Thread: Trig substitution

$\displaystyle \int \frac{x}{\sqrt{3-x^4}}dx$

$\displaystyle \int \frac{x}{\sqrt{\sqrt{3}^2-(x^2)^2}}dx$

$\displaystyle x = \sqrt{3}sinx$

$\displaystyle dx = \sqrt{3}cosx$

$\displaystyle \sqrt{3-(3sin^2\theta)^2}$

$\displaystyle \sqrt{3-(1 - 3sin^4\theta)}$

This is where i get stuck.

Any help is appreciated

Originally Posted by silencecloak

$\displaystyle \int \frac{x}{\sqrt{3-x^4}}dx$

[snip]

If you make the substitution $\displaystyle u = x^2$ you get a standard form ....

doh

i need to develop an eye for these sorts of things still

thanks