Question

Question #7e566

Answer 1

`tan^3x/3+tanx+C`

Explanation:

When integrating powers of tanx and secx it makes it easier to use suitable substitutions.

• Firstly note that:
  `tan^2x+1=sec^2x`

`int(tan^2x+1)^2dx=int(sec^2x)^2dx`

`=intsec^4x dx=intsec^2x(tan^2x+1)dx`

• Now we can use a substitution to solve further

Let `u=tan(x)`

`du=sec^2xdx`

`intsec^2x(tan^2x+1)dx=int(u^2+1)du`

`=u^3/3+u+C`

sub u back in which gives us our answer

`tan^3x/3+tanx+C`

Hope this helped, good luck :)