Question

What is $\int \tan x\tan 2x\tan 3x\mathrm{d}x$ is equal to?

Answer 1

Find $\int \tan x\tan 2x\tan 3x\mathrm{d}x$.

As

$$\begin{gathered} \tan \left(3x\right)=\tan (x+2x) \\ \Rightarrow \tan \left(3x\right)=\frac{\tan x+\tan 2x}{1-\tan x\tan 2x}\left[\because \tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}\right] \\ \Rightarrow \tan \left(3x\right)-\tan x\tan \left(2x\right)\tan \left(3x\right)=\tan x+\tan \left(2x\right) \\ \therefore \tan x\tan \left(2x\right)\tan \left(3x\right)=\tan \left(3x\right)-\tan \left(x\right)-\tan \left(2x\right) \end{gathered}$$

Now,

$$\begin{gathered} \int \tan x\tan 2x\tan 3xdx=\int \tan \left(3x\right)-\tan \left(x\right)-\tan \left(2x\right)dx \\ =\frac{1}{3}\log |\sec (3x\left)\right|-\log |\sec x|-\frac{1}{2}\log |\sec (2x\left)\right|+C \end{gathered}$$

Hence, $\int \tan x\tan 2x\tan 3xdx=\frac{1}{3}\log |\sec (3x\left)\right|-\log |\sec x|-\frac{1}{2}\log |\sec (2x\left)\right|+C$