# Integration involving ArcCos

1. Jan 19, 2010

### 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:

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

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

2. Jan 19, 2010

### blkqi

3. Jan 20, 2010

### benhou

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

4. Jan 20, 2010

### blkqi

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.

Last edited: Jan 20, 2010
5. Jan 21, 2010

### benhou

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

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

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

6. Jan 21, 2010

### benhou

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

7. Jan 21, 2010

### Char. Limit

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

8. Jan 21, 2010

### benhou

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

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

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:

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

9. Jan 21, 2010

### Char. Limit

10. Jan 22, 2010

### 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.

11. Jan 23, 2010

### 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.