Need help factoring the solution to this integral

[dlacombe13]

Homework Statement

∫ (x²+2x)cos(x) dx

The Attempt at a Solution

I will not show the entire solution because it is quite long (had to use integration by parts twice), and my problem isn't with getting the solution. My problem is the book shows one answer (which is equivalent to mine, but factored in a way that I cannot seem to achieve).

The answer I got (verified by symbolab and wolfram alpha)
x²sin(x) - 2[sin(x) - xcos(x)] + 2[xsin(x) + cos(x)] + C
Which I managed to factor (using rearranging and grouping) into:
sin(x)(x² + 2x - 2) + 2cos(x)(x + 1) + C

The book's answer is very, very close to mine, but it is irking me that I do not know how to get it exactly as such:
sin(x)(x² + 2x) + 2cos(x)(x + 1) - 2sin(x) + C

Could someone help me figure out how to turn my factored answer into the book's answer?

 

[axmls]

You just need to multiply the first sine function out to get that $-2\sin(x)$ term in there. In full detail, write $$\sin(x)(x^2 + 2x - 2) = x^2 \sin(x) + 2x \sin(x) - 2 \sin(x) = \sin(x) (x^2 + 2x) - 2 \sin(x).$$

 

[Ray Vickson]

Homework Statement

∫ (x²+2x)cos(x) dx

The Attempt at a Solution

I will not show the entire solution because it is quite long (had to use integration by parts twice), and my problem isn't with getting the solution. My problem is the book shows one answer (which is equivalent to mine, but factored in a way that I cannot seem to achieve).

The answer I got (verified by symbolab and wolfram alpha)
x²sin(x) - 2[sin(x) - xcos(x)] + 2[xsin(x) + cos(x)] + C
Which I managed to factor (using rearranging and grouping) into:
sin(x)(x² + 2x - 2) + 2cos(x)(x + 1) + C

The book's answer is very, very close to mine, but it is irking me that I do not know how to get it exactly as such:
sin(x)(x² + 2x) + 2cos(x)(x + 1) - 2sin(x) + C

Could someone help me figure out how to turn my factored answer into the book's answer?

Did you miss the fact that $\sin(x) (x^2 + 2x - 2) = \sin(x)(x^2 + 2x) - 2 \sin(x)$?

 

[dlacombe13]

Yes I certainly did... I guess what I didn't realize was that it was just the basic property a(b+c) = ab+ac but in the form a(b+c+d) = a(b+c)+ad. Thanks for pointing out what I missed!