Question

How do you integrate `int sqrttanxsec^2xdx`?

Answer 1

Let `u = tan(x)`, then `du = sec^2(x)dx`. Please see the explanation.

Explanation:

Let `u = tan(x)`, then `du = sec^2(x)dx`

`intsqrt(tan(t))sec^2dx = int(u^(1/2))du`

Integrate:

`intsqrt(tan(t))sec^2dx = 2/3u^(3/2) + C`

Reverse the substitution:

`intsqrt(tan(t))sec^2dx = 2/3(tan(x))^(3/2) + C`