Question

What's the integral of `int tanx(secx)^(3/2) dx`?

Answer 1

`inttanx(secx)^(3/2)dx = 2/3(secx)^(3/2) + C`

Explanation:

For this integral, we will use `u` substitution.
Let `u = secx`
then `du = secxtanxdx`

So we have
`inttanx(secx)^(3/2)dx = int(secx)^(1/2)*secxtanxdx=intu^(1/2)du`

Applying the formula `intx^ndx = x^(n+1)/(n+1) + C` we get

`intu^(1/2)du = u^(3/2)/(3/2)+C = 2/3u^(3/2) + C`

Finally, we substitute back in for `u`:

`2/3u^(3/2) + C = 2/3(secx)^(3/2) + C`

Thus

`inttanx(secx)^(3/2)dx = 2/3(secx)^(3/2) + C`