# How do you find the integral of f(x)=2xsin4x using integration by parts?

Feb 27, 2016

int 2x sin(4x) dx=(-1)/2x cos(4x) + 1/8 sin(4x) + C, where $C$ is the constant of integration.

#### Explanation:

Let be $f \left(x\right) = 2 x \sin \left(4 x\right)$.

We want to find int 2x sin(4x) dx. We know the integral of the function $\sin$ or $\cos$ so it would be good for us to get rid of $2 x$ by differentiating it.

Let's say $g \left(x\right) = 2 x$ and $h ' \left(x\right) = \sin \left(4 x\right)$.

So $g ' \left(x\right) = 2$ and $h \left(x\right) = \frac{- 1}{4} \cos \left(4 x\right)$.

int g(x) h'(x) dx = g(x)h(x) - int g'(x) h(x) dx using integration by parts.

int 2x sin(4x) dx = 2x*(-1)/4 cos(4x) - int 2 *(-1)/4 cos(4x) dx

=(-1)/2x cos(4x) + 1/2 int cos(4x) dx

$= \frac{- 1}{2} x \cos \left(4 x\right) + \frac{1}{2} \cdot \frac{1}{4} \sin \left(4 x\right)$

$= \frac{- 1}{2} x \cos \left(4 x\right) + \frac{1}{8} \sin \left(4 x\right)$

Therefore, int 2x sin(4x) dx = (-1)/2x cos(4x) + 1/8 sin(4x) + C, where $C$ is the constant of integration.