin }^{3}x\cos x}}$ can be simplified as:

$\begin{array}{rcl}\int \frac{dx}{\sqrt{{\sin }^{3}x\cos x}}& =& \int \frac{dx}{\sqrt{{\sin }^{4}x{\displaystyle \frac{{\sin }^{3}x\cos x}{{\sin }^{4}x}}}}\\ & =& \int \frac{dx}{{\sin }^{2}x\sqrt{{\displaystyle \frac{\cos x}{\sin x}}}}\\ & =& \int \frac{{\mathrm{cosec}}^{2}xdx}{\sqrt{\cot x}}\end{array}$

Step 2. Solve the integral using substitution.

Let $t^2=\cot x$, then $2tdt=-{\mathrm{cosec}}^{2}xdx$.

Now using the substitution $\begin{array}{rcl}& & \int \frac{{\mathrm{cosec}}^{2}xdx}{\sqrt{\cot x}}\end{array}$, the integral can be solved as

$\begin{array}{rcl}\begin{array}{l}\int \frac{{\mathrm{cosec}}^{2}xdx}{\sqrt{\cot x}}\end{array}& =& \int \frac{-2tdt}{\sqrt{t^2}}\\ & =& -2\int \frac{tdt}{t}\\ & =& -2\int dt\\ & =& -2t+C\\ & =& -2\sqrt{\cot x}+C\mathbf{[}\mathbf{\because }{\mathbf{t}}^{\mathbf{2}}\mathbf{=}\mathbf{cot}\mathbf{x}\mathbf{]}\\ & =& -\frac{2}{\sqrt{\tan x}}+C\mathbf{[}\mathbf{\because }\mathbf{cot}\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{tan}\mathbf{x}}\mathbf{]}\end{array}$

Now, as $\int \frac{dx}{\sqrt{{\sin }^{3}x\cos x}}=g\left(x\right)+C$, so the function $g\left(x\right)$ is equal to $-\frac{2}{\sqrt{\tan x}}$.

Hence, the correct option is B.