I have tried multiple substitutions and cannot get this homework problem right xD
$\displaystyle \int\dfrac{(x-1)}{\sqrt(2x-x^2)}dx$
everything in the denominator is under the square root
I have tried multiple substitutions and cannot get this homework problem right xD
$\displaystyle \int\dfrac{(x-1)}{\sqrt(2x-x^2)}dx$
everything in the denominator is under the square root
hmmm the book answer is : $\displaystyle 1-(1-x)^2$
I tried it again, and got : $\displaystyle \sqrt{1-(1-x)^2$
I dont know where i messed up, simplified out everything and this is the integral that I take
$\displaystyle 1/2\int{(1-U)^(-(1/2))} du$ cant fix the exponent, supposed to be (1-U)^(-1/2)
$\displaystyle \displaystyle \int \frac{x-1}{\sqrt{1 -(x-1)^2}} \, dx$
let $\displaystyle t = (x-1)^2$
$\displaystyle dt = 2(x-1) \, dx$
$\displaystyle \displaystyle \frac{1}{2} \int \frac{2(x-1)}{\sqrt{1 - (x-1)^2}} \, dx$
$\displaystyle \displaystyle \frac{1}{2} \int \frac{dt}{\sqrt{1 - t}}$
$\displaystyle \displaystyle -\int \frac{-1}{2\sqrt{1 - t}} \, dt$
$\displaystyle \displaystyle -\sqrt{1-t} + C$
$\displaystyle \displaystyle -\sqrt{1-(x-1)^2} + C$
$\displaystyle \displaystyle -\sqrt{2x - x^2} + C$
Hey Skeeter,
Thanks for you're time and help you put into this , very much appreciated
But as for your answer and my answer they are slightly different, but this isn't right, what I got? $\displaystyle \sqrt{1-(1-x)^2$
oh wow you are right, The derivative of my solution does not equal the original equation !! grrrr lol