Question

Evaluate the given trigonometric integral $\int{\tan x\tan 2x\tan 3xdx}$.

Answer 1

Hint: Write $\tan 3x=\tan \left( x+2x \right)$and use the formula $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$to get $\tan 3x$in terms of $\tan x\ and\ \tan 2x$.

Complete step-by-step solution -
We know $\int{\tan x=\log \left| \sec x \right|}$.
Use this and compute further.
We have to find $\int{\tan x\tan 2x\tan 3xdx}$.
Let us assume $I=\int{\tan x\tan 2x\tan 3xdx}$.
We can write $\tan \left( 3x \right)=\tan \left( x+2x \right)$.
We know, $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.
$\begin{aligned} & \Rightarrow \tan 3x=\tan \left( x+2x \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\\ & \Rightarrow \tan 3x=\dfrac{\tan x+\tan 2x}{1-\tan x\tan 2x} \\\ \end{aligned}$
On multiplying both sides by $\left( 1-\tan x\tan 2x \right)$, we will get;
$\begin{aligned} & \Rightarrow \tan 3x\left( 1-\tan x\tan 2x \right)=\dfrac{\left( \tan x+\tan 2x \right)}{\left( 1-\tan x\tan 2x \right)}\times \left( 1-\tan x\tan 2x \right) \\\ & \Rightarrow \tan 3x\left( 1-\tan x\tan 2x \right)=\tan x+\tan 2x \\\ & \Rightarrow \tan 3x-\tan x\tan 2x\tan 3x=\tan x+\tan 2x \\\ \end{aligned}$
Taking $\tan 3x$ to RHS, we will get;
$\Rightarrow -\tan x\tan 2x\tan 3x=\tan x+\tan 2x-\tan 3x$
Multiplying both sides of equation by (-1), we will get;
$\tan x\tan 2x\tan 3x=\tan 3x-\tan x-\tan 2x$
On putting this value of $\tan x\tan 2x\tan 3x$ in I, we will get,
$I=\int{\left( \tan 3x-\tan x-\tan 2x \right)}dx$
We know,

& \int{\left( f\left( x \right)+g\left( x \right)+h\left( x \right) \right)dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx+\int{h\left( x \right)dx}}}} \\\ & \Rightarrow I=\int{\tan 3xdx-}\int{\tan xdx-}\int{\tan 2xdx} \\\ \end{aligned}$$ We know, $$\begin{aligned} & \int{\tan \left( ax \right)dx=\dfrac{1}{a}}\log \left| \sec \left( ax \right) \right|+C \\\ & \Rightarrow I=\dfrac{1}{3}\log \left| \sec 3x \right|-1.\log \left| \sec x \right|-\dfrac{1}{2}\left| \sec 2x \right|+C \\\ \end{aligned}$$ Where ‘C’ is a constant of integration. Note: This is an indefinite integral, so don’t forget to add a constant of integration at last and also don’t forget to take the modulus of $\sec \left( ax \right)$ as the log of a number is defined only when the number is positive.