# Integration of sin terms

1. Oct 23, 2015

1. The problem statement, all variables and given/known data
(This is a part of the entire problem. I'm just struggling with going to the next step since it involves solving this integral.)

Integrate:

$$\int \frac {1}{\sin \theta \sqrt {R^2\sin ^2 \theta - a^2} } d\theta$$

2. Relevant equations

R and a are simply constants. Only $$\theta$$ is a variable.

3. The attempt at a solution

I have tried performing substitutions (i.e. u = \sqrt $${R^2\sin ^2\theta - a^2}$$ and in another attempt u = $$\sin \theta$$) but this seems to get me stuck in the same situation (or worsening the situation). I have also tried integration by parts, but it just seems very messy when I do it this way, and I feel like it may be the wrong approach. I might be completely missing it, but is there a particularly good substitution or method I should approach this problem with? None of the ones I've tried so far seem to work, so any suggestions would be greatly appreciated!

2. Oct 23, 2015

### HallsofIvy

Staff Emeritus
That's not going to be an easy integral but I would start with the substitution $u= R^2sin^2(\theta)$. Then $du= 2R^2 sin(\theta)cos(\theta) d\theta$. Now multiply both numerator and denominator by $2R^2 sin(\theta)cos(\theta)d\theta$. We can write the integral as $$\frac{1}{2R^2}\int \frac{2R^2sin(\theta)cos(\theta)d\theta}{sin^2(\theta)cos(\theta)\sqrt{R^2sin^2(\theta)- a^2}}$$
Now $sin^2(\theta)= u/R^2$ and $cos(\theta)= \sqrt{1- cos^2(\theta)}= \sqrt{1- \frac{u}{R^2}}$ so that becomes
$$\frac{1}{2}\int \frac{du}{u\sqrt{1- \frac{u^2}{R^2}}\sqrt{u^2- a^2}}$$

3. Oct 23, 2015

### Simon Bridge

I'd look for a trig identity.

4. Oct 23, 2015

Hi,

Thank you for the response. I actually got a result very similar to yours but this still doesn't seem like an easily solvable integrable. Am I missing something? Integration by parts doesn't exactly work and and applying a trig substitution would bring me back to where I was earlier.

5. Oct 23, 2015

Besides applying:

$$cos 2\theta = 1 - 2sin^2\theta$$

is there any other obvious choice for a better simplification? This still seems to not make the integral much better.

6. Oct 24, 2015

### geoffrey159

It seems to me that starting with a $u = \cos \theta$ change of variable, followed by a $v = \tanh^{-1} u$ would greatly simplify your integral. With a last hyperbolic change of variable, and a discussion over the sign of a constant depending of $a$ and $R$, you could solve your problem.

7. Oct 24, 2015

### Simon Bridge

Do you have any reason to think that this will turn out to be an easy integral to solve, if only you knew the trick?

8. Oct 24, 2015

### geoffrey159

Except for the first change of variable which is tricky, the next steps come quite naturally.

9. Oct 25, 2015

After looking into your approach, I resorted to using WolframAlpha to help figure out this integral. I was expecting a nicer solution since it is based on a real problem, and this definitely seems off from what I was expecting. Nonetheless, thank you for the suggestion.

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10. Oct 25, 2015

### vela

Staff Emeritus
I'd start by pulling $a$ out of the integral, leaving you with
$$\frac 1a \int \frac{1}{\sin\theta \sqrt{b^2\sin^2 \theta-1}} \,d\theta$$ where $b = R/a$. Then use the substitution $u = \csc \theta$.

Are you sure the integral you derived is correct?