cos-3-xsin-3-xdx-

Question Number 75093 by Rio Michael last updated on 07/Dec/19

$$\int{cos}^{\mathrm{3}} {xsin}^{\mathrm{3}} {xdx} \\ $$

Commented by mathmax by abdo last updated on 07/Dec/19

I =∫ cos^3 x sin^3 xdx =∫ (cosxsinx)^3 dx =(1/8) ∫ (sin(2x))^3 dx =(1/8) ∫ sin(2x)sin^2 (2x)dx =(1/8) ∫ sin(2x)(((1−cos(4x))/2))dx =(1/(16)) ∫ sin(2x)dx−(1/(16)) ∫ sin(2x)cos(4x)dx sin(2x)cos(4x) =cos((π/2)−2x)cos(4x) =(1/2){ cos((π/2)+2x) +cos((π/2)−6x)} =(1/2){sin(6x)−sin(2x)} ⇒ I =−(1/(32)) cos(2x)−(1/(32))∫( sin(6x)−sin(2x))dx =−(1/(32))cos(2x)+(1/(6×32)) cos(6x)−(1/(64)) cos(2x)+C =−(3/(64)) cos(2x) +(1/(192)) cos(6x) +C

$${I}\:=\int\:{cos}^{\mathrm{3}} {x}\:{sin}^{\mathrm{3}} {xdx}\:=\int\:\left({cosxsinx}\right)^{\mathrm{3}} \:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\:\int\:\left({sin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{3}} \:{dx}\:=\frac{\mathrm{1}}{\mathrm{8}}\:\int\:\:{sin}\left(\mathrm{2}{x}\right){sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{sin}\left(\mathrm{2}{x}\right)\left(\frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\:\int\:{sin}\left(\mathrm{2}{x}\right){dx}−\frac{\mathrm{1}}{\mathrm{16}}\:\int\:\:{sin}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right){dx} \\ $$$${sin}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right)\:={cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{cos}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{x}\right)\:+{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{6}{x}\right)\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\left(\mathrm{6}{x}\right)−{sin}\left(\mathrm{2}{x}\right)\right\}\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{32}}\:{cos}\left(\mathrm{2}{x}\right)−\frac{\mathrm{1}}{\mathrm{32}}\int\left(\:{sin}\left(\mathrm{6}{x}\right)−{sin}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{32}}{cos}\left(\mathrm{2}{x}\right)+\frac{\mathrm{1}}{\mathrm{6}×\mathrm{32}}\:{cos}\left(\mathrm{6}{x}\right)−\frac{\mathrm{1}}{\mathrm{64}}\:{cos}\left(\mathrm{2}{x}\right)+{C} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{64}}\:{cos}\left(\mathrm{2}{x}\right)\:+\frac{\mathrm{1}}{\mathrm{192}}\:{cos}\left(\mathrm{6}{x}\right)\:+{C} \\ $$$$ \\ $$

Answered by mr W last updated on 07/Dec/19

∫cos^3 xsin^3 xdx =∫cos^2 xsin^3 x d(sin x) =∫(1−sin^2 x)sin^3 x d(sin x) =∫(sin^3 x−sin^5 x) d(sin x) =((sin^4 x)/4)−((sin^6 x)/6)+C

$$\int{cos}^{\mathrm{3}} {xsin}^{\mathrm{3}} {xdx} \\ $$$$=\int{cos}^{\mathrm{2}} {xsin}^{\mathrm{3}} {x}\:{d}\left(\mathrm{sin}\:{x}\right) \\ $$$$=\int\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}\right){sin}^{\mathrm{3}} {x}\:{d}\left(\mathrm{sin}\:{x}\right) \\ $$$$=\int\left(\mathrm{sin}^{\mathrm{3}} \:{x}−\mathrm{sin}^{\mathrm{5}} \:{x}\right)\:{d}\left(\mathrm{sin}\:{x}\right) \\ $$$$=\frac{\mathrm{sin}^{\mathrm{4}} \:{x}}{\mathrm{4}}−\frac{\mathrm{sin}^{\mathrm{6}} \:{x}}{\mathrm{6}}+{C} \\ $$

Commented by Rio Michael last updated on 07/Dec/19

$${thank}\:{you}\:{sir},{so}\:{much} \\ $$

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