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

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Answer 1 · 2018-04-17T20:09:57

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

We have:

$\int (2 - 3 x) cos x d x$ $int(2-3x)cosxdx$

$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

We let:

$u = 2 - 3 x$ $u=2-3x$

$d v = cos x$ $dv=cosx$

$\Rightarrow d u = \frac{d}{d x} (2 - 3 x)$ $=>du=d/dx(2-3x)$

$\Rightarrow d u = - 3$ $=>du=-3$

$\Rightarrow v = \int cos x d x$ $=>v=intcosxdx$

$\Rightarrow v = sin x$ $=>v=sinx$

We now have:

$(2 - 3 x) \cdot sin x - \int sin x \cdot - 3 d x$ $(2-3x)*sinx-intsinx*-3dx$

$\Rightarrow (2 - 3 x) \cdot sin x + 3 \int sin x d x$ $=>(2-3x)*sinx+3intsinxdx$

Remember that:

$\int sin x d x = - cos x$ $intsinxdx=-cosx$

$\Rightarrow (2 - 3 x) \cdot sin x + 3 \cdot - cos x$ $=>(2-3x)* sinx+3* -cosx$

$\Rightarrow (2 - 3 x) \cdot sin x - 3 cos x$ $=>(2-3x)* sinx-3cosx$

Do you $C$ $C$ why this is incomplete?

$\Rightarrow (2 - 3 x) sin x - 3 cos x + C$ $=>(2-3x)sinx-3cosx+C$

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)cosxdx(2-3x) cosx dx?