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Category: Integration

sin-4-x-cos-4-xdx-

Question Number 9581 by dc630248@gmail.com last updated on 18/Dec/16 sin4x.cos4xdx Answered by mrW last updated on 18/Dec/16 $$\mathrm{sin}^{\mathrm{4}} {x}.\mathrm{cos}^{\mathrm{4}} {x}=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2sinxcosx}\right)^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}^{\mathrm{4}}…