Trig substitution problem

  • Thread starter Sympathy
  • Start date
  • #1
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Homework Statement



Evaluate the following integrals or state that they diverge. Use proper notation.

Integral from 0 to 2 of (x+1)/Square root(4-x^2)



Homework Equations






The Attempt at a Solution



I just substituted x = 2sin(theta) thus dx = 2cos(theta)

I got to the point where it is .5(integral from 0 to 2 of 2tan(theta)) + .5(integral from 0 to 2 of sec(theta)).

I think theres an easier approach at the problem.

suggestions?
 

Answers and Replies

  • #2
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Partial fractions.
 
  • #3
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I think trig subst. is a good way. The limits of integration will change since you went from x to theta.
 
  • #4
dextercioby
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So instead of going from 0 to 2, they go from 0 to [itex] \pi/2 [/itex].
 
  • #5
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Ya, that's it. Why do people have to take all the fun out of a problem? Why not leave a hint as a hint? This one-up business is boring.
 
  • #6
Gib Z
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[tex]\int\frac{x+1}{4-x^2} dx =\frac{3}{4} \int \frac{1}{2-x} dx - \frac{1}{4}\int \frac{1}{2+x} dx[/tex]. Far too simple from there, the only reason I posted even this far was because I wanted to do it as well :)
 
  • #7
Gib Z
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Shoot me in the face, i didnt see the sqrt...
 
  • #8
hello friend


ok i will help you as i can


solution

Integral from 0 to 2 of (x+1)/Square root(4-x^2) dx

let x=2sin(&)
dx=2cos(&)d&

=Integral from 0 to [PLAIN]https://www.physicsforums.com/latex_images/12/1253579-0.png [Broken] of (2sin&+1)(2cos&)/2cos& d&

=Integral from 0 to [PLAIN]https://www.physicsforums.com/latex_images/12/1253579-0.png [Broken] of (2sin&+1) d&



i think know it is easy to solve


and i hope that i could help you
 
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