Evaluate the following integral ∫√(a^2-x^2 ) dx
Put x=a sinθ∫√(a^2-a^2 〖sin〗^2 θ) a cosθ dθ=a^2 ∫〖〖cos〗^2 θ〗 dθ=a^2 ∫〖(1+cos2θ)/2 dθ〗=a^2/2 (θ+1/2 sin2θ )+c=a^2/2 (θ+sinθ cosθ )+c
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Put x=a sinθ
need an explanation for this answer? contact us directly to get an explanation for this answer∫√(a^2-a^2 〖sin〗^2 θ) a cosθ dθ
=a^2 ∫〖〖cos〗^2 θ〗 dθ
=a^2 ∫〖(1+cos2θ)/2 dθ〗=a^2/2 (θ+1/2 sin2θ )+c
=a^2/2 (θ+sinθ cosθ )+c