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Question Number 131961 by Zack_ last updated on 10/Feb/21

$$\int ({sin}^{4}x.{cos}^{4}x)dx$$

Answered by liberty last updated on 10/Feb/21

$$\mathrm{I}=\int {(\frac{1}{2}-\frac{1}{2}\cos 2\mathrm{x})}^2{(\frac{1}{2}+\frac{1}{2}\cos 2\mathrm{x})}^2\mathrm{dx}$$$$\mathrm{I}={(\frac{1}{4}-\frac{1}{4}\cos {}^{2}2\mathrm{x})}^2\mathrm{dx}$$$$\mathrm{I}=\frac{1}{16}\int {(1-{\cos }^{2}2\mathrm{x})}^2\mathrm{dx}$$$$\mathrm{I}=\frac{1}{16}\int (1-2\cos {}^{2}2\mathrm{x}+\cos {}^{4}2\mathrm{x})\mathrm{dx}$$$$\mathrm{I}=\frac{1}{16}\mathrm{x}-2\int (\frac{1}{2}+\frac{1}{2}\cos 4\mathrm{x})\mathrm{dx}+\frac{1}{16}\int \cos {}^{4}2\mathrm{x}\mathrm{dx}$$$$\mathrm{I}=-\frac{15}{16}\mathrm{x}-\frac{1}{4}\sin 4\mathrm{x}+\frac{1}{16}\int {(\frac{1}{2}+\frac{1}{2}\cos 4\mathrm{x})}^2\mathrm{dx}$$$$\mathrm{your}\mathrm{can}\mathrm{finished}\mathrm{its}$$

Answered by mr W last updated on 10/Feb/21

$$=\frac{1}{16}\int {\left(2\sin x\cos x\right)}^{4}dx$$$$=\frac{1}{16}\int {\left(\sin 2x\right)}^{4}dx$$$$=\frac{1}{64}\int {\left(2{\sin }^{2}2x\right)}^{2}dx$$$$=\frac{1}{64}\int {(1-\cos 4x)}^{2}dx$$$$=\frac{1}{64}\int (1-2\cos 4x+{\cos }^{2}4x)dx$$$$=\frac{1}{64}\int (1-2\cos 4x+\frac{1}{2}\left(2{\cos }^{2}4x\right))dx$$$$=\frac{1}{64}\int (1-2\cos 4x+\frac{1}{2}(1-\cos 8x))dx$$$$=\frac{1}{64}\int (\frac{3}{2}-2\cos 4x-\frac{1}{2}\cos 8x)dx$$$$=\frac{1}{64}(\frac{3}{2}x-\frac{1}{2}\sin 4x-\frac{1}{16}\sin 8x)+C$$

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