Unformatted text preview: 1
[sin(18x) + sin(−4x)]dx
2
1
1
=
sin(18x)dx +
[sin(−4x)]dx
2
2
= ...(Ithinkyouknowtheremainingthing ) sin(7x)cos(11x)dx = (1) Note we have sin(7x+11x) = sin(7x)cos(11x)+cos(7x)sin(11x) and sin(7x−11x) = sin(7x)cos(11x)−
cos(7x)sin(11x), so we have
1
sin(7x)cos(11x) = [sin(18x) + sin(−4x)]
2 ...
View
Full
Document
This note was uploaded on 02/14/2012 for the course MATH 334 taught by Professor Dallon during the Fall '08 term at BYU.
 Fall '08
 DALLON
 Differential Equations, Equations

Click to edit the document details