# Find constants and antiderivative problem

1. Jul 15, 2008

### Beeorz

1. The problem statement, all variables and given/known data
Find constants c1 and c2 such that F(x) = c1xsinx+c2cosx is an antiderivative of f(x) = 4xcos(x)+3sin(x)

2. Relevant equations
d/dx cosx = -sinx
d/dx sinx = cosx

3. The attempt at a solution
Don't know how to setup or handle a problem such as this. Prolly much simplier than I'm making it out to be, but heres my attempt:

(-4(x^2)sinx)/(2) + 3cosx
-2(x^2)sinx + 3cosx

Any advice would help. I know my method is wrong.

2. Jul 15, 2008

### rock.freak667

Re: Antiderivative

You don;t integrate a product like that. Try using integration by parts

$$\int u \frac{dv}{dx} dx =uv- \int v \frac{du}{dx} dx$$

3. Jul 15, 2008

### Beeorz

Re: Antiderivative

dunno if I understand that correctly...but heres what I think it means to do:

4xcosx-(cosx)(4) ??

4. Jul 15, 2008

### d_leet

Re: Antiderivative

You want to choose c1 and c2 such that F'(x) is equal to f(x), why not differentiate F(x) set the derivative equal to f(x) and see if you can figure out what the constants should be?

5. Jul 15, 2008

### Beeorz

Re: Antiderivative

F'(x) = c1xsinx+c2cosx

=c1sinx+c1xcosx-c2sinx

c1sinx+c1xcosx-c2sinx=c1xcosx+sinx(c1-c2)
c1-c2 = (4xcosx+3sinx-c1xcosx) / (sinx)

now what? doesn't seem like anything simplifies

6. Jul 15, 2008

### rock.freak667

Re: Antiderivative

c1xcosx+sinx(c1-c2) = 4xcos(x)+3sin(x)

Right?

equate coefficients now.

i.e. coefficient of sin(x) in the left side = Coefficient of sin(x) on the right side.

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