i'm trying to complete my notets from my calculus II class. my professor showed us how to do the following integral using integration by parts but i'm not following his reasoning could some one fill me in on what i'm missing. thx in advance.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int^{\pi}_03x\sin\frac{x}{2}\\{dx}[/tex]

let [tex]\ u=3x[/tex] so that [tex]\ du=3dx[/tex]

let [tex]\ dv= \sin\frac{x}{2}[/tex] so that [tex]\ v= -2\cos\frac{x}{2}[/tex]

now [tex]\int^{\pi}_03x\sin\frac{x}{2}\\{dx} = -6x\cos\frac{x}{2} \right]]^{\pi}_0\\+ 12\sin\frac{x}{2}\right]]_0^{\pi}[/tex]

what is the intermediate step that i'm missing?

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# Homework Help: Incomplet notes on integration by parts

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