I need to calculate the indefinite integral of sqrt(4-x^2). Is there an easy trick to doing this?

Answer 1

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ok, you need to do a trig substitution.

let x = 2cos theta

dx = -2sin theta dtheat

so you get:

[integral] -2sin theta sqrt(4-4cos^2 theta) dtheat

you can factor out a sqrt(4) = 2, and the -2

then 1-cos^2 theta = sin^2 theta

so sqrt(1-cos^2 theta) = sin theta

so you get [integral] -4sin^2 theta d theta

solve that then plug back in for x

Answer 2

No...

Rewrite it as ∫√1-(x/2)² dx

Let sin u = x/2 and du/dx cos u = 1/2

dx = 2 cos u

Our integral is now

2∫√1-sin² u cos u du = 2∫cos² u du

2∫cos² u du = ∫ 1 + cos(2u) du

= u + sin(2u)/2 = u + cos u sin u

from our org. subsitutions

sin u = x/2

u = arcsin (x/2)

cos u = √4-x² --> draw a triangle if you do not get this one

so the final answer is:

arcsin(x/2) + (x/2)√4-x²