Question

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

Answer 1

`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.

Answer 2

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`