# How do you integrate int xsin(4x) by integration by parts method?

Jul 25, 2016

$= - \frac{1}{4} x \cos \left(4 x\right) + \frac{1}{16} \sin \left(4 x\right) + C$

#### Explanation:

$\int x \sin \left(4 x\right) \setminus \mathrm{dx}$

$= \int x \frac{d}{\mathrm{dx}} \left(- \frac{1}{4} \cos \left(4 x\right)\right) \setminus \mathrm{dx}$

which by IBP:
$= - \frac{1}{4} x \cos \left(4 x\right) + \frac{1}{4} \int \frac{d}{\mathrm{dx}} \left(x\right) \cdot \cos \left(4 x\right) \setminus \mathrm{dx}$

$= - \frac{1}{4} x \cos \left(4 x\right) + \frac{1}{4} \int \cos \left(4 x\right) \setminus \mathrm{dx}$

$= - \frac{1}{4} x \cos \left(4 x\right) + \frac{1}{4} \cdot \frac{1}{4} \sin \left(4 x\right) + C$

$= - \frac{1}{4} x \cos \left(4 x\right) + \frac{1}{16} \sin \left(4 x\right) + C$

Jul 25, 2016

$- \frac{1}{4} x \cos \left(4 x\right) + \frac{1}{16} \sin \left(4 x\right) + C$

#### Explanation:

Before integrating by parts, we can first use substitution to get rid of the $4 x$ in the sine function.

Let color(purple)(t=4x. This implies that color(green)(dt=4dx and that color(blue)(x=t/4. Thus:

$\texttt{\int} x \sin \left(4 x\right) \mathrm{dx} = \frac{1}{4} \int \textcolor{g r e e n}{4} \textcolor{b l u e}{x} \sin \left(\textcolor{p u r p \le}{4 x}\right) \textcolor{g r e e n}{\mathrm{dx}} = \frac{1}{4} \int \frac{t}{4} \sin \left(t\right) \mathrm{dt} = \frac{1}{16} \int t \sin \left(t\right) \mathrm{dt}$

Now is a better time to use integration by parts, which takes the form:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

So, for intcolor(red)tcolor(brown)(sin(t)dt, let:

$\textcolor{red}{u = t} \text{ "=>" } \mathrm{du} = \mathrm{dt}$

$\textcolor{b r o w n}{\mathrm{dv} = \sin \left(t\right) \mathrm{dt}} \text{ "=>" "intdv=intsin(t)dt" "=>" } v = - \cos \left(t\right)$

This gives us, following $\int u \mathrm{dv} = u v - \int v \mathrm{du}$:

$\int t \sin \left(t\right) \mathrm{dt} = - t \cos \left(t\right) - \int \left(- \cos \left(t\right)\right) \mathrm{dt}$

$= - t \cos \left(t\right) + \int \cos \left(t\right) \mathrm{dt}$

$= - t \cos \left(t\right) + \sin \left(t\right)$

Since $t = 4 x$:

$\int t \sin \left(t\right) \mathrm{dt} = - 4 x \cos \left(4 x\right) + \sin \left(4 x\right)$

And since $\int x \sin \left(4 x\right) \mathrm{dx} = \frac{1}{16} \int t \sin \left(t\right) \mathrm{dt}$, divide the expression by $16$:

$\int x \sin \left(4 x\right) \mathrm{dx} = - \frac{1}{4} x \cos \left(4 x\right) + \frac{1}{16} \sin \left(4 x\right) + C$