# Integration by Substitution

1. Jan 28, 2007

### User Name

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

Solve the differential equation.

dy/dx = 4x + 4x/square root of (16-x^2)

2. Relevant equations

Substituting using U...

3. The attempt at a solution

I'm not sure if that's what I am supposed to do, but I tried using the U substitution...

4x + 4x/square root of (16-x^2)
u = 16 - x^2
du = -2x
-2du = 4x (I multiplied both sides by 2 and brought the negative over)

-2 (integration sign [the S]) du + du/square root of u

Take the anti-derivative...

-2 x (u)^(-1/2) + C (changed square root into a power)

-2 x (16-x^2) + C

Unfortunately, that is not the answer they give me. I'm kinda stuck.. am I going in the right direction?

2. Jan 29, 2007

### morson

Everything fell apart right after you made the substitution. Stating that "du = -2x" is incorrect. Rather, du = -2x dx. Then, you find an expression for "dx", sub that in, and integrate with respect to u.

Edit: It's easier to split the integral into a sum of two integrals, so it's the integral of 4x + the other integral.

Last edited: Jan 29, 2007
3. Jan 29, 2007

### Schrodinger's Dog

dy/dx = 4x + 4x/square root of (16-x^2)

I'd imagine that partial fractions would help here.

It would be alot easier, I'll have a play with it later. Time for work. Just a thought though.

$$\int 4x+=>$$

$$2x^2+C$$

$$\int \frac {4x}{\sqrt(16-x^2)}=>$$

Actually from here just substitute du from here.

Last edited: Jan 29, 2007
4. Jan 29, 2007

### HallsofIvy

Staff Emeritus
No, not partial fractions with that "x" in the numerator.

Let u= 16- x2. Then du= -2x dx so that -2du= 4xdx.