#inte^(3x)sin(4x)dx#?

2 Answers
Mar 5, 2018

#1/25e^(3x)(3sin(4x)-4cos(4x))+C#

Explanation:

According to Integration by Parts:

#intuvdx#, where #u# and #v# are functions, is given by:

#uintvdx-intu'(intvdx)dx#

Here, #u=e^(3x)# and #v=sin(4x)#, so we can say:

#inte^(3x)sin(4x)dx=e^(3x)intsin(4x)dx-int(d/dx(e^(3x)))(intsin(4x)dx)dx#

Here, we must calculate, first:

#intsin(4x)dx#. According to Integration by Substitution:

#intf(g(x))g'(x)dx=intf(u)du#, where #g(x)=u#.

Here, #g(x)=4x#. The derivative of #4x# is #4#, so we can rewrite the integral as:

#1/4intsin(u)du#

#=1/4(-cos(u))#, but as #u=4x#, we input:

#=-1/4cos(4x)#. We can input these into our original integral:

#e^(3x)*-1/4cos(4x)-int(d/dx(e^(3x)))(-1/4cos(4x))dx#

#d/dxe^(3x)=3e^(3x)#, so we can rewrite:

#-1/4e^(3x)cos(4x)-int-3/4e^(3x)cos(4x)dx#

#-1/4e^(3x)cos(4x)-(-3/4inte^(3x)cos(4x)dx)#

Well, look at what we get. Integrate by parts again:

We'll concentrate on the integral.

#e^(3x)intcos(4x)dx-int(d/dx(e^(3x)))(intcos(4x)dx)dx#

#intcos(4x)dx=1/4sin(4x)#, as we saw earlier. Do the exact same thing, but #intcos(u)du=sin(u)#, while our other integral gave us #-cos(u)#. So we now have:

#1/4e^(3x)sin(4x)-int3/4e^(3x)sin(x)dx#

#1/4e^(3x)sin(4x)-3/4inte^(3x)sin(x)dx#

We're back. A never ending loop, doesn't it seem like? But remember, altogether, we have:

#-1/4e^(3x)cos(4x)-(-3/4(1/4e^(3x)sin(4x)-3/4inte^(3x)sin(x)dx))#

#3/4((e^(3x)sin(4x))/4-(3inte^(3x)sin(x)dx)/4)-(e^(3x)cos(4x))/4#

#(3e^(3x)sin(4x)-9inte^(3x)sin(x)dx-4e^(3x)cos(4x))/16#

We really don't seem to be going anywhere with this. But remember, all of what we just wrote can be equated to #inte^(3x)sin(4x)dx#. So:

#inte^(3x)sin(4x)dx=(3e^(3x)sin(4x)-9inte^(3x)sin(x)dx-4e^(3x)cos(4x))/16#, or:

#16inte^(3x)sin(4x)dx=3e^(3x)sin(4x)-9inte^(3x)sin(x)dx-4e^(3x)cos(4x)#

We see two integrals of #e^(3x)sin(4x)#! A break! Simple algebra is all that's left to do.

#16inte^(3x)sin(4x)dx+9inte^(3x)sin(4x)dx=3e^(3x)sin(4x)-4e^(3x)cos(4x)#

#25inte^(3x)sin(4x)dx=e^(3x)(3sin(4x)-4cos(4x))#

And finally, we have:

#inte^(3x)sin(4x)dx=1/25e^(3x)(3sin(4x)-4cos(4x))#

Done. Hallelujah.

See below.

Explanation:

Using de Moivre's identity

#e^(ix) = cos x + i sin x#

#int\ e^(3x)sin(4x)dx = "Im"[int\ e^(3x)(cos(4x)+isin(4x))dx]#

but

#int\ e^(3x)(cos(4x)+isin(4x))dx = int\ e^(3x)e^(i4x) dx = int\ e^((3+4i)x) dx = 1/(3+4i)e^((3+4i)x)#

then

#1/(3+4i)e^((3+4i)x) = (3-4i)/25e^(3x)(cos(4x)+i sin(4x)) =#

#= e^(3x)/25((3 cos(4x)+4 sin(4x))+i(3 sin(4x)-4 cos(4x)))#

and finally

#int\ e^(3x)sin(4x)dx =e^(3x)/25(3 sin(4x)-4 cos(4x))+C_0#