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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

$\displaystyle\int\frac{dx}{\sqrt{2x-x^2}}$

i know that i have to change the integrand into this form $\displaystyle\frac{du}{\sqrt{a^2-u^2}}$ can you please show me how. thanks!

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

$\displaystyle\int\frac{dx}{\sqrt{2x-x^2}}$

i know that i have to change the integrand into this form $\displaystyle\frac{du}{\sqrt{a^2-u^2}}$ can you please show me how. thanks!

[tex]\displaystyle \begin{align*} \int{\frac{dx}{\sqrt{2x - x^2}}} &= \int{\frac{dx}{\sqrt{-\left( x^2 - 2x \right) } }} \\ &= \int{\frac{dx}{\sqrt{- \left[ x^2 -2x + (-1)^2 - (-1)^2 \right] }}} \\ &= \int{\frac{dx}{\sqrt{ - \left[ \left( x - 1 \right) ^2 - 1 \right] }}} \\ &= \int{ \frac{dx}{\sqrt{ 1 - \left( x - 1 \right) ^2 } }} \end{align*}[/tex]

$\displaystyle\int\frac{dx}{\sqrt{2x-x^2}}$

i know that i have to change the integrand into this form $\displaystyle\frac{du}{\sqrt{a^2-u^2}}$ can you please show me how. thanks!

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