Integration: completing the square and inverse trig functions

[JOhnJDC]

Homework Statement

Find \int(x+2)dx/sqrt(3+2x-x²)

Homework Equations

\intdu/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:

\int(x+2)dx/sqrt(3+2x-x²) = \int(u+3)du/sqrt(a²-u²)

= \intudu/sqrt(a²-u²) + 3\intdu/sqrt(a²-u²)

This is where I get confused. I know that 3\intdu/sqrt(a²-u²) = 3sin^-1u/a

But, according to my book, \intudu/sqrt(a²-u²) = -sqrt(a²-u²) and I don't understand why.

Can someone offer an explanation? Thanks.

 

[annoymage]

This is where I get confused. I know that 3\intdu/sqrt(a²-u²) = 3sin^-1u/a

But, according to my book, \intudu/sqrt(a²-u²) = -sqrt(a²-u²) and I don't understand why.

Can someone offer an explanation? Thanks.

using substitution

cos \theta = \frac{u}{a}

 

[JOhnJDC]

using substitution

cos \theta = \frac{u}{a}

Are you saying that I should substitute u=a sin theta to obtain:

\intudu/sqrt(a²-u²) = \int(a sin theta)(a cos theta)(d theta)/(a cos theta) = a cos theta

So, u = a cos theta and cos theta = u/a, but u/a doesn't equal -sqrt(a²-u²)

What am I missing?

 

[songoku]

This is where I get confused. I know that 3\intdu/sqrt(a²-u²) = 3sin^-1u/a

But, according to my book, \intudu/sqrt(a²-u²) = -sqrt(a²-u²) and I don't understand why.

Can someone offer an explanation? Thanks.

see the difference :
3\intdu/sqrt(a²-u²) = 3sin^-1u/a

and

\intu du/sqrt(a²-u²) = -sqrt(a²-u²)

to obtain the second one, use : v = a²-u²

 

[Mentallic]

Just take the derivative of both answers and you'll see why.

 

[JOhnJDC]

I see it now. Thanks for your help.

\int(a²-u²)^-1/2udu

Substitute v=a²-u², dv=-2udu, udu=-dv/2

= -1/2\intv^-1/2dv = -v^-1/2 = -sqrt(a²-u²)