(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find [tex]\int[/tex](x+2)dx/sqrt(3+2x-x^{2})

2. Relevant equations

[tex]\int[/tex]du/sqrt(a^{2}- u^{2}) = sin^{-1}u/a

3. The attempt at a solution

I began by completing the square:

3+2x-x^{2}= 4 -(x^{2}-2x+1)

So, 4-(x-1)^{2}= a^{2}-u^{2}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-x^{2}) = [tex]\int[/tex](u+3)du/sqrt(a^{2}-u^{2})

= [tex]\int[/tex]udu/sqrt(a^{2}-u^{2}) + 3[tex]\int[/tex]du/sqrt(a^{2}-u^{2})

This is where I get confused. I know that 3[tex]\int[/tex]du/sqrt(a^{2}-u^{2}) = 3sin^{-1}u/a

But, according to my book, [tex]\int[/tex]udu/sqrt(a^{2}-u^{2}) = -sqrt(a^{2}-u^{2}) and I don't understand why.

Can someone offer an explanation? Thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Integration: completing the square and inverse trig functions

**Physics Forums | Science Articles, Homework Help, Discussion**