Question

The value of $\int \frac{\cot x}{\sqrt{5+9{\cot }^{2}x}}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{1}{2}{\sin }^{-1}\left(\frac{2\sin x}{3}\right)+C$
• B. $\frac{1}{2}{\sin }^{-1}\left(\frac{3\sin x}{2}\right)+C$
• C. $\frac{1}{3}{\sin }^{-1}\left(\frac{3\sin x}{2}\right)+C$
• D. $\frac{1}{3}{\sin }^{-1}\left(\frac{2\sin x}{3}\right)+C$

Answer 1

The correct option is A $\frac{1}{2}{\sin }^{-1}\left(\frac{2\sin x}{3}\right)+C$

$I=\int \frac{\cot x}{\sqrt{5+9{\cot }^{2}x}}dx$
$=\int \frac{\cos x}{\sqrt{5{\sin }^{2}x+9{\cos }^{2}x}}dx$
$=\int \frac{\cos x}{\sqrt{9-4{\sin }^{2}x}}dx$
Put $\sin x=t\Rightarrow \left(\cos x\right)dx=dt$
$\therefore I=\int \frac{1}{\sqrt{9-4t^2}}dt$
$=\frac{1}{2}\int \frac{1}{\sqrt{\frac{9}{4}-t^2}}dt$
$=\frac{1}{2}{\sin }^{-1}\left(\frac{2t}{3}\right)+C$
$=\frac{1}{2}{\sin }^{-1}\left(\frac{2\sin x}{3}\right)+C$