# Homework Help: How do I solve this integral?

1. Aug 4, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Hi there. Well, the problem statement says: Calculate the arc length of the cardioid $$\rho=a(1+\cos\theta)$$.

2. Relevant equations
So I used the formula for the arc length in the polar form:

$$\int_a^b \! \sqrt{(\rho)^2+(\displaystyle\frac{d\rho}{d\theta})^2} \, d\theta$$

3. The attempt at a solution
Then I get
$$\int_0^{2\pi} \! \sqrt{(a+a\cos\theta)^2+(-a\sin\theta)^2} \, d\theta$$

$$=\int_0^{2\pi} \! \sqrt{a^2+2a^2\cos\theta+a^2\cos^2\theta+a^2\sin^2\theta} \, d\theta$$

$$=\int_0^{2\pi} \! \sqrt{a^2(2\cos\theta+\cos^2\theta+\sin^2\theta)} \, d\theta$$

$$=\int_0^{2\pi} \! |a|\sqrt{2\cos\theta+\cos^2\theta+\sin^2\theta} \, d\theta=|a|\int_0^{2\pi} \!\sqrt{2\cos\theta+1} \, d\theta$$

I can't solve this integral, I don't know how to. Any help?

Last edited: Aug 4, 2010
2. Aug 4, 2010

### Staff: Mentor

In the equation above, you're missing a term.

3. Aug 4, 2010

### Telemachus

Sorry:
$$\int_0^{2\pi} \! \sqrt{a^2+2a^2\cos\theta+a^2\cos^2\theta+a^2\sin^2\theta} \, d\theta$$

Thats it.

4. Aug 4, 2010

### Staff: Mentor

That's it. With some simplification, the quantity in the radical is 2a2 + 2a2cos(theta). I'm not sure there's a nice, neat closed form antiderivative for this problem. Are you sure you have to actually evaluate the integral, or do you just need to set it up?

5. Aug 4, 2010

### Dick

You can write 1+cos(theta) as a perfect square using a double (or half) angle formula.

6. Aug 4, 2010

### Telemachus

Mm it asked for the arc length, didn't say just set it up. Mathematica gives 4 Sqrt[3] Abs[a] EllipticE[4/3] as the solution, and I don't know what the hell "EllipticE" means, less how to arrive to that result :P

The other I had of the same kind I think I've solved it right, but it was easier, it asks for the second revolution of the spiral $$\rho=e^{2\theta}$$

So I did $$\rho'=2e^{2theta}$$

$$\int_{2\pi}^{4\pi} \! \sqrt{(e^{2\theta})^2+(2e^{2\theta})^2} \, d\theta$$

$$\int_{2\pi}^{4\pi} \! \sqrt{e^{4\theta}+(4e^{4\theta}} \, d\theta$$

$$\int_{2\pi}^{4\pi} \! \sqrt{5e^{4\theta}} \, d\theta$$

$$\int_{2\pi}^{4\pi} \! \sqrt{5e^{4\theta}} \, d\theta$$

$$\sqrt{5}\int_{2\pi}^{4\pi} \! e^{2\theta} \, d\theta$$

$$\sqrt{5}(\dysplaystyle\frac{e^{8\pi}-e^{4\pi}}{2})$$

Is that right?

7. Aug 4, 2010

### Staff: Mentor

Mathematica is saying that the first integral is an elliptic integral (see here).

Your work for the second integral looks fine.

8. Aug 4, 2010

### Telemachus

I think I can use the half angle formula (I watched at wikipedia after your post).
This:
$$\cos \frac{\theta}{2} = \pm\, \sqrt{\frac{1 + \cos\theta}{2}}$$

Am I right? thanks Dick

Thanks Mark

9. Aug 4, 2010

### Dick

That's the one.

10. Aug 4, 2010

### Telemachus

Thanks both of you. See you later!

11. Aug 4, 2010

### Telemachus

mm It seems it won't help, I've tried factorizing the 2, but...

$$|a|\int_0^{2\pi} \!\sqrt{2\cos\theta+1} \, d\theta=\sqrt{2}|a|\int_0^{2\pi} \!\sqrt{\cos\theta+\dyslpaystyle\frac{1}{2}}$$

I thought I was on the right way, but I've missed that the denominator 2 its only for the 1 inside the square root.

And the double angle formula I think it's not gonna work neither

\begin{align} \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &= \color{red} 2 \cos^2 \theta - 1 \color{black} \\ &= 1 - 2 \sin^2 \theta \\ &= \frac{1 - \tan^2 \theta} {1 + \tan^2 \theta} \end{align}

It seems to me that Mark were right :P

Last edited: Aug 4, 2010
12. Aug 4, 2010

### vela

Staff Emeritus
You're using the integrand with the mistake in it. (You can tell it's wrong because the radicand is negative for some values of theta.) Use the correct one you found later, and you'll be able to integrate it using Dick's hint.

13. Aug 4, 2010

### Telemachus

I don't think so, I can't see the mistake. I've already corrected the one that Mark marked :P

14. Aug 4, 2010

### Staff: Mentor

The integral should look like this:
$$=\sqrt{2}|a|\int_0^{2\pi} \sqrt{1 + cos\theta} d\theta$$

Use Dick's hint.

15. Aug 4, 2010

### vela

Staff Emeritus
And that was your corrected one.
Now you are using the original one again, which has the mistake.

16. Aug 4, 2010

### Staff: Mentor

vela,
This is the original integral:
$$\int_0^{2\pi} \! \sqrt{(a+a\cos\theta)^2+(-a\sin\theta)^2} \, d\theta$$

$$\sqrt{2a^2(1 + cos(\theta))}$$
$$\sqrt{2}|a|\int_0^{2\pi} \! \sqrt{1+\cos\theta} \, d\theta$$ Dividing and multiplying by two I can get something to apply the half angle formula as Dick said before.