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We use the technique of trigonometric substitution.

∫ √(1-x^2) dx , { let x= sinΦ ,then dx= cosΦ dΦ

∫ √(1-sin^2Φ ) cosΦ dΦ =

Substitute 1-sin^2Φ = cos^2Φ

∫ √(cos^2Φ ) cosΦ dΦ =

∫ cosΦ cosΦ dΦ =

∫ cos^2Φ dΦ =

Substitute cos^2Φ= (1+cos(2Φ ) )/2

∫ (1+cos(2Φ ) )/2 dΦ =

∫ [1/2+cos(2Φ)/2 ]dΦ =

Φ/2 + sin(2Φ)/4 +C =

Substitue sin2Φ= 2sinΦcosΦ

Φ/2 + sinΦcosΦ/2 + C

Now draw a triangle with radius 1,and opposite x. ( because sinΦ=x/1 ). Then adjacent is √(1-x^2) and Φ=arcsin(x).

Φ/2 + sinΦcosΦ/2 + C =

arcsin(x)/2 + x√(1-x^2) /2 + C =

(1/2) ( arcsin(x) + x√(1-x^2) ) + C

This is correct: http://www.wolframalpha.com/input/?i=%E2%88%AB+%E2...

Let, x=sinα

Differentiating it dx=cosαdα

Now we put the value of x and dx in the given equation

∫▒cos2αdα

=½∫▒2cos2αdα

=1/2∫▒〖(1+cos2α)〗

=1/2+1/4sin2α+k

Now we get

½+1/4sin(2sin-1x)+k

Ans: ½+1/4sin(2sin-1x)+k

use trig-sub and the u * v - int v * du

first use the trig-sub via [(a^2) - (x^2)]^n dx

a = a; x = a* sin(θ); dx = a*cos(θ) dθ

as a = 1, x = sin(θ); dx = cos(θ) dθ

int sqrt[(1^2) - {[sin(θ)]^2}] * cos(θ) dθ

int sqrt{[cos(θ)]^2} * cos(θ) dθ

int [cos(θ)]^2 dθ

int (1) - [sin(θ)]^2 dθ

u = -sin(θ); dv = sin(θ)

du = -cos(θ) dθ; v = -cos(θ)

θ + sin(θ) * cos(θ) - int [cos(θ)]^2 dθ = int [cos(θ)]^2 dθ

θ + sin(θ) * cos(θ) = 2 * int [cos(θ)]^2 dθ

{(θ) + [sin(θ)] * [cos(θ)]} / 2

Now use the Pythagorean Theorem

sin = opp/hyp; sin = x / 1; adjacent = sqrt[(1) - (x^2)]; opposite = x; hypotenuse = 1

sin(θ) = x; θ = arcsin(x)

{[arcsin(x)] + (x) * sqrt[(1) - (x^2)]} / 2

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