Trig substitution

**silencecloak** · Mar 5th 2009, 07:26 PM

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

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

$x = \sqrt{3}sinx$

$dx = \sqrt{3}cosx$

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

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

This is where i get stuck.

Any help is appreciated

**mr fantastic** · Mar 5th 2009, 08:05 PM

Originally Posted by silencecloak

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

[snip]

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

**silencecloak** · Mar 5th 2009, 08:12 PM

doh

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

thanks

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