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

Answer: (A)

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