intenzxboi Messages 98 Reaction score 0 Thread starter Feb 1, 2009 #1 Homework Statement [tex]\int[/tex](sin (e^(-2x))) / e^(2x) The Attempt at a Solution so i set u=e^2x du=(e^2x)(2) dx im kinda stuck on how to get rid of the e^-2x
Homework Statement [tex]\int[/tex](sin (e^(-2x))) / e^(2x) The Attempt at a Solution so i set u=e^2x du=(e^2x)(2) dx im kinda stuck on how to get rid of the e^-2x
arildno Science Advisor Homework Helper Gold Member Dearly Missed Messages 10,165 Reaction score 138 Feb 1, 2009 #2 Set u=e^-2x instead. Therefore, du/dx=-2u, that is dx=-du/2u Use this to simplify your integrand!
intenzxboi Messages 98 Reaction score 0 Feb 1, 2009 #3 k using u=e^-2x du=-2 (e^-2x) dx so... -1/2 [tex]\int[/tex] sin u du -1/2 (-cos u) + c is that right?
k using u=e^-2x du=-2 (e^-2x) dx so... -1/2 [tex]\int[/tex] sin u du -1/2 (-cos u) + c is that right?
Dick Science Advisor Homework Helper Messages 26,254 Reaction score 623 Feb 1, 2009 #4 Sure. But there is a simpler way to write (-1/2)*(-cos(u)).
Dick Science Advisor Homework Helper Messages 26,254 Reaction score 623 Feb 1, 2009 #6 cos(u)/2+C. Use more parentheses. cos u /2 can also be read cos(u/2).