# A troublesome integral

1. May 29, 2013

### muzialis

All,

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. May 29, 2013

### VantagePoint72

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
3. May 29, 2013

### muzialis

LastOneStanding,
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
$$a(cos(2x)-4log(cos(x)))$$
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...
Thanks

4. May 29, 2013

### VantagePoint72

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.

5. May 29, 2013

### muzialis

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.