integrate xcos4x

does anyone know how to integrate xcos4x please show your working :smile:

thanks

Reply 1

I don't have time to do it myself, but I think you should be able to easily do it using Integration by Parts...a quick serach online would find you the method if you don't know it/can't remember it.

Reply 2

Gaz031

Yes it's integration by parts. Integration by parts may be stated as:
INT v(du/dx) = UV - INT u(dv/dx)

Reply 3

manps

Result is
1/16(cos4x) + ¼x(sin4x) + c

Reply 4

username9816

Margerie Dawes

does anyone know how to integrate xcos4x please show your working :smile:

thanks

Q.) Find Int. xcos4x dx.
Let u = x --> du/dx = 1
Let dv/dx = cos4x --> v = (sin4x)/4

--> Int. xcos4x dx. = x/4(sin4x) - Int. (sin4x)/4 dx.
= x/4(sin4x) - [(-cos4x)/16] + k
= x/4(sin4x) + cos4x/16 + k
= [4xsin4x + cos4x]/16 + k