Here's the question you clicked on:
quakeash
what is the integral of 2/ (2x-x^2)^1/2 my guesses is sin^-1 (whatever) +c maybe ?
the integral of x/ (2x-x^2)^1/2
try completing squares .. 2x - x^2 = 1 - (1-x)^2
and let 1 - x = sin u, x = 1 + sin u
not sure though ... haven't checked!!
the complete square u gave me is wrong its 1-(x-1)^2 i guess right ?
@quakeash , would you mind writing your question on drawing board again? Click the draw button below and rewrite your question on that
|dw:1337271739470:dw| Is this your question ??
LOL ... it wasn't trig substitution, http://www.wolframalpha.com/input/?i=integrate+x%2F%28sqrt%282x+-+x^2%29%29
It is pretty simple. First follow what @experimentX said |dw:1337271981227:dw|
|dw:1337272062537:dw|
soory mistake in the lat step
It should be |dw:1337272154314:dw|
the solution . since the integral of dx/(a^2 - x^2) = sin^-1 (x/a) + c then my question after completing square would be .. 2/(1-(x-1)^2)^1/2 and its sin ^-1 (x-1/1) +c /////////////// case closed :D
Now substitute back x -1 in place of t . You will get your answer