Odyssey
Oct4-04, 09:44 PM
Hi, I carried out the integration until the very end...I don't know how to convert the variable back to the original one. :confused:
\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! :smile:
\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! :smile: