Big trigonometric substitution problem.
Okay, it's the last problem I have to do in the section and it stumped me.
It's the integral of ((4-x)/x)^(1/2)dx
The book says to let x=u^2.
The answer is 4*arcsin(x^(1/2)/2) + x^(1/2) * (4-x)^(1/2) + C.
No matter what I try I am not getting close to this answer. I know I'm supposed to get it into two pieces, one with form integral du/(a^2-u^2) .
I start by letting u = x^(1/2). u^2=x. du= 1/2*x^(-1/2)
So this gives me 2*integral square root of (4-u^2)/u^2 du.
I then have to substitute. Since the integral is in the form of the square root of a^2-x^2 I use u=a*sin(t) or U=2*sin(t) and du=2*cos(t).
This gives me the integral of 2*((4-4sin^2(t))/(4sin^2(t))^(1/2) * 2*cos(t).
I think I have it right down to this point. Then when I retry the problem I keep trying different things from here on out.
Sorry for the TI-89 format. I'm going to try to learn LaTex in the next week or so.
Thanks for your help!