Don't forget the constants of integration.
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A = ∫ √(a2-x2) dx
Let x = a sin(θ)
dx = a cos(θ) dθ
A = ∫ a2cos(θ)√(1-sin2(θ)) dθ
A = ∫ a2cos2(θ) dθ
A = (a2/2) ∫ (cos(2θ)+1) dθ
A = (a2/2)(sin(2θ)/2 + θ) + C1
A = (a2/2)(sin(θ)cos(θ) + θ) + C1
A = (x/2)√(a2-x2) + a2sin-1(x/a)/2 + C1
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B = ∫ √(a2-x2) dx
Let x = a cos(θ)
dx = -a sin(θ) dθ
B = ∫ -a2sin(θ)√(1-cos2(θ)) dθ
B = ∫ -a2sin2(θ) dθ
B = -(a2/2) ∫ (1-cos(2θ)) dθ
B = -(a2/2)(θ-sin(2θ)/2) + C2
B = (a2/2)(sin(θ)cos(θ)-θ) + C2
B = (x/2)√(a2-x2) - a2cos-1(x/a)/2 + C2
Book works out integral of sqtr(a^2-x^2)dx as a^2/2*arcsin(x/a)+x/2*sqrt(a^2-x^2) by taking x=a sin theta, and advises to work it out using x=a cos theta. I did and got the first term as
-a^2/2*arccos(x/a)and the same second term. This means arcsin x/a=-arccos x/a which can be true for a particular case only and not in general for all values. Where am I going wrong? Kindly advise.
2 answers
Thank you very much.