Question

Evaluate the following integral:
$\int {\tan x.\tan 2x.\tan 3xdx} $
${\text{A}}{\text{. - }}\dfrac{1}{3}\ln |\cos (3x)| + \dfrac{1}{2}\ln |\cos (2x)| + \ln |\cos (x)| + C$
${\text{B}}{\text{. - }}\ln |\sec x| - \dfrac{1}{2}\ln |\sec (2x)| + \dfrac{1}{2}\ln |\sec 3x| + C$
${\text{C}}{\text{. }}\ln |\cos (x)| + \ln |\sec (2x)| + \ln |\cos (3x)| + C$
${\text{D}}{\text{. - }}\dfrac{1}{3}\ln |\sec (3x)| + \dfrac{1}{2}\ln |\sec (2x)| + \ln |\sec (x)| + C$

Answer 1

Hint- Use the formula, $\tan (3x) = \dfrac{{\tan x + \tan 2x}}{{1 - \tan x.\tan 2x}}$ to evaluate the given integral. Try to convert the given terms using this formula and then solve.

Complete step-by-step answer:
We have been given in the question to evaluate the integral, \[\int {\tan x.\tan 2x.\tan 3xdx} \].
Now we know that $\tan (3x) = \dfrac{{\tan x + \tan 2x}}{{1 - \tan x.\tan 2x}}$.
Using the above formula, we can write-
$\tan (3x).(1 - \tan x.\tan 2x) = \tan x + \tan 2x$
Solving it further, we get-
$
  \tan (3x) - \tan x.\tan (2x).\tan (3x) = \tan x + \tan 2x \\
  \therefore \tan x.\tan (2x).\tan (3x) = \tan (3x) - \tan (x) - \tan (2x) \to (1) \\
$
Now, the given integral is, \[\int {\tan x.\tan 2x.\tan 3xdx} \]
Using the equation (1), we can write the integral as-
$\int {\tan (3x) - \tan (x) - \tan (2x).dx} $
Let $I = \int {\tan (3x) - \tan (x) - \tan (2x).dx} $
Now, as we know $\int {\tan (x).dx = \ln |\sec x|} $, using this result to evaluate the above integral.
\[
  I = \int {\tan (3x) - \tan (x) - \tan (2x).dx} \\
   \Rightarrow I = \int {\tan (3x).dx - } \int {\tan (x).dx} - \int {\tan (2x).dx} \\
   \Rightarrow I = \dfrac{1}{3}\ln |\sec (3x)| - \ln |\sec x| - \dfrac{1}{2}\ln |\sec (2x)| + C \\
 \]
Now we can write,
\[ = \dfrac{1}{3}\ln |\dfrac{1}{{\cos (3x)}}| - \ln |\dfrac{1}{{\cos x}}| - \dfrac{1}{2}\ln |\dfrac{1}{{\cos (2x)}}| + C\]
Also, we know- $\ln \dfrac{a}{b} = \ln a - \ln b$ using this we get-
\[
  I = \dfrac{1}{3}(\ln (1) - \ln |\cos (3x|) - (\ln (1) - \ln |\cos x|) - \dfrac{1}{2}(\ln (1) - \ln |\cos (2x)| + C \\
  \because \ln (1) = 0 \\
   \Rightarrow I = - \dfrac{1}{3}(\ln |\cos (3x)|) + \ln |\cos x| + \dfrac{1}{2}(\ln |\cos (2x)|) + C \\
 \]
Therefore, the above integral evaluates to-\[\int {\tan x.\tan 2x.\tan 3xdx} = - \dfrac{1}{3}\ln |\cos (3x)| + \dfrac{1}{2}\ln |\cos (2x)| + \ln |\cos (x)| + C\]
Hence, the correct option is option ${\text{A}}{\text{. - }}\dfrac{1}{3}\ln |\cos (3x)| + \dfrac{1}{2}\ln |\cos (2x)| + \ln |\cos (x)| + C$.

Note- Whenever such types of questions appear, try solving step by step. As mentioned in the solution, first write the standard formula of $\tan (3x)$, and then try to express $\tan x.\tan 2x.\tan 3x$in terms of $\tan (3x) - \tan (x) - \tan (2x)$, which will be easier to integrate. And then evaluate the integral by using the formula $\int {\tan (x).dx = \ln |\sec x|} $.