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Question Number 20044 by mondodotto@gmail.com last updated on 20/Aug/17

Answered by $@ty@m last updated on 20/Aug/17

$$=\frac{1}{2}\int 2sin4xcos2xdx$$$$=\frac{1}{2}\int (sin6x+sin2x)dx$$$$=\frac{1}{2}\int sin6xdx+\frac{1}{2}\int sin2xdx$$$$=\frac{-cos6x}{12}-\frac{cos2x}{4}+C$$

Answered by mrW1 last updated on 21/Aug/17

$$\int \sin 4\mathrm{x}\cos 2\mathrm{x}\mathrm{dx}$$$$=\int 2\sin 2\mathrm{x}\cos 2\mathrm{x}\cos 2\mathrm{x}\mathrm{dx}$$$$=\int \sin 2\mathrm{x}{\cos }^{2}2\mathrm{x}\mathrm{d}2\mathrm{x}$$$$=-\int {\cos }^{2}2\mathrm{x}\mathrm{dcos}2\mathrm{x}$$$$=-\frac{{\cos }^{3}2\mathrm{x}}{3}+\mathrm{C}$$

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