- #1
loserspearl
- 5
- 0
∫(x2 + 7x) cosx dx
If I make v = (x2 + 7x) and du = cosx dx I get
((x2 + 7x) sinx)/2
If I make v = cosx and du = (x2 + 7x) dx I get
((x3/3 + 7x2/2) cosx)/2
using the form X=Y-X to X=Y/2
Neither are correct, what did I do wrong?
If I make v = (x2 + 7x) and du = cosx dx I get
((x2 + 7x) sinx)/2
If I make v = cosx and du = (x2 + 7x) dx I get
((x3/3 + 7x2/2) cosx)/2
using the form X=Y-X to X=Y/2
Neither are correct, what did I do wrong?