Menu Close

cos-3-2x-sin-3-3x-dx-

Tinku Tara 0 Comments
Pin




Question Number 91961 by jagoll last updated on 04/May/20
∫ cos^3 (2x) sin^3 (3x) dx
$$\int\:\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{2x}\right)\:\mathrm{sin}\:^{\mathrm{3}} \left(\mathrm{3x}\right)\:\mathrm{dx}\: \\ $$
Commented by mathmax by abdo last updated on 04/May/20
cos(2x)sin(3x) =cos(2x)cos((π/2)−3x) =(1/2){cos(−x+(π/2))+cos(5x−(π/2))} =(1/2){ sinx +sin(5x)} ⇒ cos^3 (2x)sin^3 (3x) =(1/8){sinx +sin(5x)}^3 =(1/8){ sin^3 x +3sin^2 x sin(5x)+3sinxsin^2 (5x) +sin^3 (5x)} sin^3 x =sin^2 x sinx =((1−cos(2x))/2)sinx =(1/2)sinx −(1/2) cos(2x)sinx but cos(2x)sinx =cos(2x)cos((π/2)−x) =(1/2){ cos(x+(π/2))+cos(3x−(π/2))} =(1/2){−sinx +sin(3x)} ⇒ sin^3 x =(1/2)sin(x)−(1/4)(−sinx +sin(3x)) =(3/4)sinx−(1/4)sin(3x) ⇒ 8I = ∫ ((3/4)sinx−(1/4)sin(3x))dx +3∫ sin^2 x sin(5x)dx +3 ∫ sinx sin^2 5x +∫ ((3/4)sin(5x) −(1/4)sin(15x))dx also sin^2 x sin(5x) =(1/2)(1−cos(2x))sin(5x) =(1/2)sin(5x)−(1/2)cos(2x)sin(5x) but cos(2x)sin(5x) =cos(2x)cos((π/2)−5x) =(1/2)(cos(−3x+(π/2))+cos(7x−(π/2))) =(1/2)(sin(3x)+sin(7x)) now its eazy to solve the integral...
$${cos}\left(\mathrm{2}{x}\right){sin}\left(\mathrm{3}{x}\right)\:={cos}\left(\mathrm{2}{x}\right){cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{3}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(−{x}+\frac{\pi}{\mathrm{2}}\right)+{cos}\left(\mathrm{5}{x}−\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{sinx}\:+{sin}\left(\mathrm{5}{x}\right)\right\}\:\Rightarrow \\ $$$${cos}^{\mathrm{3}} \left(\mathrm{2}{x}\right){sin}^{\mathrm{3}} \left(\mathrm{3}{x}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\left\{{sinx}\:+{sin}\left(\mathrm{5}{x}\right)\right\}^{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left\{\:{sin}^{\mathrm{3}} {x}\:+\mathrm{3}{sin}^{\mathrm{2}} {x}\:{sin}\left(\mathrm{5}{x}\right)+\mathrm{3}{sinxsin}^{\mathrm{2}} \left(\mathrm{5}{x}\right)\:+{sin}^{\mathrm{3}} \left(\mathrm{5}{x}\right)\right\} \\ $$$${sin}^{\mathrm{3}} {x}\:={sin}^{\mathrm{2}} {x}\:{sinx}\:=\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{sinx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sinx}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{cos}\left(\mathrm{2}{x}\right){sinx}\:\:{but}\:\:{cos}\left(\mathrm{2}{x}\right){sinx}\:={cos}\left(\mathrm{2}{x}\right){cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{cos}\left({x}+\frac{\pi}{\mathrm{2}}\right)+{cos}\left(\mathrm{3}{x}−\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{−{sinx}\:+{sin}\left(\mathrm{3}{x}\right)\right\}\:\Rightarrow \\ $$$${sin}^{\mathrm{3}} {x}\:=\frac{\mathrm{1}}{\mathrm{2}}{sin}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(−{sinx}\:+{sin}\left(\mathrm{3}{x}\right)\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{sinx}−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{x}\right)\:\Rightarrow \\ $$$$\mathrm{8}{I}\:=\:\int\:\left(\frac{\mathrm{3}}{\mathrm{4}}{sinx}−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{x}\right)\right){dx}\:+\mathrm{3}\int\:{sin}^{\mathrm{2}} {x}\:{sin}\left(\mathrm{5}{x}\right){dx} \\ $$$$+\mathrm{3}\:\int\:{sinx}\:{sin}^{\mathrm{2}} \mathrm{5}{x}\:\:+\int\:\:\left(\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\mathrm{5}{x}\right)\:−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{15}{x}\right)\right){dx} \\ $$$${also}\:\:{sin}^{\mathrm{2}} {x}\:{sin}\left(\mathrm{5}{x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){sin}\left(\mathrm{5}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{5}{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{2}{x}\right){sin}\left(\mathrm{5}{x}\right)\:{but}\: \\ $$$${cos}\left(\mathrm{2}{x}\right){sin}\left(\mathrm{5}{x}\right)\:={cos}\left(\mathrm{2}{x}\right){cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{5}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(−\mathrm{3}{x}+\frac{\pi}{\mathrm{2}}\right)+{cos}\left(\mathrm{7}{x}−\frac{\pi}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({sin}\left(\mathrm{3}{x}\right)+{sin}\left(\mathrm{7}{x}\right)\right)\:{now}\:{its}\:{eazy}\:{to}\:{solve}\:{the}\:{integral}… \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *