# integration giving inverse trig

#### paulmdrdo

##### Active member
hey guys can you help solve this problem.

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

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!

#### Prove It

##### Well-known member
MHB Math Helper
hey guys can you help solve this problem.

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

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!
\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*}

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