How do you integrate

square root of (4-x^2) ?

Pleasee help!

put x = 2 sin(t)

then dx = 2 cos(t) dt

sqrt[4 - x^2] dx = 2 cos^2(t) dt

Then, you use:

cos^2(t) = 1/2[cos(2t) + 1]

Correection:

sqrt[4 - x^2] dx = 4 cos^2(t) dt

To integrate the square root of (4 - x^2), you can use trigonometric substitution.

First, let's express (4 - x^2) in terms of trigonometric functions. Recognize that (4 - x^2) can be rewritten as (2^2 - x^2). This resembles the identity for the difference of squares, which can be factored as (2 - x)(2 + x).

Next, let's substitute x with 2sinθ, where θ is the angle. This substitution is valid because the derivative of sinθ gives us dx.

Thus, (2 - x)(2 + x) becomes (2 - 2sinθ)(2 + 2sinθ), which simplifies to 4(1 - sin^2θ). Using the Pythagorean identity sin^2θ + cos^2θ = 1, we get 4cos^2θ.

Now, we can integrate the expression √(4 - x^2) using the trigonometric substitution:

∫√(4 - x^2) dx
= ∫2cosθ * |dx|
= 2∫cosθ * |dx|

To determine |dx| in terms of dθ, we differentiate x = 2sinθ with respect to θ, which gives us dx = 2cosθ dθ. Therefore, |dx| = 2cosθ dθ.

Substituting |dx| = 2cosθ dθ into the integral expression, we have:

= 2∫cosθ * 2cosθ dθ
= 4∫cos^2θ dθ

Integrating cos^2θ can be done using the formula for the integral of cos^2θ, which is (∫cos^2θ dθ = (θ/2) + (sin2θ/4) + C). So, applying this formula to our integral, we get:

= 4((θ/2) + (sin2θ/4)) + C

Finally, substitute back θ = sin⁻¹(x/2) into the expression:

= 4((sin⁻¹(x/2)/2) + (sin2(sin⁻¹(x/2))/4)) + C

Note: C represents the constant of integration, and sin⁻¹(x/2) denotes the inverse sine function.