How Do You Integrate Complex Trigonometric Functions?

[TheCanadian]

Homework Statement

(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:

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

Homework Equations

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

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!

 

[HallsofIvy]

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
$$\frac{1}{2}\int \frac{du}{u\sqrt{1- \frac{u^2}{R^2}}\sqrt{u^2- a^2}}$$

 

[Simon Bridge]

I'd look for a trig identity.

 

[TheCanadian]

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
$$\frac{1}{2}\int \frac{du}{u\sqrt{1- \frac{u^2}{R^2}}\sqrt{u^2- a^2}}$$

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.

 

[TheCanadian]

I'd look for a trig identity.

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.

 

[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.

 

[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?

 

[geoffrey159]

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?

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

 

[TheCanadian]

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
$$\frac{1}{2}\int \frac{du}{u\sqrt{1- \frac{u^2}{R^2}}\sqrt{u^2- a^2}}$$

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.

 

[vela]

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 $u = \csc \theta$.

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.

Are you sure the integral you derived is correct?