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

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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.