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Question Number 139544 by melanie last updated on 28/Apr/21

Answered by bemath last updated on 28/Apr/21

$$\mathrm{Tanzalin}\mathrm{formula}$$$$\begin{array}{|cc|}\hline \mathrm{u}\left(\mathrm{diff}\right)& \mathrm{dv}\left(\mathrm{integrate}\right)\\ 4\mathrm{x}& \cos (2-3\mathrm{x})\\ 4& -\frac{1}{3}\sin (2-3\mathrm{x})\\ 0& -\frac{1}{9}\cos (2-3\mathrm{x})\\ \hline\end{array}$$$$\mathrm{I}=-\frac{4\mathrm{x}\sin (2-3\mathrm{x})}{3}+\frac{4\cos (2-3\mathrm{x})}{9}+\mathrm{c}$$

Answered by MJS_new last updated on 28/Apr/21

$$\int 4x\cos (2-3x)dx=$$$$\mathrm{by}\mathrm{parts}$$$$u^{\prime }=\cos (2-3x)\to u=-\frac{1}{3}\sin (2-3x)$$$$v=4x\to v^{\prime }=4$$$$=-\frac{4}{3}x\sin (2-3x)+\frac{4}{3}\int \sin (2-3x)dx=$$$$=-\frac{4}{3}x\sin (2-3x)+\frac{4}{9}\cos (2-3x)+C$$

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