How do you find the integral of xsin(6x) dx?

1 Answer
Nov 10, 2015

$- \frac{1}{6} x \cos \left(6 x\right) + \frac{1}{36} \cdot \sin \left(6 x\right)$

Explanation:

Use "integration by parts".

The formula is
$\int f \left(x\right) g ' \left(x\right) \text{d" x = f(x) g(x) - int f'(x) g(x) "d} x$

So, in your product $x \cdot \sin \left(6 x\right)$ you need to determine which factor you would like to integrate and which factor you would like to differentiate.

As you have a $\sin$ function as one of the arguments, I would argue that it is usually a good bet to pick it as the one to integrate since integrating it doesn't make it more complicated.

So, let's say $f \left(x\right) = x$ and $g ' \left(x\right) = \sin \left(6 x\right)$.

As next, let's differentiate $f \left(x\right)$ and integrate $g ' \left(x\right)$ in order to compute $f ' \left(x\right)$ and $g \left(x\right)$, respectively.

$f \left(x\right) = x \implies f ' \left(x\right) = 1$
$g ' \left(x\right) = \sin \left(6 x\right) \implies g \left(x\right) = - \frac{1}{6} \cos \left(6 x\right)$

Now, it's time to apply the formula:

$\int x \sin \left(6 x\right) \text{d} x$

= x * (- 1/6 cos(6x)) - int 1 * (-1/6) cos(6x)) "d"x

$= - \frac{1}{6} x \cos \left(6 x\right) + \frac{1}{6} \int \cos \left(6 x\right) \text{d} x$

$= - \frac{1}{6} x \cos \left(6 x\right) + \frac{1}{6} \cdot \frac{1}{6} \cdot \sin \left(6 x\right)$

$= - \frac{1}{6} x \cos \left(6 x\right) + \frac{1}{36} \cdot \sin \left(6 x\right)$