Question

Need to find the indefinite integral of::
Sqrt(1-x^2)dx

** Any way to do that without trig substitution? **
[Tried integrating by parts, did not work]

Answer 1

try trig sub

Answer 2

spure it off

Answer 3

without trig sub.

Answer 4

are you expecting it to solve without trig sub?

Answer 5

The thing is, that our lectorer gave us this question before we learnt trig subs. (Maybe by mistake?).

Answer 6

there might be other ways .. but I don't know yet.

Answer 7

I=∫▒〖√((1-x^2 ) ) dx〗=∫▒√((1-x^2 ) )×1dx
Integrate by parts
=√((1-x^2 ) ) x-∫▒d/dx √((1-x^2 ) ) xdx
=x√(1-x^2 )-∫▒〖(-2x)/(2√(1-x^2 )) xdx〗
=x√(1-x^2 ) -∫▒(1-x^2-1)/√(1-x^2 ) dx
=x√(1-x^2 ) -∫▒√(1-x^2 ) dx+∫▒〖1/√(1-x^2 ) dx〗
=x√(1-x^2 ) –I +sin^(-1)⁡x+c
2I=x√(1-x^2 )+ sin^(-1)⁡x+c
I=(x√(1-x^2 ))/2+1/2 sin^(-1)⁡x+C

Answer 8

interesting ... this is just opposite of differentiation.