Question

` ∫  tan 2x tan 3x  tan 5x    dx  `

Answer 1

\[We\ know\ that, \]
\[ \tan 5x = \tan \left( 2x + 3x \right)\]
\[ \Rightarrow \tan 5x = \frac{\tan 2x + \tan 3x}{1 - \tan 2x \tan 3x}\]
\[ \Rightarrow \tan 5x - \tan 2x \tan 3x \tan 5x = \tan 2x + \tan 3x\]
\[ \Rightarrow \tan 2x \tan 3x \tan 5x = \tan 5x - \tan 2x - \tan 3x\]
\[ \therefore \int\tan 2x \tan 3x \tan 5x = \int\left( \tan 5x - \tan 2x - \tan 3x \right)dx\]
\[ = \frac{1}{5} \ln \left| \sec 5x \right| - \frac{1}{2} \ln \left| \sec 2x \right| - \frac{1}{3} \ln \left| \sec 3x \right| + C\]