integral substitution

[jeca86]

the integral of (x/square root of (1-x^4))*dx
use x= square root of (sintheta)

I get really confused with the square roots. can someone help me set up the problem?

 

[Unco]

Letting \(\displaystyle \mbox{x^2 = sin(\theta) \Rightarrow 2x dx = cos(\theta) d\theta}\)
(and \(\displaystyle \mbox{ x^4 = \sin^2{(\theta)}}\))

\(\displaystyle \L\mbox{ \int \frac{x}{\sqrt{1 - x^4}} dx = \frac{1}{2} \int \frac{2x}{\sqrt{1 - x^4}} dx = \frac{1}{2} \int \frac{\cos{(\theta)}}{\sqrt{1 - \sin^2{(\theta})}} d\theta}\)

 

[soroban]

Hello, jeca86!

\(\displaystyle \L\int \frac{x}{\sqrt{1\,-\,x^4}}\,dx\)

Let \(\displaystyle u\,=\.x^2\;\;\Rightarrow\;\;du\,=\,2x\,dx\;\;\Rightarrow\;\;dx\,=\,\frac{du}{2x}\)

Substitute: \(\displaystyle \L\,\int\frac{x}{\sqrt{1\,-\,u^2}}\,\left(\frac{du}{2x}\right) \:= \:\frac{1}{2}\int\frac{du}{\sqrt{1\,-\,u^2}}\)

Do you recognize the arcsine form?

 

[Unco]

The instructions say to use x = sqrt(sin(theta)), Soroban. It's a certainty that if I kept my original post as you have, you would post what I now have.