Trigonometric Substitution Problem - Calculus 2

[khatche4]

Hey there
This is a trig substitution for my Calculus 2 class and I really have NO idea how to get started...

\int\frac{4}{\sqrt{3-2x^2}}dx

My professor has yet to go over how to evaluate trigonometric substitutions with coefficients in front of variables.

 

[Hitman2-2]

how about factoring out the 2?

 

[khatche4]

How would you factor out the two?

Because \sqrt{3-2x^2} is not the same as 2*\sqrt{\frac{3}{2}-x^2}

 

[Hitman2-2]

How would you factor out the two?

Because \sqrt{3-2x^2} is not the same as 2*\sqrt{\frac{3}{2}-x^2}

Yes, but \sqrt{3-2x^2} = \sqrt{2}*\sqrt{\frac{3}{2}-x^2}. I may not have been clear in my previous post.

 

[khatche4]

Oh! Duh! Thank you!
I'll give it a try.