Question

If $\int \frac{\sqrt{\cot x}}{\sin x\cos x}dx=P\sqrt{\cot x}+Q$ , then the value of $P$ is

• A. $-2$
• B. $2$
• C. $3$
• D. $-3$

Answer 1

The correct option is A

$-2$

Explanation for the correct option:

Step 1: Simplify the given function

$\begin{array}{c}\int \frac{\sqrt{\cot x}}{\sin x\cos x}dx=\int \frac{\sqrt{\cot x}}{\frac{\sin x}{\cos x}\cos x\cdot \cos x}dx\\ =\int \frac{\sqrt{\cot x}}{\tan x\cdot {\cos }^{2}x}dx\\ =\int \frac{\sqrt{\cot x}}{\tan x}\cdot {\sec }^{2}xdx\\ =\int \frac{1}{{\left(\tan x\right)}^{\frac{3}{2}}}\cdot {\sec }^{2}xdx............\left(i\right)\end{array}$

Step 2:Solve by making suitable substitution

Let $\tan x=m$,

$\begin{array}{l}\tan x=m\\ \Rightarrow {\sec }^{2}xdx=dm\end{array}$

Substitute in the above equation (i) we get,

$\begin{array}{c}\int \frac{\sqrt{\cot x}}{\sin x\cos x}dx=\int \frac{1}{{\left(m\right)}^{\frac{3}{2}}}\cdot dm\\ =\int m^{-\frac{3}{2}}\cdot dm\\ =\frac{-2}{\sqrt{m}}+c\end{array}$

Put $m=\tan x$, then

$\begin{array}{c}\int \frac{\sqrt{\cot x}}{\sin x\cos x}dx=\frac{-2}{\sqrt{\tan x}}+C\\ =-2\sqrt{\cot x}+C\end{array}$

Step 3: Compare with the given relation, then

$\begin{array}{c}\therefore P\sqrt{\cot x}+Q=-2\sqrt{\cot x}+C\\ \Rightarrow P=-2\end{array}$

Hence, the correct option is (A).