Hello I was wondering if anyone could give me a hand with this. Integrate $\displaystyle \int\frac{dx}{\sqrt{2x-x^2}}$.
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Originally Posted by slevvio Hello I was wondering if anyone could give me a hand with this. Integrate $\displaystyle \int\frac{dx}{\sqrt{2x-x^2}}$. Complete the square in the denominator $\displaystyle = \int\frac{dx}{\sqrt{1 + -1 + 2x-x^2}}$ $\displaystyle = \int \frac{dx}{\sqrt{1 - (x - 1)^2)}}$ Let y = x - 1, so dy = dx $\displaystyle = \int \frac{dy}{\sqrt{1 - y^2}}$ Does this look familiar? -Dan
thanks!
$\displaystyle \int\frac{dx}{\sqrt{2x-x^2}} = sin^{-1}(x-1) + c$ ?
Originally Posted by slevvio $\displaystyle \int\frac{dx}{\sqrt{2x-x^2}} = sin^{-1}(x-1) + c$ ?
:d
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