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Question Number 86405 by Power last updated on 28/Mar/20

Commented by MJS last updated on 28/Mar/20

$${\sin }^{2}3x{\sin }^{3}2x=$$$$=\frac{1}{16}(\sin 12x-3\sin 8x-2\sin 6x+3\sin 4x+6\sin 2x)$$$$\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{super}\mathrm{easy}$$

Commented by jagoll last updated on 28/Mar/20

$$\mathrm{wkwkwkwk}$$

Answered by jagoll last updated on 28/Mar/20

$$\frac{1}{4}{\left(2\sin 3\mathrm{x}\sin 2\mathrm{x}\right)}^2=\frac{1}{4}[\cos \mathrm{x}-\cos 5\mathrm{x}]$$$$\int \frac{1}{4}(\sin 2\mathrm{x}\cos \mathrm{x}-\sin 2\mathrm{x}\cos 5\mathrm{x})\mathrm{dx}$$$$\mathrm{it}\mathrm{now}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

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