integrate: square root of (4-x^2)
2 answers
I think its 2-x^1, but I can't be sure as I learned it in precalculus.
No to matt's answer. It requires integration by substitution and one other trick. Let u = 4 - x^2.
x = sqrt(4-u) dx = -(1/2)/sqrt(4-u)
That makes the integral that of
-(1/2) u^(1/2)/sqrt(4-u)
Perhaps that can be integrated by parts.
The answer (from my table of integrals) is (1/2)[x*sqrt(4-x^2) -4 sin^-1(x/2)]
I just don't have time to do it. perhaps Reiny or Damon will come to the resuce
x = sqrt(4-u) dx = -(1/2)/sqrt(4-u)
That makes the integral that of
-(1/2) u^(1/2)/sqrt(4-u)
Perhaps that can be integrated by parts.
The answer (from my table of integrals) is (1/2)[x*sqrt(4-x^2) -4 sin^-1(x/2)]
I just don't have time to do it. perhaps Reiny or Damon will come to the resuce
1 answer