Question

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

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

Answer 1

Answer: (A)

$\int \frac{cotx}{\sqrt{5+9co{t}^{2}x}}dx$
$=\int \frac{cosx}{\sqrt{5si{n}^{2}x+9co{s}^{2}x}}dx$
$=\int \frac{cosx}{\sqrt{5+4co{s}^{2}x}}dx=\int \frac{cosx}{\sqrt{9-4si{n}^{2}x}}dx$
Put $sinx=t$
$\int \frac{dt}{\sqrt{9-4t^2}}=\frac{1}{2}\int \frac{dt}{\sqrt{\frac{9}{4}-t^2}}=\frac{1}{2}{\left(sin\right)}^{-1}\left(\frac{2t}{3}\right)$
$+C=\frac{1}{2}{\left(sin\right)}^{-1}\left(\frac{2sinx}{3}\right)+C$ .