Can You Solve These Integrals Involving ArcCos?

[benhou]

I was trying to derive the formula for the moment of inertia of a rectangular prism at its centre of mass. And ended up with this:

\frac{1}{2}M(a/2)^{2}+2\frac{M}{ab}\int^{b/2}_{a/2}arccos(\frac{a^{2}}{2r^{2}}-1)rdr+4\frac{M}{ab}\int^{\sqrt{a^{2}+b^{2}}}_{b/2}(\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr

Due to the limitation of my knowledge in Calculus, I was stuck here. Can anyone help me solve these two integral:

\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr

\int (\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr

 

[blkqi]

Looks like you could first reduce a bit your integrands (at least one of them): http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities

Then if you could just get the derivative you use partial integration.

 

[benhou]

Still have no idea. First,by "reduce", do you mean to expand the first integral ? Second, I have not learned partial integration.

 

[blkqi]

Still have no idea. First,by "reduce", do you mean to expand the first integral ? Second, I have not learned partial integration.

By reduce I mean simplify. Do it by trig identities (wikipedia has a great list of these):

\arccos\alpha \pm \arccos\beta = \arccos(\alpha\beta \mp \sqrt{(1-\alpha^2)(1-\beta^2)})

\arcsin(x)+\arccos(x)=\pi/2

For taking the integral of these trig functions, here is the trick: first learn the derivatives, then use integration by parts

\int u\, \frac{dv}{dx}\; dx=uv-\int v\, \frac{du}{dx} \; dx\!

Notice that if you know the derivative of u, \frac{du}{dx}, you can evalute the integral of u wrt x, \int u \; dx, using the derivative instead.

 

[benhou]

It seems impossible now. Since we know the derivative of arccos but not the integral of it, eg. for the first integral: <br /> \int arccos(\frac{a^{2}}{2r^{2}}-1)rdr<br />

which
u=arccos(\frac{a^{2}}{2r^{2}}-1)
\frac{dv}{dr}=r

then,from <br /> \int u\, \frac{dv}{dx}\; dx=uv-\int v\, \frac{du}{dx} \; dx\!<br />
and \frac{d}{dx}arccos(u)=\frac{-1}{\sqrt{1-u^{2}}}\frac{du}{dx}

we have
\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr=arccos(\frac{a^{2}}{2r^{2}}-1)\frac{r^{2}}{2}+\int \frac{1}{\sqrt{1-(\frac{a^{2}}{2r^{2}}-1)^{2}}}[\frac{d}{dr}(\frac{a^{2}}{2r^{2}}-1)]\frac{r^{2}}{2}dr

when I simplified it:
\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr=arccos(\frac{a^{2}}{2r^{2}}-1)\frac{r^{2}}{2}-[arcsin\frac{a}{2r}](ar)-a\int arcsin\frac{a}{2r}dr

Now I have to integrate ArcSin. Isn't this impossible? I am close to dying. I kind of started on the second integral, and it seem like it would get twice as long as this.

 

[benhou]

Does anyone know how to derive the formula for the moment of inertia of a rectangular prism?

 

[Char. Limit]

Wait, where did that arcsin come in? It looks like you had everything algebraic before...

 

[benhou]

Oopsss, I misused one formula. Now I corrected it. From this one:

<br /> \int arccos(\frac{a^{2}}{2r^{2}}-1)rdr=arccos(\frac{a^{2}}{2r^{2}}-1)\frac{r^{2}}{2}+\int \frac{1}{\sqrt{1-(\frac{a^{2}}{2r^{2}}-1)^{2}}}[\frac{d}{dr}(\frac{a^{2}}{2r^{2}}-1)]\frac{r^{2}}{2}dr<br />

By using this formula:

\int\frac{1}{\sqrt{u^{2}\pm a^{2}}}du=ln\left|u+\sqrt{u^{2}\pm a^{2}}\right|+C

and further integration by parts, I got this:

\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr=arccos(\frac{a^{2}}{2r^{2}}-1)\frac{r^{2}}{2}-\frac{ar}{2}ln\left|r+\sqrt{r^{2}-(a/2)^{2}}\right|+\frac{a}{2}\int ln\left|r+\sqrt{r^{2}-(a/2)^{2}}\right|dr

 

[Char. Limit]

Well, take a look at what W-A says about that particular logarithmic integral:

http://www.wolframalpha.com/input/?i=Integral+(log(abs(r+(r^2+(a/2)^2)^(1/2))))+dr

 

[benhou]

Damn! I made a major mistake! The two rdr should be r^{3}dr.
I should make sure before I start next time. Thanks for both of your help though. I appreciate it.

 

[Char. Limit]

Don't give up now! Just work through it carefully. Try showing (and checking) each step. I will not allow you to give up!

Joking, nothing I can say or do would require you to do anything. Still, you should have that attitude for yourself, sometimes.