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 $theta=sin^-1(t^2)$ .

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

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