Question

The value of $\int \frac{\cot x}{\sqrt{5+9 \cot ^2 x}} d x$ 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)

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