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Am I understanding trigonometric solution correctly?

  1. Sep 28, 2011 #1
    More of a general inquiry..

    I was given some homework to do on trigonometric substitution. It looks to me like the goal is always to get an integral of 1 by itself, then replacing theta that results from integrating with it's x equivalent?

    On my homework, as long as I made the right choice for substituting, I always got the integral of cos/cos or similar, which was 1, integrated was theta, which was found by solving my substitution for theta.
     
  2. jcsd
  3. Sep 28, 2011 #2
    The point of a trig substitution is to get rid of square roots or radicals. Or something squared plus something. In your case it became one but that doesn't happen all the time.
     
  4. Sep 28, 2011 #3
    Okay, I've been working with a few more and I often get results like:

    sin(arcsin t)
    cos(arcsin t)

    And results like that. Is there a good way to remember the algebraic equivalent for those?
     
  5. Sep 28, 2011 #4

    Mark44

    Staff: Mentor

    For the first, sin(arcsin t) = t, subject to possible restrictions of the domain of the sine function.

    For the second one, I personally don't think it's worthwhile to clutter my brain with a formula. Instead, sketch a right triangle one of whose acute angles represents arcsin(t). So you have a right triangle with [itex]\theta[/itex] as one of the acute angles.

    Since [itex]\theta[/itex] = arcsin(t), then t = sin([itex]\theta[/itex]). You can label the opposite side as t, and the hypotenuse as 1. What's the length of the other side (the adjacent side)? IOW, what is cos([itex]\theta[/itex])? That will be the same as cos(arcsin(t)).
     
  6. Sep 29, 2011 #5
    Square root one minus x squared?
     
  7. Sep 29, 2011 #6

    Mark44

    Staff: Mentor

    You were asking about cos(arcsin(t)). There wasn't an x anywhere in sight.
     
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