Question Number 67235 by prof Abdo imad last updated on 24/Aug/19
$${find}\:\:\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} \:{x}^{\mathrm{2}} \left\{{cosx}−{sinx}\right\}^{\mathrm{3}} {dx} \\ $$
Commented by mathmax by abdo last updated on 01/Sep/19
![let I =∫_(−(π/3)) ^(π/3) x^2 {cosx −sinx}^3 and J=∫_(−(π/6)) ^(π/6) x^2 {cosx +sinx}^3 dx we have I +J =∫_(−(π/3)) ^(π/3) x^2 (cosx −sinx +cosx +sinx)((cosx−sinx)^2 −(cosx−sinx)(cosx +sinx)+(cosx +sinx)^2 }dx =∫_(−(π/3)) ^(π/3) 2x^2 cosx{1−2cosxsinx −(cos^2 x−sin^2 x) +1+2cosx sinx}dx =∫_(−(π/3)) ^(π/3) 2x^2 cosx{2 −cos(2x)}dx =4 ∫_0 ^(π/3) x^2 cosx(2−cos(2x))dx =8 ∫_0 ^(π/3) x^2 cosxdx −4∫_0 ^(π/3) cosx.cos(2x)dx by parts ∫_0 ^(π/3) x^2 cosxdx=[x^2 sinx]_0 ^(π/3) −∫_0 ^(π/3) 2xsinx dx =−2{[−xcosx]_0 ^(π/3) −∫_0 ^(π/3) (−cosx)dx} =−2{−(π/6) +[sinx]_0 ^(π/3) }=−2{−(π/6)+((√3)/2)} =(π/3)−(√3) ∫_0 ^(π/3) cosx cos(2x)dx =(1/2)∫_0 ^(π/3) (cos(3x)+cosx)dx=(1/6)[sin(3x)]_0 ^(π/3) +(1/2)[sinx]_0 ^(π/3) =(1/2)((√3)/2) =((√3)/4) ⇒ I+J =8((π/3)−(√3))−(√3) =((8π)/3)−9(√3) I−J =∫_(−(π/3)) ^(π/3) x^2 {(cosx−sinx)^3 −(cosx+sinx)^3 }dx =∫_(−(π/3)) ^(π/3) x^2 (−2sinx)(1−2cosx sinx +cos(2x) +1+2cosxsinx}dx =−2∫_(−(π/3)) ^(π/3) x^2 sinx{2+cos(2x)}dx =0 because the function is odd ⇒I−J =0 ⇒I =J ⇒ ⇒2I =((8π)/3)−9(√3) ⇒I =((4π)/3)−(9/2)(√3)](Q67892.png)
$${let}\:{I}\:=\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} {x}^{\mathrm{2}} \left\{{cosx}\:−{sinx}\right\}^{\mathrm{3}} \:{and}\:{J}=\int_{−\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} {x}^{\mathrm{2}} \left\{{cosx}\:+{sinx}\right\}^{\mathrm{3}} {dx} \\ $$$${we}\:{have}\:{I}\:+{J}\:=\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} {x}^{\mathrm{2}} \left({cosx}\:−{sinx}\:+{cosx}\:+{sinx}\right)\left(\left({cosx}−{sinx}\right)^{\mathrm{2}} \right. \\ $$$$\left.−\left({cosx}−{sinx}\right)\left({cosx}\:+{sinx}\right)+\left({cosx}\:+{sinx}\right)^{\mathrm{2}} \right\}{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} \mathrm{2}{x}^{\mathrm{2}} {cosx}\left\{\mathrm{1}−\mathrm{2}{cosxsinx}\:−\left({cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}\right)\:+\mathrm{1}+\mathrm{2}{cosx}\:{sinx}\right\}{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} \:\mathrm{2}{x}^{\mathrm{2}} {cosx}\left\{\mathrm{2}\:−{cos}\left(\mathrm{2}{x}\right)\right\}{dx} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} {x}^{\mathrm{2}} {cosx}\left(\mathrm{2}−{cos}\left(\mathrm{2}{x}\right)\right){dx}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} {x}^{\mathrm{2}} {cosxdx}\:−\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:{cosx}.{cos}\left(\mathrm{2}{x}\right){dx} \\ $$$${by}\:{parts}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:{x}^{\mathrm{2}} {cosxdx}=\left[{x}^{\mathrm{2}} {sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\mathrm{2}{xsinx}\:{dx} \\ $$$$=−\mathrm{2}\left\{\left[−{xcosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\left(−{cosx}\right){dx}\right\} \\ $$$$=−\mathrm{2}\left\{−\frac{\pi}{\mathrm{6}}\:+\left[{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \right\}=−\mathrm{2}\left\{−\frac{\pi}{\mathrm{6}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}\:=\frac{\pi}{\mathrm{3}}−\sqrt{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} {cosx}\:{cos}\left(\mathrm{2}{x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \left({cos}\left(\mathrm{3}{x}\right)+{cosx}\right){dx}=\frac{\mathrm{1}}{\mathrm{6}}\left[{sin}\left(\mathrm{3}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\left[{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow\:{I}+{J}\:=\mathrm{8}\left(\frac{\pi}{\mathrm{3}}−\sqrt{\mathrm{3}}\right)−\sqrt{\mathrm{3}}\:=\frac{\mathrm{8}\pi}{\mathrm{3}}−\mathrm{9}\sqrt{\mathrm{3}} \\ $$$${I}−{J}\:=\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} {x}^{\mathrm{2}} \left\{\left({cosx}−{sinx}\right)^{\mathrm{3}} −\left({cosx}+{sinx}\right)^{\mathrm{3}} \right\}{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} {x}^{\mathrm{2}} \left(−\mathrm{2}{sinx}\right)\left(\mathrm{1}−\mathrm{2}{cosx}\:{sinx}\:+{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{1}+\mathrm{2}{cosxsinx}\right\}{dx} \\ $$$$=−\mathrm{2}\int_{−\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{3}}} {x}^{\mathrm{2}} {sinx}\left\{\mathrm{2}+{cos}\left(\mathrm{2}{x}\right)\right\}{dx} \\ $$$$=\mathrm{0}\:\:{because}\:{the}\:{function}\:{is}\:{odd}\:\Rightarrow{I}−{J}\:=\mathrm{0}\:\:\Rightarrow{I}\:={J}\:\Rightarrow \\ $$$$\Rightarrow\mathrm{2}{I}\:=\frac{\mathrm{8}\pi}{\mathrm{3}}−\mathrm{9}\sqrt{\mathrm{3}}\:\Rightarrow{I}\:=\frac{\mathrm{4}\pi}{\mathrm{3}}−\frac{\mathrm{9}}{\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$ \\ $$