How do you find the integral of #int t/sqrt(1-t^4)dt#?

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Answer 1 · 2017-01-15T20:35:33

#1/2sin^-1(t^2)+C#

#I=intt/sqrt(1-t^4)dt#

We will use the substitution #t^2=sintheta#. Thus #2tdt=costhetad theta# and #t^4=sin^2theta#. Then:

#I=1/2int(2tdt)/sqrt(1-t^4)#

#I=1/2intcostheta/sqrt(1-sin^2theta)d theta#

Using #sin^2theta+cos^2theta=1# we see that #sqrt(1-sin^2theta)=costheta#.

#I=1/2intcostheta/costhetad theta#

#I=1/2intd theta#

#I=1/2theta+C#

From the original substitution #t^2=sintheta# we see that #theta=sin^-1(t^2)#.

#I=1/2sin^-1(t^2)+C#