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Question Number 125837 by joki last updated on 14/Dec/20
prove that :  ∫(√(a^2 −x^2 ))dx=_ (x/2)(√(a^2 −x^2 ))+(a^2 /2)sin^(−1) ((x/a))+c
provethat:a2x2dx=x2a2x2+a22sin1(xa)+c
Answered by Dwaipayan Shikari last updated on 14/Dec/20
Take  x=asinθ   ,  1=acosθ(dθ/dx)  ∫(√(a^2 −x^2 )) dx = ∫acosθ(√(a^2 −a^2 sin^2 θ))  dθ  =a^2 ∫cos^2 θ dθ =(a^2 /2)∫1+cos2θ dθ  =((a^2 θ)/2)+(a^2 /4)sin2θ +C  θ=sin^(−1) (x/a)    sin2θ =2(x/a^2 )(√(a−x^2 ))  So ∫(√(a^2 −x^2 )) dx = ((a^2 sin^(−1) (x/a))/2)+(x/2)(√(a^2 −x^2 )) +C
Takex=asinθ,1=acosθdθdxa2x2dx=acosθa2a2sin2θdθ=a2cos2θdθ=a221+cos2θdθ=a2θ2+a24sin2θ+Cθ=sin1xasin2θ=2xa2ax2Soa2x2dx=a2sin1xa2+x2a2x2+C

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