integrate with limits of 2 and -2 (x^3(cos x/2) +1/2) (sqrt 4-x^2)dx

Answer 1

To integrate the given function with limits of 2 and -2, we first need to find the antiderivative of the function.

Let's denote the function as f(x):

f(x) = x^3(cos(x/2) + 1/2) * √(4 - x^2)

To find the antiderivative of this function, we will need to apply the product rule followed by the chain rule.

By applying integration by parts, the antiderivative of the function f(x) becomes:

∫[x^3(cos(x/2) + 1/2)√(4 - x^2)dx]
0 ∫ [x^3*sin⁡(x/2)√(4 - x^2)dx + 1/2∫√(4 - x^2)dx]
Now, we need to evaluate the indefinite integrals:

1. Integral of x^3*sin⁡(x/2)√(4 - x^2)dx
Let's set u = x^3 and dv = sin(x/2)√(4 - x^2)dx
Then, du = 3x^2 dx and v = -2(4 - x^2)^(1/2)

The integral becomes:
= -2x^3(4 - x^2)^(1/2)sin(x/2) - ∫(-2)(4 - x^2)^(1/2) * 3x^2 dx
= -2x^3(4 - x^2)^(1/2)sin(x/2) + 6∫x^2(4 - x^2)^(1/2) dx

2. Integral of 1/2∫√(4 - x^2)dx
This integral can be solved by using trigonometric substitution. Let x = 2sin(θ), then dx = 2cos(θ) dθ and 4 - x^2 = 4cos^2(θ).

The integral becomes:
1/2∫√(4 - x^2)dx
= 1/2∫2cos(θ) * 2cos(θ) dθ
= 2∫cos^2(θ)dθ
Letting t = sin(θ) and dt = cos(θ)dθ, the integral simplifies to:
= 2∫(1 - t^2) dt
= 2(t - t^3/3) + C
= 2sin(θ) - 2sin^3(θ)/3 + C
Substitute back for θ:
= 2sin(arcsin(x/2)) - 2sin^3(arcsin(x/2))/3 + C
= 2x/2 - 2(x/2)^3/3 + C
= x - x^3/6 + C

Substitute the indefinite integrals back into the original function:

= -2x^3(4 - x^2)^(1/2)sin(x/2) + 6∫x^2(4 - x^2)^(1/2) dx + x - x^3/6 + C

Now, we can evaluate the definite integral with limits of -2 and 2.

= [-2(2)^3(4 - 2^2)^(1/2)sin(2/2) + 6∫(2)^2(4 - 2^2)^(1/2)dx + 2 - 2^3/6]
- [-2(-2)^3(4 - (-2)^2)^(1/2)sin(-2/2) + 6∫(-2)^2(4 - (-2)^2)^(1/2)dx - 2 - (-2)^3/6]

= [-64(0)sin(1) + 6∫16(4)dx + 2 - 8/6]
- [64(0)sin(-1) + 6∫16(4)dx - 2 - 8/6]

= [0 + 6(64)(4)x + 2 - 4/3]
- [0 + 6(64)(4)x - 2 + 4/3]

= [6(64)(4)(2) + 2 - 4/3]
- [6(64)(4)(-2) - 2 + 4/3]

= [384 + 2 - 4/3] - [-384 - 2 + 4/3]
= 382 + 2 - 4/3 + 384 + 2 - 4/3

= 766 - 8/3

= 766 2/3

Therefore, the value of the definite integral with limits of 2 and -2 is 766 2/3.