# stuck on integration

• Jun 6th 2008, 06:56 PM
khuezy
stuck on integration
[(x^2)/(squareroot:(2x - x^2) ]dx

do I do u-substitution?
• Jun 6th 2008, 07:01 PM
galactus
$\int\frac{x^{2}}{\sqrt{2x-x^{2}}}dx$

You could let $x=u+1, \;\ dx=du$

Make the subs and get:

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

Now, we can let $u=sin(t), \;\ du=cos(t)dt$

Make the subs and it whittles down to:

$\int{(1+sin(t))^{2}}dt$

$\int{sin^{2}(t)}dt+2\int{sin(t)}dt+\int{dt}$

$\frac{-1}{2}sin(t)cos(t)-2cos(t)+\frac{3}{2}t$

Resub for $t=sin^{-1}(u)$

That gives us:

$\frac{3sin^{-1}(u)}{2}-\frac{u\sqrt{1-u^{2}}}{2}-2\sqrt{1-u^{2}}$

Resub for $u=x-1$ and get:

$\frac{3}{2}sin^{-1}(x-1)-\frac{x\sqrt{2x-x^{2}}}{2}-\frac{3}{2}\sqrt{2x-x^{2}}$
• Jun 6th 2008, 07:05 PM
Mathstud28
Quote:

Originally Posted by khuezy
[(x^2)/(squareroot:(2x - x^2) ]dx

do I do u-substitution?

Ok
we have $\int\frac{x^2}{\sqrt{2x-x^2}}dx$

Completing the square we get

$\int\frac{x^2}{\sqrt{1-(x-1)^2}}$

Now letting $x-1=\sin(\theta)$

so $x=\sin(\theta)-1\Rightarrow{dx=\cos(\theta)}$

Giving us

$\int\frac{(\sin(\theta)-1)^2}{\sqrt{1-\sin^2(\theta)}}d\theta$

Can you go from there?
• Jun 6th 2008, 07:26 PM
khuezy
thanks
I got the integral of (sin(t) +1)^2
then that becomes sin^2(t) + 2sin(t) + 1

the integral of sin^2(t)
is 1/2[1+cos(2t)
how do I make that in terms of 't'?
• Jun 6th 2008, 07:45 PM
khuezy
How did you get to there?
You coul
How did you get to there?

$\frac{-1}{2}sin(t)cos(t)-2cos(t)+\frac{3}{2}t$

how does sin^2(t) go to 1/2sintcost?