Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Calculus Help?
Ive been having some trouble solving these problems. Any help would be absolutely appreciated.
I figured out #3.
Figured out 1 & 3.
1 Answer
- 1 year ago
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
