$\displaystyle \int {t}{\sqrt{2t-t^2}}dt$
this integral has a long answer.... i need advice on what to do first... then when im stuck on sumthing ill ask for another help here...
thanks
Complete the square in the radical:
$\displaystyle \int {t}{\sqrt{2t - t^2}}dt$
$\displaystyle = \int t\sqrt{1 + (-1 + 2t - t^2)}dt$
$\displaystyle = \int t\sqrt{1 - (1 - 2t + t^2)}dt$
$\displaystyle = \int t\sqrt{1 - (t - 1)^2}dt$
Now let $\displaystyle x = t - 1 \implies dx = dt$
$\displaystyle = \int (x + 1)\sqrt{1 - x^2}dx$
$\displaystyle = \int x\sqrt{1 - x^2}dx + \int \sqrt{1 - x^2} dx$
Try these.
-Dan