Question

The value of $\int \left(\tan 3x\tan 6x\tan 9x\right)dx$ is
(where $C$ is constant of integration)

• A. $\frac{\ln |\sec 9x|}{9}-\frac{\ln |\sec 3x|}{3}+C$
• B. $\frac{\ln |\sec 6x|}{6}-\frac{\ln |\sec 3x|}{3}+C$
• C. $\frac{\ln |\sec 9x|}{9}-\frac{\ln |\sec 6x|}{6}+C$
• D. $\frac{\ln |\sec 9x|}{9}-\frac{\ln |\sec 6x|}{6}-\frac{\ln |\sec 3x|}{3}+C$

Answer 1

The correct option is D $\frac{\ln |\sec 9x|}{9}-\frac{\ln |\sec 6x|}{6}-\frac{\ln |\sec 3x|}{3}+C$

We know,
$9x=6x+3x$
Taking $\tan$ on both sides, we get

$\Rightarrow \tan 9x=\frac{\tan 6x+\tan 3x}{1-\tan 6x\tan 3x}$

$\Rightarrow \tan 9x-\tan 9x\tan 6x\tan 3x=\tan 6x+\tan 3x$

$\therefore \tan 9x-\tan 6x-\tan 3x=\tan 3x\cdot \tan 6x\cdot \tan 9x$

$\Rightarrow I=\int (\tan 9x-\tan 6x-\tan 3x)dx$

$\Rightarrow I=\frac{\ln |\sec 9x|}{9}-\frac{\ln |\sec 6x|}{6}-\frac{\ln |\sec 3x|}{3}+C$