i started of by making u=2x, du=x, which will cancel out the x and give me how would i go from here
Follow Math Help Forum on Facebook and Google+
Originally Posted by simsima_1 i started of by making u=2x, du=x, which will cancel out the x and give me how would i go from here if u=2x then du=2dx not x sorry try this trig fomula so you would get now we have the integral by parts we get...
View Tag Cloud
LinkBack URL
About LinkBacksOriginally Posted by simsima_1
![\frac{1}{2}\int x \sin(4x)dx=\frac{1}{2}[-\frac{x}{4}\cos(4x)+\frac{1}{16}\sin(4x)]=-\frac{x}{8}\cos(4x)+\frac{1}{32}\sin(4x)+C](http://latex.codecogs.com/png.latex?\frac{1}{2}\int x \sin(4x)dx=\frac{1}{2}[-\frac{x}{4}\cos(4x)+\frac{1}{16}\sin(4x)]=-\frac{x}{8}\cos(4x)+\frac{1}{32}\sin(4x)+C)
