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Question Number 195254 by Spillover last updated on 02/Aug/23
∫^(spillover) ((sin^2 xcos^2 x)/((sin^5 x+cos^3 xsin^2 x+sin^3 xcos^2 x+cos^5 x)^2 ))dx
$$\int^{\boldsymbol{{spillover}}} \frac{\mathrm{sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}{\left(\mathrm{sin}\:^{\mathrm{5}} {x}+\mathrm{cos}\:^{\mathrm{3}} {x}\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:^{\mathrm{3}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{5}} {x}\right)^{\mathrm{2}} }{dx} \\ $$
Answered by Spillover last updated on 02/Aug/23
let a=sin x b=cos x_(spillover) let a=sin x b=cos x_(spillover) a^2 =sin^2 x b^2 =cos^2 x a^2 +b^2 =sin^2 x + cos^2 x=1_(spillover) sin^5 x+cos^3 xsin^2 x+sin^3 xcos^2 x+cos^5 x)^2 a^5 +b^3 a^2 +a^3 b^2 +b^5 _(spillover) =a^2 (a^3 +b^3 )+b^2 (a^3 +b^3 )=(a^2 +b^2 )(a^3 +b^3 ) a^5 +b^3 a^2 +a^3 b^2 +b^5 =(a^2 +b^2 )(a^3 +b^3 ) but a^2 +b^2 =sin^2 x + cos^2 x=1 a^5 +b^3 a^2 +a^3 b^2 +b^5 =(a^3 +b^3 ) (a^3 +b^3 )=sin^3 x+cos^3 x_(spillover) ∫((sin^2 xcos^2 x)/((sin^3 x+cos^3 x)^2 ))dx ⇒ ∫ ((tan^2 xsec^2 x)/((1+tan^3 x)^2 ))dx let u=1+tan^3 x du=3sec^2 xtan^2 xdx_(spillover) ∫ ((tan^2 xsec^2 x)/u^2 ).(du/(3sec^2 xtan^2 x))=∫(1/3)(du/u^2 )=−(1/(3u))+c ∫(1/3)(du/u^2 )=−(1/(3u))+D =−(1/(3(1+tan^3 x)))+D
$${let}\:\:\:\:\:\:\underset{{spillover}} {\:\:\:{a}=\mathrm{sin}\:{x}\:\:\:\:\:\:\:{b}=\mathrm{cos}\:{x}}\:\:\: \\ $$$${let}\:\:\:\:\:\:\underset{{spillover}} {\:\:\:{a}=\mathrm{sin}\:{x}\:\:\:\:\:\:\:{b}=\mathrm{cos}\:{x}}\:\:\: \\ $$$${a}^{\mathrm{2}} =\mathrm{sin}\:^{\mathrm{2}} {x}\:\:\:\:\:{b}^{\mathrm{2}} =\mathrm{cos}\:^{\mathrm{2}} {x}\:\:\:\:\:\:\:\underset{{spillover}} {{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{sin}\:^{\mathrm{2}} {x}\:\:+\:\mathrm{cos}\:^{\mathrm{2}} {x}=\mathrm{1}} \\ $$$$\left.\mathrm{sin}\:^{\mathrm{5}} {x}+\mathrm{cos}\:^{\mathrm{3}} {x}\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:^{\mathrm{3}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{5}} {x}\right)^{\mathrm{2}} \\ $$$$\underset{{spillover}} {{a}^{\mathrm{5}} +{b}^{\mathrm{3}} {a}^{\mathrm{2}} +{a}^{\mathrm{3}} {b}^{\mathrm{2}} +{b}^{\mathrm{5}} }={a}^{\mathrm{2}} \left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)+{b}^{\mathrm{2}} \left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)=\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right) \\ $$$${a}^{\mathrm{5}} +{b}^{\mathrm{3}} {a}^{\mathrm{2}} +{a}^{\mathrm{3}} {b}^{\mathrm{2}} +{b}^{\mathrm{5}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right) \\ $$$${but}\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{sin}\:^{\mathrm{2}} {x}\:\:+\:\mathrm{cos}\:^{\mathrm{2}} {x}=\mathrm{1} \\ $$$${a}^{\mathrm{5}} +{b}^{\mathrm{3}} {a}^{\mathrm{2}} +{a}^{\mathrm{3}} {b}^{\mathrm{2}} +{b}^{\mathrm{5}} =\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right) \\ $$$$\underset{{spillover}} {\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)=\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{cos}\:^{\mathrm{3}} {x}} \\ $$$$\int\frac{\boldsymbol{\mathrm{sin}}\:^{\mathrm{2}} \boldsymbol{{x}\mathrm{cos}}\:^{\mathrm{2}} \boldsymbol{{x}}}{\left(\boldsymbol{\mathrm{sin}}\:^{\mathrm{3}} \boldsymbol{{x}}+\boldsymbol{\mathrm{cos}}\:^{\mathrm{3}} \boldsymbol{{x}}\right)^{\mathrm{2}} }\boldsymbol{{dx}}\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\int\:\:\frac{\boldsymbol{\mathrm{tan}}\:^{\mathrm{2}} \boldsymbol{{x}\mathrm{sec}}\:^{\mathrm{2}} \boldsymbol{{x}}}{\left(\mathrm{1}+\boldsymbol{\mathrm{tan}}\:^{\mathrm{3}} \boldsymbol{{x}}\right)^{\mathrm{2}} }\boldsymbol{{dx}} \\ $$$${let}\:\:\underset{\boldsymbol{{spillover}}} {{u}=\mathrm{1}+\mathrm{tan}\:^{\mathrm{3}} {x}\:\:\:\:\:\:{du}=\mathrm{3sec}\:^{\mathrm{2}} {x}\mathrm{tan}\:^{\mathrm{2}} {xdx}} \\ $$$$\int\:\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}\mathrm{sec}\:^{\mathrm{2}} {x}}{{u}^{\mathrm{2}} }.\frac{{du}}{\mathrm{3sec}\:^{\mathrm{2}} {x}\mathrm{tan}\:^{\mathrm{2}} {x}}=\int\frac{\mathrm{1}}{\mathrm{3}}\frac{{du}}{{u}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{3}{u}}+{c} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{3}}\frac{{du}}{{u}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{3}{u}}+{D} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{3}} {x}\right)}+{D} \\ $$$$ \\ $$

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