hello,

i've been in this question for ages and can't seem to crack it.

I need to substitute a suitable circular of hyperbolic fn so that I may find

a) integr[(a^2-x^2)^1/2]

b) integr[(a^2+x^2)^1/2]

c) ntegr[(x^2-a^2)^1/2]

I am looking at the first, and I think x=asinu => dx = acosu du

so (a^2-x^2)^1/2 = (acosu)^2 = [(a^2)/2] [cos2u+1]

and integr[cos2u+1] = [(sin2u)/2] + u + c_1

and subbing for u (this is one of my problems) I get my solution

= [(a^2)/2] {[(sin2arcsin(x/a))/2] + arcsin(x/a)} + c

I can also get it to

= (ax/2)[(pi/2) -a ) + (a^2/2)arcsin(x/a) +c

but i really want to get my book's answer which is

(1/2) { [ x(a^2 - x^2)^(1/2) - (a^2) arcsin(x/a)]}

please help me, I feel so fed up with trying this one question i have honestly been at it for hours!