Question

$\int (12\sin x+5\cos x{)}^{-1}dx$ is [$c$ is an arbitrary constant]

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

Answer 1

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

Let $\alpha ={\tan }^{-1}\frac{5}{12}$
$\Rightarrow \sin \alpha =\frac{5}{13}$ and $\cos \alpha =\frac{12}{13}$
$\Rightarrow \int \frac{dx}{12\sin x+5\cos x}=\int \frac{dx}{13\sin (x+\alpha )}$
$=\frac{1}{13}\int \frac{\frac{1}{2}{\sec }^2\frac{x+\alpha }{2}}{\tan \frac{x+\alpha }{2}}dx=\frac{1}{13}\log \left|\tan (\frac{x}{2}+\frac{\alpha }{2})\right|$
$=\frac{1}{13}\log \left|\tan (\frac{x}{2}+\frac{1}{2}{\tan }^{-1}\frac{5}{12})\right|+c$