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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
spilloversin2xcos2x(sin5x+cos3xsin2x+sin3xcos2x+cos5x)2dx
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
leta=sinxb=cosxspilloverleta=sinxb=cosxspillovera2=sin2xb2=cos2xa2+b2=sin2x+cos2x=1spilloversin5x+cos3xsin2x+sin3xcos2x+cos5x)2a5+b3a2+a3b2+b5spillover=a2(a3+b3)+b2(a3+b3)=(a2+b2)(a3+b3)a5+b3a2+a3b2+b5=(a2+b2)(a3+b3)buta2+b2=sin2x+cos2x=1a5+b3a2+a3b2+b5=(a3+b3)(a3+b3)=sin3x+cos3xspilloversin2xcos2x(sin3x+cos3x)2dxtan2xsec2x(1+tan3x)2dxletu=1+tan3xdu=3sec2xtan2xdxspillovertan2xsec2xu2.du3sec2xtan2x=13duu2=13u+c13duu2=13u+D=13(1+tan3x)+D

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