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.