How do you integrate #(e^x)sin4x dx#?

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Answer 1 · 2015-04-29T08:32:29

This integral is a cyclic one and has to be done two times with the theorem of the integration by parts, that says:

#intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx#

We can assume that #f(x)=sin4x# and #g'(x)dx=e^xdx#,

#f'(x)dx=cos4x*4dx#

#g(x)=e^x#,

so:

#I=e^xsin4x-inte^xcos4x*4dx=#

#=e^xsin4x-4inte^xcos4xdx=(1)#.

And now...again:

#f(x)=cos4x# and #g'(x)dx=e^xdx#,

#f'(x)dx=-sin4x*4dx#

#g(x)=e^x#.

So:

#(1)=e^xsin4x-4[e^xcos4x-inte^x(-sin4x*4)dx]=#

#=e^xsin4x-4e^xcos4x-16inte^xsin4xdx#.

So, finally, we can write:

#I=e^xsin4x-4e^xcos4x-16IrArr#

#17I=e^xsin4x-4e^xcos4xrArr#

#I=e^x/17(sin4x-4cos4x)+c#.