Jackson Duncan

2022-06-25

How to integrate $\int \frac{du}{u\sqrt{c-2u}}$

Josie Stephenson

Let x=2u,dx=2du.
$\int \frac{du}{u\sqrt{c-2u}}=\int \frac{2du}{2u\sqrt{c-2u}}=\int \frac{dx}{x\sqrt{c-x}}$
Let $v=\sqrt{c-x},dv=\frac{-dx}{2\sqrt{c-x}},x=c-{v}^{2}$
$\int \frac{dx}{x\sqrt{c-x}}=\int \frac{-2dv}{c-{v}^{2}}=-\frac{2}{c}\int \frac{dv}{1-\left(\frac{v}{\sqrt{c}}{\right)}^{2}}=-\frac{2}{c}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\left(\frac{v}{\sqrt{c}}\right)+{\zeta }_{0}$
$\zeta =-\frac{2}{c}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\left(\frac{v}{\sqrt{c}}\right)+{\zeta }_{0}=-\frac{2}{c}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\left(\frac{\sqrt{c-2u}}{\sqrt{c}}\right)+{\zeta }_{0}$
$u=\frac{c}{2}\left(1-{\mathrm{tanh}}^{2}\left(-\frac{c}{2}\left(\zeta -{\zeta }_{0}\right)\right)\right)=\frac{c}{2}{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}\left(-\frac{c}{2}\left(\zeta -{\zeta }_{0}\right)\right)=\frac{c}{2}{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}\left(\frac{c}{2}\left(\zeta -{\zeta }_{0}\right)\right)$

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