Question

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

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

Answer 1

Answer: (A)

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