Question

$\int \frac{dx}{5\sin x+12\cos x}$ is equal to
(where $C$ is integration constant)

• A. $\frac{1}{13}\ln \left|\frac{4+6\tan \frac{x}{2}}{9-6\tan \frac{x}{2}}\right|+C$
• B. $\frac{1}{26}\ln \left|\frac{4-6\tan \frac{x}{2}}{9+6\tan \frac{x}{2}}\right|+C$
• C. $\frac{1}{13}\ln \left|\frac{4-6\tan \frac{x}{2}}{9+6\tan \frac{x}{2}}\right|+C$
• D. $\frac{1}{26}\ln \left|\frac{4+6\tan \frac{x}{2}}{9-6\tan \frac{x}{2}}\right|+C$

Answer 1

The correct option is A $\frac{1}{13}\ln \left|\frac{4+6\tan \frac{x}{2}}{9-6\tan \frac{x}{2}}\right|+C$

$I=\int \frac{dx}{5\sin x+12\cos x}$
Let, $\tan \frac{x}{2}=t$
$\Rightarrow x=2{\tan }^{-1}t,dx=\frac{2dt}{1+t^2},\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$
So, integral becomes
$I=\int \frac{\frac{2}{1+t^2}}{5\cdot \frac{2t}{1+t^2}+12\cdot \frac{1-t^2}{1+t^2}}dt=\int \frac{2dt}{10t+12-12t^2}=\frac{2}{12}\int \frac{dt}{1-(t^2-\frac{10}{12}t)}=\frac{1}{6}\int \frac{dt}{1+\frac{25}{144}-{(t-\frac{5}{12})}^2}=\frac{1}{6}\int \frac{dt}{{\left(\frac{13}{12}\right)}^2-{(t-\frac{5}{12})}^2}=\frac{1}{6}\cdot \frac{1}{2\cdot \frac{13}{12}}\ln \left|\frac{\frac{13}{12}+t-\frac{5}{12}}{\frac{13}{12}-t+\frac{5}{12}}\right|+C=\frac{1}{13}\ln \left|\frac{\frac{4}{6}+t}{\frac{9}{6}-t}\right|+C=\frac{1}{13}\ln \left|\frac{4+6t}{9-6t}\right|+C=\frac{1}{13}\ln \left|\frac{4+6\tan \frac{x}{2}}{9-6\tan \frac{x}{2}}\right|+C$