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Question Number 175410 by Stephan last updated on 29/Aug/22

$$$$$$A=\int 9x\cos 2xdx$$$$$$

Answered by MikeH last updated on 29/Aug/22

$$\mathrm{let}\{\begin{array}{l}u=x\Rightarrow du=dx\\ dv=\cos 2xdx\Rightarrow v=\frac{1}{2}\sin 2x\end{array}$$$$A=9[\frac{x\sin 2x}{2}-\frac{1}{2}\int \sin 2xdx]$$$$A=\frac{9}{2}(x\sin 2x+\frac{1}{2}\cos 2x)+k$$

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