Calculus problem....please help?

Question

can someone please integrate the following equation for me:

(4 - x^2)^1/2

Answer 1

Integral (  (4 - x^2)^(1/2) dx )

Let x = 2sin(t).

dx = 2cos(t) dt

Integral (  (4 - 4sin^2(t))^(1/2) 2cos(t) dt )

2 * Integral ( (4 - 4sin^2(t))^(1/2) cos(t) dt )

2 * Integral ( 2(1 - sin^2(t))^(1/2) cos(t) dt )

2 * 2 * Integral ( (1 - sin^2(t))^(1/2) cos(t) dt )

4 * Integral (  [cos^2(t)]^(1/2) cos(t) dt )

4 * Integral ( cos(t) cos(t) dt )

4 * Integral (  cos^2(t) dt )

Use the half angle formula, cos^2(t) = (1/2)(1 + cos(2t))

4 * Integral (  [1/2] (1 + cos(2t)) dt )

4 * [1/2] * Integral ( (1 + cos(2t)) dt )

2 * Integral ( (1 + cos(2t)) dt )

Which we can now integrate easily.

2 * [ t + (1/2)sin(2t) ] + C

2t + sin(2t) + C

Use the double angle identity sin(2t) = 2sin(t)cos(t)

2t + 2sin(t)cos(t) + C

To substitute back in terms of x, remember that

x = 2sin(t).  This means

sin(t) = x/2 = opp/hyp

opp = x

hyp = 2, so by Pythagoras,

adj = sqrt(4 - x^2)

That means sin(t) = opp/hyp = x/2

cos(t) = adj/hyp = sqrt(4 - x^2)/2

Also, t = arcsin(x/2), so this answer:

2t + 2sin(t)cos(t) + C

Changes to this answer:

2arcsin(x/2) + 2[ x/2 ] [ sqrt(4 - x^2) / 2 ] + C

2arcsin(x/2) + (1/2) x sqrt(4 - x^2) + C