how to integrate (1-x^2)^1/2?

∫ (1-x^2)^1/2 dx

Put x= sint dx = cos t

=∫ (1- (sint)^2 )^1/2 * cos t dt

= ∫ cos ^2 t dt

=∫ 1 + cos 2t dt

= t + (sin 2t) /2 + C

(Remember sin 2t = 2 sin t cos t = 2 x √(1-x^2)

Resubstitute

= arc sin(x) + x √(1-x^2) + C

(arc sin means sin inverse)

(Ans)

∫ √(1-x²) dx
let sin(θ) = x ; cos(θ) dθ = dx
= ∫ cos²(θ) dθ
= 1/2 ∫ 1 + cos(2θ) dθ
= 1/2 (θ + sin(θ) cos(θ)) + c
= 1/2 (asin(x) + x√(1-x²)) + c

Answer: 1/2 (asin(x) + x√(1-x²)) + c

x=sint
dx = cost dt

Int. (1-x^2)^1/2 dx
=Int. (1-sin^2t)^1/2 cost dt
=Int. cos^2t dt
=Int. (1+cos2t)/2 dt
=Int. dt/2 + Int. cos2t /2 dt
= t/2 + sin2t/4 + c
= invsin(x)/2 + sint cost/2 + c
= invsin(x)/2 + x*(1-x^2)^(1/2) /2 + c

use trig sub to do this problem.

(1-x^2)^1/2 dx It is in a form of (a^2-x^2)^1/2

Let x=a sinx where a=1

So:

x= sin(u) dx=cos(u)du (1-x^2)^1/2 = cosx (this happens when you
substitute x and using trig identities

cos(u)(cos(u))du = (cos(u))^2 dx

(cos(u))^2= (1+cos2u)/2 (an trig identity)

integrate (1+cos2u)/2= 1/2 du + (cos2u)/2du

0.5 u + 0.25 sin 2u + C
0.5u + 0.25 sin(u)cos(u) +C (trig identity)

remember that x=sin(u) solve for u:
u= arcsin x

So 0.5arcsinx + 0.25(x/1)(1-x^2)^1/2 +C (remember you have a triangle to represent x=sin(u) where hypotenuse is 1 and the opposite side is x and the adjacent (1-x^2)^1/2

so your answer is: 0.5 arcsin x + (0.25x)(1-x^2)^1/2 + C

well the integral to that function is one you just have to memorize (ie the integral of 1/(x^2+1) is arctanx). arcsin(x) is the integral of the function you gave.

here's a video if you need more help: http://video.google.com/videoplay?docid=-2191970111156304404