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find-ax-x-2-dx-




Question Number 101531 by mhmd last updated on 03/Jul/20
find ∫(√(ax−x^2 ))dx
findaxx2dx
Answered by Mr.D.N. last updated on 03/Jul/20
  I = ∫(√(ax−x^2 ))dx     = ∫(√(−(x^2 −ax)))   dx    =  ∫  (√(((a/2))^2 −(x−(a/2))^2 ))     dx    =  (((x−(a/2))(√(((a/2))^2 −(x−(a/2))^2 )))/2) + ((((a/2))^2 )/2)sin^(−1) (((x−(a/2)))/(a/2))+C  = ((((2x−a)/2)(√(ax−x^2 )))/2) + (a^2 /8) sin^(−1) ((2x−a)/a)+C  = (((2x−a)(√(ax−x^2 )))/4) + (a^2 /8) sin^(−1)  (((2x−a))/a) +C//.
I=axx2dx=(x2ax)dx=(a2)2(xa2)2dx=(xa2)(a2)2(xa2)22+(a2)22sin1(xa2)a2+C=2xa2axx22+a28sin12xaa+C=(2xa)axx24+a28sin1(2xa)a+C//.
Commented by mhmd last updated on 03/Jul/20
thank you sir
thankyousir

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