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Question Number 131068 by EDWIN88 last updated on 01/Feb/21

$$\int \sin {}^{2}x\cos 4xdx=?$$

Answered by Ar Brandon last updated on 01/Feb/21

$$\mathcal{I}=\int {\sin }^2\mathrm{xcos}4\mathrm{xdx}=\frac{1}{2}\int (1+\cos 2\mathrm{x})\cos 4\mathrm{xdx}$$$$=\frac{1}{2}\int [\cos 4\mathrm{x}+\frac{1}{2}(\cos 6\mathrm{x}+\cos 2\mathrm{x})]\mathrm{dx}$$$$=\frac{\sin 4\mathrm{x}}{8}+\frac{\sin 6\mathrm{x}}{24}+\frac{\sin 2\mathrm{x}}{8}+\mathcal{C}$$

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