Question

$\int \tan x\tan 2x\tan 3xdx=$

• A. $\frac{1}{3}\log \sec 3x-\frac{1}{2}\log \sec 2x+\log \sec x+c$
• B. $\frac{1}{3}\log \sec 3x+\frac{1}{2}\log \sec 2x-\log \sec x+c$
• C. $\frac{1}{3}\log \sec 3x-\frac{1}{2}\log \sec 2x-\log \sec x+c$
• D. $\frac{1}{3}\log \sec 3x+\frac{1}{2}\log \sec 2x+\log \sec x+c$

Answer 1

The correct option is C $\frac{1}{3}\log \sec 3x-\frac{1}{2}\log \sec 2x-\log \sec x+c$

$\int \tan x\tan 2x\tan 3xdx$

Now,

$\tan 3x=\tan (2x+x)\tan 3x=\frac{\tan 2x+\tan x}{1-\tan 2x\tan 3x}\tan 3x-\tan 2x-\tan x=\tan 3x\tan 2x\tan x$

$\int \tan 3x\tan 2x\tan xdx=\int \tan 3xdx-\int \tan 2xdx-\int \tan xdx$

We know that $\int \tan axdx=\frac{1}{a}\log \left(\sec x\right)+C$

$=\frac{1}{3}\log \left[\sec 3x\right]-\frac{1}{2}\log \left[\sec 2x\right]-\log \left[\sec x\right]+C$