Evaluate: `inttan^3xdx`

Explanation

Here `∫tan^3xdx`

=`∫tanx.tan^2xdx=∫tanx(sec^2x−1)dx`

=`∫tanx.sec^2xdx−∫tanxdx`


In the first integral, put u=tanx so that

`du/dx=sec^2x`

=`∫u.du−log∣secx∣+c=u^2/2​−log∣secx∣+c`

=`tan^2x/2​−log∣secx∣+c`