# Integral of t \sqrt (2t - t^2)

• Aug 2nd 2007, 04:10 AM
Integral of t \sqrt (2t - t^2)
$\displaystyle \int {t}{\sqrt{2t-t^2}}dt$
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
• Aug 2nd 2007, 04:51 AM
galactus
One recourse may be to let $\displaystyle t=u+1, \;\ t-1=u, \;\ du=dt$

This gives:

$\displaystyle \int\sqrt{1-u^{2}}(u+1)du$

Now, perhaps try a trig sub.

Just a quick thought from glancing at it.
• Aug 2nd 2007, 05:02 AM
topsquark
Quote:

$\displaystyle \int {t}{\sqrt{2t-t^2}}dt$
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
• Aug 2nd 2007, 08:00 AM
curvature
Quote:

Originally Posted by topsquark
Complete the square in the radical:
$\displaystyle \int {t}{\sqrt{2t - t^2}}dt$

$\displaystyle = \int t\sqrt{1 + (-1 + 2t - t^2)}dt$

$\displaystyle = \int t\sqrt{1 - (1 - 2t + t^2)}dt$

$\displaystyle = \int t\sqrt{1 - (t - 1)^2}dt$

Then