Question & Answer
QUESTION

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

ANSWER Verified Verified
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, $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.
$\begin{align}
  & \Rightarrow \tan 3x=\tan \left( x+2x \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\
 & \Rightarrow \tan 3x=\dfrac{\tan x+\tan 2x}{1-\tan x\tan 2x} \\
\end{align}$
On multiplying both sides by $\left( 1-\tan x\tan 2x \right)$, we will get;
$\begin{align}
  & \Rightarrow \tan 3x\left( 1-\tan x\tan 2x \right)=\dfrac{\left( \tan x+\tan 2x \right)}{\left( 1-\tan x\tan 2x \right)}\times \left( 1-\tan x\tan 2x \right) \\
 & \Rightarrow \tan 3x\left( 1-\tan x\tan 2x \right)=\tan x+\tan 2x \\
 & \Rightarrow \tan 3x-\tan x\tan 2x\tan 3x=\tan x+\tan 2x \\
\end{align}$
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,
\[\begin{align}
  & \int{\left( f\left( x \right)+g\left( x \right)+h\left( x \right) \right)dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx+\int{h\left( x \right)dx}}}} \\
 & \Rightarrow I=\int{\tan 3xdx-}\int{\tan xdx-}\int{\tan 2xdx} \\
\end{align}\]
We know,
\[\begin{align}
  & \int{\tan \left( ax \right)dx=\dfrac{1}{a}}\log \left| \sec \left( ax \right) \right|+C \\
 & \Rightarrow I=\dfrac{1}{3}\log \left| \sec 3x \right|-1.\log \left| \sec x \right|-\dfrac{1}{2}\left| \sec 2x \right|+C \\
\end{align}\]
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.