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Homework Help: Trigonometric Substitution Problem w/ Sin Substitution

  1. Nov 19, 2016 #1
    1. The problem statement, all variables and given/known data

    ∫(√(64 - x^2)) / x dx

    I must solve this using a sin substitution.

    2. Relevant equations

    x = 8sinΘ
    dx = 8cosΘ dΘ
    Θ = arcsin(x/8)
    Pythagorean Identities

    3. The attempt at a solution

    (After substitution)
    = ∫8cosΘ * (√(64 - 64sin^2Θ)) / 8sinΘ dΘ
    = ∫(cosΘ * (√(64(1 - sin^2Θ))) / sinΘ dΘ
    = ∫(cosΘ * (√(64cos^2Θ)) / sinΘ dΘ
    = ∫8cos^2Θ / sinΘ dΘ

    At this point, I didn't see any easy substitutions, like the rest of the problem set. So I thought there was a good chance I made an error. I checked to see if the answer to this integral was the same as the answer to the original integral, and saw that it wasn't. So somewhere up to this point, I made a mistake. After looking over my work, I don't see what I did wrong. Maybe it's just the integral calculator I'm using, I'm not sure.

    Either way, I don't really know how to proceed with this problem. Sorry if the formatting is hard to read. I don't know the bbcode for posting these symbols properly and am in a rush at the moment.
  2. jcsd
  3. Nov 19, 2016 #2

    Simon Bridge

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    Science Advisor
    Homework Helper

    This reads: $$\int \frac{\sqrt{64-x^2}}{x}\; dx$$
    ... is that correct?

    If so, then sub ##x=8\sin\theta## does get you ##dx = 8\cos\theta\; d\theta##
    To give: $$8\int \frac{\cos^2\theta}{\sin\theta}\; d\theta$$ ... which is as far as you've got.

    You need to use trig identities to turn this into integrals you know how to do ... how about ##\cos^2\theta = 1-\sin^2\theta## ?
    The integrand is also ##\cos\theta / \tan\theta## and other forms you may have some luck figuring out.
    Basically you need to find a big table of trig identities and integrals.
  4. Nov 19, 2016 #3
    That is correct. And thank you, I was able to work out the solution. I don't know why, but I kept insisting to myself that this simplified integral needed a u substitution when in reality all I had to do was simplify it further with some trig identities. Sometimes when I do problems in a rush I make mistakes like this.
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