How do you find the integral of #int t/sqrt(1-t^4)dt#?
1 Answer
Jan 15, 2017
Explanation:
#I=intt/sqrt(1-t^4)dt#
We will use the substitution
#I=1/2int(2tdt)/sqrt(1-t^4)#
#I=1/2intcostheta/sqrt(1-sin^2theta)d theta#
Using
#I=1/2intcostheta/costhetad theta#
#I=1/2intd theta#
#I=1/2theta+C#
From the original substitution
#I=1/2sin^-1(t^2)+C#