Use integration by parts to evaluate the integral...

Question

∫( 2x³+4x+√6 ) sin(5x) dx

 

I struggle with the more difficult ones, please show all working!

Answer 1

∫( 2x³+4x+√6 ) sin(5x) dx=

|u=2x³+4x+√6, du=(6x²+4)dx,  dv=sin(5x) dx, v=-(1/5)cos(5x)|=

=-(1/5)(2x³+4x+√6)cos(5x)+(2/5)∫(3x²+2)cos(5x)dx=

=|u=3x²+2, du=6xdx, dv = cos(5x)dx,  v=(1/5)sin(5x)|=

=-(1/5)(2x³+4x+√6)cos(5x)+(2/25)(3x²+2)sin(5x)dx-(12/25)∫x sin(5x)dx=

=|u=x, du=dx, dv = sin(5x)dx, v=-(1/5)cos(5x)|=

=-(1/5)(2x³+4x+√6)cos(5x)+(2/25)(3x²+2)sin(5x)dx+(12/125)xcos(5x)-(12/125)∫cos(5x)dx=

=-(1/5)(2x³+4x+√6)cos(5x)+(2/25)(3x²+2)sin(5x)dx+(12/125)xcos(5x)--(12/625)sin(5x)+C=

 

=-(1/125)(50x³+100x+25√6-12x)cos(5x)+(2/625)(75x²+50-6)sin(5x)+C=

=-(1/125)(50x³+88x+25√6)cos(5x)+(2/625)(75x²+44)sin(5x)+C.