- #1
Alem2000
- 117
- 0
I can't seem to get the correct answer but I keep getting very close the question is [tex]\int_{0}^{\pi/2}xcos(x)dx[/tex]
this is what I did
[tex]u=x [/tex]
[tex] v=1/2(sin(2x))dx[/tex]
[tex]du=dx [/tex]
[tex] dv=cos(2x)dx[/tex]
and then i get [tex]=1/2(xsin(2x))\right_{0}^{\pi/2}-1/2\int_{0}^{\pi/2}sin(2x)dx[/tex]
then [tex]1/2xsin(2x)\end_{0}^{\pi/2}-1/2(-1/2cos(2x0)\end_{o}^{\pi/2}[/tex]
OKay!...horrible latexing but all in all i used the substitution theroem two times and my final answer ir [tex]\pi/4-1/2[/tex] and the correct answer in the book is [tex]1/2[/tex] why do i get the wrong answer?
this is what I did
[tex]u=x [/tex]
[tex] v=1/2(sin(2x))dx[/tex]
[tex]du=dx [/tex]
[tex] dv=cos(2x)dx[/tex]
and then i get [tex]=1/2(xsin(2x))\right_{0}^{\pi/2}-1/2\int_{0}^{\pi/2}sin(2x)dx[/tex]
then [tex]1/2xsin(2x)\end_{0}^{\pi/2}-1/2(-1/2cos(2x0)\end_{o}^{\pi/2}[/tex]
OKay!...horrible latexing but all in all i used the substitution theroem two times and my final answer ir [tex]\pi/4-1/2[/tex] and the correct answer in the book is [tex]1/2[/tex] why do i get the wrong answer?
Last edited: