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

1. May 29, 2010

### JOhnJDC

1. The problem statement, all variables and given/known data

Find $$\int$$(x+2)dx/sqrt(3+2x-x2)

2. Relevant equations

$$\int$$du/sqrt(a2 - u2) = sin-1u/a

3. 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)

= $$\int$$udu/sqrt(a2-u2) + 3$$\int$$du/sqrt(a2-u2)

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

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

Can someone offer an explanation? Thanks.

2. May 29, 2010

### annoymage

using substitution

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

3. May 29, 2010

### JOhnJDC

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

$$\int$$udu/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?

4. May 30, 2010

### songoku

see the difference :
3$$\int$$du/sqrt(a2-u2) = 3sin-1u/a

and

$$\int$$u du/sqrt(a2-u2) = -sqrt(a2-u2)

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

5. May 30, 2010

### Mentallic

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

6. May 30, 2010

### 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$$\int$$v-1/2dv = -v-1/2 = -sqrt(a2-u2)