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

1 Answer
Nov 12, 2015

#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#