# How do you find the integral of (2-3x) cosx dx?

Apr 17, 2018

$\left(2 - 3 x\right) \sin x - 3 \cos x + C$

#### Explanation:

We have:

$\int \left(2 - 3 x\right) \cos x \mathrm{dx}$

The integration by parts states that:

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

We let:

$u = 2 - 3 x$

$\mathrm{dv} = \cos x$

$\implies \mathrm{du} = \frac{d}{\mathrm{dx}} \left(2 - 3 x\right)$

$\implies \mathrm{du} = - 3$

$\implies v = \int \cos x \mathrm{dx}$

$\implies v = \sin x$

We now have:

$\left(2 - 3 x\right) \cdot \sin x - \int \sin x \cdot - 3 \mathrm{dx}$

$\implies \left(2 - 3 x\right) \cdot \sin x + 3 \int \sin x \mathrm{dx}$

Remember that:

$\int \sin x \mathrm{dx} = - \cos x$

$\implies \left(2 - 3 x\right) \cdot \sin x + 3 \cdot - \cos x$

$\implies \left(2 - 3 x\right) \cdot \sin x - 3 \cos x$

Do you $C$ why this is incomplete?

$\implies \left(2 - 3 x\right) \sin x - 3 \cos x + C$

That is the answer!Just note that my way of solving is the simplest, but not the only solution.