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

Mar 5, 2018

$\frac{1}{25} {e}^{3 x} \left(3 \sin \left(4 x\right) - 4 \cos \left(4 x\right)\right) + C$

#### Explanation:

According to Integration by Parts:

$\int u v \mathrm{dx}$, where $u$ and $v$ are functions, is given by:

$u \int v \mathrm{dx} - \int u ' \left(\int v \mathrm{dx}\right) \mathrm{dx}$

Here, $u = {e}^{3 x}$ and $v = \sin \left(4 x\right)$, so we can say:

$\int {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} = {e}^{3 x} \int \sin \left(4 x\right) \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} \left({e}^{3 x}\right)\right) \left(\int \sin \left(4 x\right) \mathrm{dx}\right) \mathrm{dx}$

Here, we must calculate, first:

$\int \sin \left(4 x\right) \mathrm{dx}$. According to Integration by Substitution:

$\int f \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = \int f \left(u\right) \mathrm{du}$, where $g \left(x\right) = u$.

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

$\frac{1}{4} \int \sin \left(u\right) \mathrm{du}$

$= \frac{1}{4} \left(- \cos \left(u\right)\right)$, but as $u = 4 x$, we input:

$= - \frac{1}{4} \cos \left(4 x\right)$. We can input these into our original integral:

${e}^{3 x} \cdot - \frac{1}{4} \cos \left(4 x\right) - \int \left(\frac{d}{\mathrm{dx}} \left({e}^{3 x}\right)\right) \left(- \frac{1}{4} \cos \left(4 x\right)\right) \mathrm{dx}$

$\frac{d}{\mathrm{dx}} {e}^{3 x} = 3 {e}^{3 x}$, so we can rewrite:

$- \frac{1}{4} {e}^{3 x} \cos \left(4 x\right) - \int - \frac{3}{4} {e}^{3 x} \cos \left(4 x\right) \mathrm{dx}$

$- \frac{1}{4} {e}^{3 x} \cos \left(4 x\right) - \left(- \frac{3}{4} \int {e}^{3 x} \cos \left(4 x\right) \mathrm{dx}\right)$

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

We'll concentrate on the integral.

${e}^{3 x} \int \cos \left(4 x\right) \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} \left({e}^{3 x}\right)\right) \left(\int \cos \left(4 x\right) \mathrm{dx}\right) \mathrm{dx}$

$\int \cos \left(4 x\right) \mathrm{dx} = \frac{1}{4} \sin \left(4 x\right)$, as we saw earlier. Do the exact same thing, but $\int \cos \left(u\right) \mathrm{du} = \sin \left(u\right)$, while our other integral gave us $- \cos \left(u\right)$. So we now have:

$\frac{1}{4} {e}^{3 x} \sin \left(4 x\right) - \int \frac{3}{4} {e}^{3 x} \sin \left(x\right) \mathrm{dx}$

$\frac{1}{4} {e}^{3 x} \sin \left(4 x\right) - \frac{3}{4} \int {e}^{3 x} \sin \left(x\right) \mathrm{dx}$

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

$- \frac{1}{4} {e}^{3 x} \cos \left(4 x\right) - \left(- \frac{3}{4} \left(\frac{1}{4} {e}^{3 x} \sin \left(4 x\right) - \frac{3}{4} \int {e}^{3 x} \sin \left(x\right) \mathrm{dx}\right)\right)$

$\frac{3}{4} \left(\frac{{e}^{3 x} \sin \left(4 x\right)}{4} - \frac{3 \int {e}^{3 x} \sin \left(x\right) \mathrm{dx}}{4}\right) - \frac{{e}^{3 x} \cos \left(4 x\right)}{4}$

$\frac{3 {e}^{3 x} \sin \left(4 x\right) - 9 \int {e}^{3 x} \sin \left(x\right) \mathrm{dx} - 4 {e}^{3 x} \cos \left(4 x\right)}{16}$

We really don't seem to be going anywhere with this. But remember, all of what we just wrote can be equated to $\int {e}^{3 x} \sin \left(4 x\right) \mathrm{dx}$. So:

$\int {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} = \frac{3 {e}^{3 x} \sin \left(4 x\right) - 9 \int {e}^{3 x} \sin \left(x\right) \mathrm{dx} - 4 {e}^{3 x} \cos \left(4 x\right)}{16}$, or:

$16 \int {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} = 3 {e}^{3 x} \sin \left(4 x\right) - 9 \int {e}^{3 x} \sin \left(x\right) \mathrm{dx} - 4 {e}^{3 x} \cos \left(4 x\right)$

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

$16 \int {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} + 9 \int {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} = 3 {e}^{3 x} \sin \left(4 x\right) - 4 {e}^{3 x} \cos \left(4 x\right)$

$25 \int {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} = {e}^{3 x} \left(3 \sin \left(4 x\right) - 4 \cos \left(4 x\right)\right)$

And finally, we have:

$\int {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} = \frac{1}{25} {e}^{3 x} \left(3 \sin \left(4 x\right) - 4 \cos \left(4 x\right)\right)$

Done. Hallelujah.

Mar 5, 2018

See below.

#### Explanation:

Using de Moivre's identity

${e}^{i x} = \cos x + i \sin x$

$\int \setminus {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} = \text{Im} \left[\int \setminus {e}^{3 x} \left(\cos \left(4 x\right) + i \sin \left(4 x\right)\right) \mathrm{dx}\right]$

but

$\int \setminus {e}^{3 x} \left(\cos \left(4 x\right) + i \sin \left(4 x\right)\right) \mathrm{dx} = \int \setminus {e}^{3 x} {e}^{i 4 x} \mathrm{dx} = \int \setminus {e}^{\left(3 + 4 i\right) x} \mathrm{dx} = \frac{1}{3 + 4 i} {e}^{\left(3 + 4 i\right) x}$

then

$\frac{1}{3 + 4 i} {e}^{\left(3 + 4 i\right) x} = \frac{3 - 4 i}{25} {e}^{3 x} \left(\cos \left(4 x\right) + i \sin \left(4 x\right)\right) =$

$= {e}^{3 x} / 25 \left(\left(3 \cos \left(4 x\right) + 4 \sin \left(4 x\right)\right) + i \left(3 \sin \left(4 x\right) - 4 \cos \left(4 x\right)\right)\right)$

and finally

$\int \setminus {e}^{3 x} \sin \left(4 x\right) \mathrm{dx} = {e}^{3 x} / 25 \left(3 \sin \left(4 x\right) - 4 \cos \left(4 x\right)\right) + {C}_{0}$