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Question Number 131961 by Zack_ last updated on 10/Feb/21
∫(sin^4 x.cos^4 x)dx
$$\int\left({sin}^{\mathrm{4}} {x}.{cos}^{\mathrm{4}} {x}\right){dx} \\ $$
Answered by liberty last updated on 10/Feb/21
I=∫ ((1/2)−(1/2)cos 2x)^2 ((1/2)+(1/2)cos 2x)^2 dx I= ((1/4)−(1/4)cos^2 2x)^2 dx I=(1/(16))∫(1−cos^2 2x)^2 dx I=(1/(16))∫(1−2cos^2 2x+cos^4 2x)dx I=(1/(16))x−2∫((1/2)+(1/2)cos 4x)dx+(1/(16))∫cos^4 2x dx I=−((15)/(16))x−(1/4)sin 4x+(1/(16))∫((1/2)+(1/2)cos 4x)^2 dx your can finished its
$$\mathrm{I}=\int\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{I}=\:\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:^{\mathrm{2}} \mathrm{2x}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{16}}\int\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\mathrm{2x}\right)^{\mathrm{2}} \:\mathrm{dx} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{16}}\int\left(\mathrm{1}−\mathrm{2cos}\:^{\mathrm{2}} \mathrm{2x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{2x}\right)\mathrm{dx} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{16}}\mathrm{x}−\mathrm{2}\int\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4x}\right)\mathrm{dx}+\frac{\mathrm{1}}{\mathrm{16}}\int\mathrm{cos}\:^{\mathrm{4}} \mathrm{2x}\:\mathrm{dx} \\ $$$$\mathrm{I}=−\frac{\mathrm{15}}{\mathrm{16}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{4x}+\frac{\mathrm{1}}{\mathrm{16}}\int\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4x}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{your}\:\mathrm{can}\:\mathrm{finished}\:\mathrm{its} \\ $$
Answered by mr W last updated on 10/Feb/21
=(1/(16))∫(2 sin x cos x)^4 dx =(1/(16))∫(sin 2x)^4 dx =(1/(64))∫(2 sin^2 2x)^2 dx =(1/(64))∫(1−cos 4x)^2 dx =(1/(64))∫(1−2cos 4x+cos^2 4x)dx =(1/(64))∫(1−2cos 4x+(1/2)(2 cos^2 4x))dx =(1/(64))∫(1−2cos 4x+(1/2)(1−cos 8x))dx =(1/(64))∫((3/2)−2cos 4x−(1/2)cos 8x)dx =(1/(64))((3/2)x−(1/2)sin 4x−(1/(16))sin 8x)+C
$$=\frac{\mathrm{1}}{\mathrm{16}}\int\left(\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\right)^{\mathrm{4}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\int\left(\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{4}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int\left(\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}\right)^{\mathrm{2}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int\left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}{x}\right)^{\mathrm{2}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int\left(\mathrm{1}−\mathrm{2cos}\:\mathrm{4}{x}+\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int\left(\mathrm{1}−\mathrm{2cos}\:\mathrm{4}{x}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int\left(\mathrm{1}−\mathrm{2cos}\:\mathrm{4}{x}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{8}{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int\left(\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{2cos}\:\mathrm{4}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{8}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\left(\frac{\mathrm{3}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{4}{x}−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{8}{x}\right)+{C} \\ $$

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