# Trigonometric Substitution Problem - Calculus 2

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?

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.

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