Asked by ms

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.
Answered by Graham
Don't forget the constants of integration.
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A = ∫ √(a<sup>2</sup>-x<sup>2</sup>) dx

Let x = a sin(θ)
dx = a cos(θ) dθ

A = ∫ a<sup>2</sup>cos(θ)√(1-sin<sup>2</sup>(θ)) dθ

A = ∫ a<sup>2</sup>cos<sup>2</sup>(θ) dθ

A = (a<sup>2</sup>/2) ∫ (cos(2θ)+1) dθ

A = (a<sup>2</sup>/2)(sin(2θ)/2 + θ) + C<sub>1</sub>

A = (a<sup>2</sup>/2)(sin(θ)cos(θ) + θ) + C<sub>1</sub>

A = (x/2)√(a<sup>2</sup>-x<sup>2</sup>) + a<sup>2</sup>sin<sup>-1</sup>(x/a)/2 + C<sub>1</sub>

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B = ∫ √(a<sup>2</sup>-x<sup>2</sup>) dx

Let x = a cos(θ)
dx = -a sin(θ) dθ

B = ∫ -a<sup>2</sup>sin(θ)√(1-cos<sup>2</sup>(θ)) dθ

B = ∫ -a<sup>2</sup>sin<sup>2</sup>(θ) dθ

B = -(a<sup>2</sup>/2) ∫ (1-cos(2θ)) dθ

B = -(a<sup>2</sup>/2)(θ-sin(2θ)/2) + C<sub>2</sub>

B = (a<sup>2</sup>/2)(sin(θ)cos(θ)-θ) + C<sub>2</sub>

B = (x/2)√(a<sup>2</sup>-x<sup>2</sup>) - a<sup>2</sup>cos<sup>-1</sup>(x/a)/2 + C<sub>2</sub>
Answered by Anonymous
Thank you very much.
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