Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A troublesome integral

  1. May 29, 2013 #1

    I would like to compute the integral $$ C \int_{-a}^{0} \sqrt {\frac{-2x}{a+x}}\mathrm{d}x$$.
    If anyone is interested in its physical meaning they can check section 9.4.3 at http://solidmechanics.org/text/Chapter9_4/Chapter9_4.htm#Sect9_4_1.

    I am struggling to make any progress, yet the result is presented without computations on a number of textbook, hinting at the fact it should be relatively simple to reproduce...
    The result is, if the cionstant C is taken as $$C = \frac{K^{2} 2 (1-\nu^{2})}{E \sqrt{2} \pi}$$, $$ a \frac{1-\nu^{2}}{E} K^{2}$$.

    Thank you ever so much
  2. jcsd
  3. May 29, 2013 #2
    Take ##x = -a\sin^2\theta## with ##\theta## going from ##\pi/2## to ##\pi##. Then ##-2x = 2a\sin^2\theta##, ##a+x = a\cos^2\theta##, and ##dx = -2a\sin\theta\cos\theta d\theta##. Plug it all in and you should be able to do the resulting integral.

    (Hint: aside from recognizing the form of the integral, knowing that there are factors of ##\pi## in the final answer is usually a tip-off that you should use trigonometric substitution).
    Last edited: May 29, 2013
  4. May 29, 2013 #3
    thank you ever so much.
    So, after the change of variable you suggested I still would not know where to start to look for an analytical solution, but while I work on it I used Wolfram, which yields the indefinite integral
    Now, $$cos(\frac{1}{2}\pi )$$ is zero, so one term will diverge upon trying to compute the definite integral betwen the limits of integration $$\pi$$ and $$\frac{\pi}{2}$$, I am not sure I am right here...
  5. May 29, 2013 #4
    Show your steps. You're making a mistake with the substitution because that is not the correct antiderivative. Go slowly and don't rely on WolframAlpha. The integrand you get from this substitution is very simple and, if you're really stuck on how to do it, can be looked up in any standard table of integrals like this one.
  6. May 29, 2013 #5
    I got it, only my laziness prevented me to notice the argument was a squared tangent leading to mssive simplification, after which it was smooth crusing.
    Thank you ever so much hope sometime to be able to exchange the favour.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: A troublesome integral
  1. The Integral (Replies: 17)

  2. An integral (Replies: 1)

  3. Integral of (Replies: 3)

  4. On Integration (Replies: 4)

  5. An integral (Replies: 2)