Question

The value of $\int \frac{1}{3\sin x-\cos x+3}dx$ is

• A. $\log \left(\frac{\tan \frac{x}{2}+1}{2\tan \frac{x}{2}+1}\right)+C$
• B. $\frac{1}{2}\log \left(\frac{2\tan \frac{x}{2}+1}{\tan \frac{x}{2}+1}\right)+C$
• C. $\log \left(\frac{2\tan \frac{x}{2}+1}{\tan \frac{x}{2}+1}\right)+C$
• D. $2\log \left(\frac{2\tan \frac{x}{2}+1}{\tan \frac{x}{2}+1}\right)+C$

Answer 1

Answer: (C)

$\log \left(\frac{2\tan \frac{x}{2}+1}{\tan \frac{x}{2}+1}\right)+C$

Let

$I=\int \frac{dx}{3\sin x-\cos x+3}$

$\{\because \sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}and\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\}$

$I=\int \frac{dx}{3\left\{\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right\}-\left\{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right\}+3}$

$=\int \frac{(1+{\tan }^2\frac{x}{2})dx}{6\tan \frac{x}{2}-1+{\tan }^2\frac{x}{2}+3+3{\tan }^2\frac{x}{2}}$

$=\int \frac{{\sec }^2\frac{x}{2}}{4{\tan }^2\frac{x}{2}+6\tan \frac{x}{2}+2}dx$

(Let $t=\tan \frac{x}{2},dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx)$
$=\int \frac{dt}{2t^2+3t+1}$

$=\int \frac{dt}{(t+1)(2t+1)}$

$=\int \{\frac{-1}{(t+1)}+\frac{2}{(2t+1)}\}dt$ (by partial fraction)

$=-\log (t+1)+\frac{2}{2}\log (2t+1)+C$

$=\log \frac{(2t+1)}{(t+1)}+C$

$=\log \left(\frac{2\tan \frac{x}{2}+1}{\tan \frac{x}{2}+1}\right)+C$