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

[tex]\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^2-u^2}} [/tex]

[tex] Let u = a\sin{\Theta}[/tex]

[tex] du = a\cos{\Theta}d\Theta[/tex]

The integral becomes...

[tex]\int_{R_{0}}^{R(\Theta)}\frac{a\cos{\Theta}d\Theta}{a\cos{\Theta}a\sin{\Theta}}[/tex]

[tex]\frac{1}{a}\int_{R_{0}}^{R(\Theta)}\csc{\Theta}d{\Theta}[/tex]

[tex]\csc{\Theta}d{\Theta} = \ln {|\csc{\Theta}-\cot{\Theta}|}+C[/tex]

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 [tex]\Theta=\sin^{-1}{u/a}[/tex], but if I plug the [tex]\sin^{-1}{u/a}[/tex] into Theta, the expression becomes super messy and I really don't know what to do with it.

Please help, thanks in advance!

[tex]\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^2-u^2}} [/tex]

[tex] Let u = a\sin{\Theta}[/tex]

[tex] du = a\cos{\Theta}d\Theta[/tex]

The integral becomes...

[tex]\int_{R_{0}}^{R(\Theta)}\frac{a\cos{\Theta}d\Theta}{a\cos{\Theta}a\sin{\Theta}}[/tex]

[tex]\frac{1}{a}\int_{R_{0}}^{R(\Theta)}\csc{\Theta}d{\Theta}[/tex]

[tex]\csc{\Theta}d{\Theta} = \ln {|\csc{\Theta}-\cot{\Theta}|}+C[/tex]

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 [tex]\Theta=\sin^{-1}{u/a}[/tex], but if I plug the [tex]\sin^{-1}{u/a}[/tex] into Theta, the expression becomes super messy and I really don't know what to do with it.

Please help, thanks in advance!

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