Find constants and antiderivative problem

Beeorz
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Homework Statement


Find constants c1 and c2 such that F(x) = c1xsinx+c2cosx is an antiderivative of f(x) = 4xcos(x)+3sin(x)

Homework Equations


d/dx cosx = -sinx
d/dx sinx = cosx

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 here's my attempt:

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

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


dunno if I understand that correctly...but here's what I think it means to do:

4xcosx-(cosx)(4) ??
 


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?
 


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
 


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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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