# How do you find the integral of intxcos(5x)dx?

Mar 23, 2017

Use integration by parts.
$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

#### Explanation:

Let $u = x$, then $\mathrm{du} = \mathrm{dx} , \mathrm{dv} = \cos \left(5 x\right) \mathrm{dx} , \mathmr{and} v = \frac{1}{5} \sin \left(5 x\right)$

Substituting these values into the integration by parts equation:

$\int x \cos \left(5 x\right) \mathrm{dx} = \frac{x}{5} \sin \left(5 x\right) - \frac{1}{5} \int \sin \left(5 x\right) \mathrm{dx}$

The remaining integral on the right is trivial:

$\int x \cos \left(5 x\right) \mathrm{dx} = \frac{x}{5} \sin \left(5 x\right) + \frac{1}{25} \cos \left(5 x\right) + C$

Mar 23, 2017

$\frac{1}{25} \left(5 x \sin 5 x + \cos 5 x\right) + C$

#### Explanation:

Use integration by parts $\int u \mathrm{dv} = u \cdot v - \int v \mathrm{du}$

Let $u = x , \mathrm{du} = \mathrm{dx}$
Let $\mathrm{dv} = \cos 5 x \mathrm{dx} , v = \int \cos 5 x \mathrm{dx}$

Let $w = 5 x , \mathrm{dw} = 5 \mathrm{dx}$
$\int \cos 5 x \mathrm{dx} = \frac{1}{5} \int \cos w \mathrm{dw} = \frac{1}{5} \sin 5 x$

Fill in the integration by parts:
$\int x \cos 5 x \mathrm{dx} = \frac{1}{5} x \sin 5 x - \frac{1}{5} \int \sin 5 x \mathrm{dx}$

To complete the last integration piece:
Let $w = 5 x , \mathrm{dw} = 5 \mathrm{dx}$

$- \frac{1}{5} \int \sin 5 x \mathrm{dx} = - \frac{1}{5} \cdot \frac{1}{5} \int 5 \sin 5 x \mathrm{dx} = - \frac{1}{25} \int \sin w \mathrm{dw} = - \frac{1}{25} \left(- \cos 5 x\right) = \frac{1}{25} \cos 5 x$

The final solution simplified:
$\int x \cos 5 x \mathrm{dx} = \frac{1}{5} x \sin 5 x + \frac{1}{25} \cos 5 x + C = \frac{1}{25} \left(5 x \sin 5 x + \cos 5 x\right) + C$