# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \tan 2 x \tan 3 x \tan 5 x d x$

Solution:

We know $\tan 5 x=\tan (2 x+3 x)$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$\therefore \tan (2 x+3 x)=\frac{\tan 2 x+\tan 3 x}{1-\tan 2 x \tan 3 x}$

$\therefore \tan (5 x)=\frac{\tan 2 x+\tan 3 x}{1-\tan 2 x \tan 3 x}$

$\Rightarrow \tan (5 x)(1-\tan 2 x \cdot \tan 3 x)=\tan (2 x)+\tan (3 x)$

$\Rightarrow \tan (5 x)-\tan 2 x \cdot \tan 3 x \cdot \tan 5 x=\tan (2 x)+\tan (3 x)$

$\Rightarrow \tan (5 x)-\tan (2 x)-\tan (3 x)=\tan 2 x \cdot \tan 3 x \cdot \tan 5 x$

Substituting the above result in given equation we get

$\Rightarrow \int \tan 5 \mathrm{x}-\tan 3 \mathrm{x}-\tan 2 \mathrm{x} \mathrm{dx}$

$\Rightarrow \int \tan 5 \mathrm{x} \mathrm{dx}-\int \tan 3 \mathrm{x} \mathrm{dx}-\int \tan 2 \mathrm{x} \mathrm{dx}$

$\Rightarrow \frac{-1}{5} \ln |\cos 5 x|-\frac{(-1)}{3} \ln |\cos 3 x|-\frac{(-1)}{2} \ln |\cos 2 x|+c .$

$\Rightarrow \frac{-1}{5} \ln |\cos 5 x|+\frac{1}{3} \ln |\cos 3 x|+\frac{1}{2} \ln |\cos 2 x|+c .$