Calculus Help?

1) The problem is Int(x*sin2x)dx
Integrate by parts
Int(u *dv) = uv - int(v * du)
We need to indentify the parts now.
u = x
dv = sin2x
v = int(sin2x)dx = -1/2 * cos(2x)
du =1
Now plug it all into the equation Int(u *dv) = uv - int(v * du)
Int(x*sin2x)dx = -1/2*x*cos(2x) - int(-1/2 *cos(2x))
                       = -1/2*x*cos(2x) + 1/4 sin(2x)
                       =1/4(sin(2x) - 2x*cos(2x)) + c
I think I did this right

2) I'm not really sure how to do this, it's been too long

3) Int((x-6)/(x(x-3)) dx
First, simplify the integrand using partial fraction decomposition, and you end up with
Int(2/x - 1/(x-3))dx
= Int(2/x) dx - Int((x-3)^-1)dx
= 2 Int x^-1 dx - int ((x-3)^-1)dx
= 2 ln x - ln (x-3) + c

4) Not quite sure...Show more