How do you integrate #int tan^3xsec^2x# using substitution?
1 Answer
Mar 14, 2018
Use the substitution
Explanation:
Let
#I=inttan^3xsec^2dx#
Apply the substitution
#I=intu^3du#
Integrate directly:
#I=1/4u^4+C#
Reverse the substitution:
#I=1/4tan^4x+C#