Question

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

• A. $\frac{1}{13}\ln \left|\tan (\frac{x}{2}+\frac{1}{2}{\tan }^{-1}\frac{5}{12})\right|+C$
• B. $-\frac{1}{13}\ln \left|\tan (\frac{x}{2}+\frac{1}{2}{\tan }^{-1}\frac{5}{12})\right|+C$
• C. $\frac{1}{13}\ln \left|\tan (\frac{x}{2}-\frac{1}{2}{\tan }^{-1}\frac{12}{5})\right|+C$
• D. $-\frac{1}{13}\ln \left|\tan (\frac{x}{2}-\frac{1}{2}{\tan }^{-1}\frac{12}{5})\right|+C$

Answer 1

The correct option is A $\frac{1}{13}\ln \left|\tan (\frac{x}{2}+\frac{1}{2}{\tan }^{-1}\frac{5}{12})\right|+C$

Let $I=\int \frac{dx}{12\sin x+5\cos x}$
$\Rightarrow I=\frac{1}{13}\int \frac{dx}{\frac{12}{13}\sin x+\frac{5}{13}\cos x}$
$\cos \theta =\frac{12}{13},\sin \theta =\frac{5}{13},\tan \theta =\frac{5}{12}$

$\Rightarrow I=\frac{1}{13}\int \frac{dx}{\sin x\cos \theta +\cos x\sin \theta }$

$\Rightarrow I=\frac{1}{13}\int \mathrm{cosec}(x+\theta )dx$
$\Rightarrow I=\frac{1}{13}\ln \left|\tan (\frac{x}{2}+\frac{\theta }{2})\right|+C$

$\Rightarrow I=\frac{1}{13}\ln \left|\tan (\frac{x}{2}+\frac{1}{2}{\tan }^{-1}\frac{5}{12})\right|+C$
where, $\theta ={\tan }^{-1}\frac{5}{12}$