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

Don't forget the constants of integration.
-------------

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

-------------------------
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
Thank you very much.
Similar Questions
  1. How do you find:the Integral of arcsin(1 / (sqrt x^2 - 1) ) dx ?? (The integral of arcsin of one over the squareroot of x
    1. answers icon 0 answers
    1. answers icon 1 answer
  2. Hello! I have this problem:x(dx)/sqrt(9-x^2) I was wondering why I can't use trig substitution and substitute sqrt(9-x^2) for
    1. answers icon 1 answer
  3. Hi.In an integration solution, the integral of (1/(sqrt (8-u squared)) is written as arcsin(u/sqrt 8), but I don't see how they
    1. answers icon 2 answers
more similar questions