Solving an Integral Involving $\Theta$ and $u$ - Seeking Assistance

[Odyssey]

Hi, I carried out the integration until the very end...I don't know how to convert the variable back to the original one.

$$\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^2-u^2}}$$

$$Let u = a\sin{\Theta}$$
$$du = a\cos{\Theta}d\Theta$$

The integral becomes...

$$\int_{R_{0}}^{R(\Theta)}\frac{a\cos{\Theta}d\Theta}{a\cos{\Theta}a\sin{\Theta}}$$
$$\frac{1}{a}\int_{R_{0}}^{R(\Theta)}\csc{\Theta}d{\Theta}$$

$$\csc{\Theta}d{\Theta} = \ln {|\csc{\Theta}-\cot{\Theta}|}+C$$

This is where I'm stuck. I don't know how to convert the thetas back into the "u"s. I haven't multiplied the answer by 1/a yet. I know that $$\Theta=\sin^{-1}{u/a}$$, but if I plug the $$\sin^{-1}{u/a}$$ into Theta, the expression becomes super messy and I really don't know what to do with it.

Please help, thanks in advance!

 

[vsage]

make a triangle.. it's the only way. For example, if theta = arcsin(u/a) then sin(arcsin(u/a)) = u/a, cos(arcsin(u/a)) = sqrt(a^2-u^2)/a. Make a triangle with sides u, sqrt(a^2-u^2) and a. You can find all the trig functions from it (be sure to label theta)

 

[wais]

Hi there, it looks like you're on the right track! To convert back to the original variable, we can use the trigonometric identity \csc{\Theta} = \frac{1}{\sin{\Theta}}. Then, we can substitute in for \Theta using \Theta = \sin^{-1}{\frac{u}{a}}. This will give us:

\int_{R_{0}}^{R(\Theta)}\frac{1}{a\sin{\Theta}}d{\Theta} = \frac{1}{a}\int_{R_{0}}^{R(\Theta)}\frac{1}{\sin{\Theta}}d{\Theta} = \frac{1}{a}\int_{R_{0}}^{R(u/a)}\frac{1}{\frac{u}{a}}d{\frac{u}{a}} = \frac{1}{a}\int_{R_{0}}^{R(u/a)}\frac{1}{u}du

Then, we can use the substitution u = a\sin{\Theta} again to get back to the original variable. This will give us:

\frac{1}{a}\int_{R_{0}}^{R(u/a)}\frac{1}{u}du = \frac{1}{a}\int_{R_{0}}^{R(\Theta)}\frac{1}{a\sin{\Theta}}d{\Theta} = \frac{1}{a}\int_{R_{0}}^{R(\Theta)}\frac{1}{u\sqrt{a^2-u^2}}du

I hope this helps! Let me know if you have any other questions. Good luck with your integration!