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

Question

2902 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T20:09:57

$(2-3x)sinx-3cosx+C$

We have:

$int(2-3x)cosxdx$

$intudv=uv-intvdu$

We let:

$u=2-3x$

$dv=cosx$

$=>du=d/dx(2-3x)$

$=>du=-3$

$=>v=intcosxdx$

$=>v=sinx$

We now have:

$(2-3x)*sinx-intsinx*-3dx$

$=>(2-3x)*sinx+3intsinxdx$

Remember that:

$intsinxdx=-cosx$

$=>(2-3x)* sinx+3* -cosx$

$=>(2-3x)* sinx-3cosx$

Do you $C$ why this is incomplete?

$=>(2-3x)sinx-3cosx+C$

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

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