- #1
JOhnJDC
- 35
- 0
Homework Statement
Find [tex]\int[/tex](x+2)dx/sqrt(3+2x-x2)
Homework Equations
[tex]\int[/tex]du/sqrt(a2 - u2) = sin-1u/a
The Attempt at a Solution
I began by completing the square:
3+2x-x2 = 4 -(x2-2x+1)
So, 4-(x-1)2 = a2-u2 and a=2 and u=(x-1)
Further, since x=(u+1), dx=du and (x+2)=(u+3)
Substituting, I got:
[tex]\int[/tex](x+2)dx/sqrt(3+2x-x2) = [tex]\int[/tex](u+3)du/sqrt(a2-u2)
= [tex]\int[/tex]udu/sqrt(a2-u2) + 3[tex]\int[/tex]du/sqrt(a2-u2)
This is where I get confused. I know that 3[tex]\int[/tex]du/sqrt(a2-u2) = 3sin-1u/a
But, according to my book, [tex]\int[/tex]udu/sqrt(a2-u2) = -sqrt(a2-u2) and I don't understand why.
Can someone offer an explanation? Thanks.