Trigonometric Substitution Problem - Calculus 2

  • Thread starter khatche4
  • Start date
  • #1
22
0
Hey there
This is a trig substitution for my Calculus 2 class and I really have NO idea how to get started...

[tex]\int\frac{4}{\sqrt{3-2x^2}}dx[/tex]

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

Answers and Replies

  • #2
Hitman2-2
how about factoring out the 2?
 
  • #3
22
0
How would you factor out the two?

Because [tex]\sqrt{3-2x^2}[/tex] is not the same as 2*[tex]\sqrt{\frac{3}{2}-x^2}[/tex]
 
  • #4
Hitman2-2
How would you factor out the two?

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

Yes, but [tex] \sqrt{3-2x^2} = \sqrt{2}*\sqrt{\frac{3}{2}-x^2} [/tex]. I may not have been clear in my previous post.
 
  • #5
22
0
Oh! Duh! Thank you!!!
I'll give it a try.
 

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