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Mathematical Integration.. Can anyone identify the fault here.. :)

∫[-2,2] (x^3 cos((x/2)+(1/2))) * √(4-x^2) dx  

Lemme apply integration by parts. Let’s choose:

u = x^3 dv = cos((x/2)+(1/2)) * √(4-x^2) dx

du/dx = 3x^2 v = ∫cos((x/2)+(1/2)) * √(4-x^2) dx

Lets solve v using the substitution u = (x/2)+(1/2):

v = ∫cos(u) * √(4-(2u-1)^2) * 2 du

Let z = 2u-1, so that dz/2 = du:

v = ∫cos((z+1)/2) * √(4-z^2) dz

Let’s evaluate this integral using the substitution z = 2sin(t):

v = ∫cos(sin(t)+1)/2) * √(4-4sin^2(t)) * 2cos(t) dt
= 2 ∫cos(sin(t)+1)/2) * 2cos(t) * 2cos(t) dt
= 8 ∫cos(t)cos(sin(t)/2 + π/4) dt
= 8 ∫cos(t)cos(sin(t)/2)cos(π/4) – sin(t)sin(sin(t)/2)sin(π/4) dt
= 8 ∫cos(t)cos(sin(t)/2)cos(π/4) dt (since sin(π/4) = cos(π/4) = 1/√2 and ∫sin(t)sin(sin(t)/2) dt = 0)
= 4√2 ∫cos(t)cos(sin(t)/2) dt

Assumption: Let w = sin(t)/2, so that dw = cos(t)/2 dt:

v = 8√2 ∫cos(2w) dw
= 8√2 sin(2w)
= 8√2 sin(sin(t)/2)

Therefore:

∫[-2,2] (x^3 cos((x/2)+(1/2))) * √(4-x^2) dx
= uv|[-2,2] – ∫[-2,2] v du/dx dx
= x^3 * 8√2 sin((x/2)+(1/2)) * √(4-x^2)|[-2,2] – ∫[-2,2] 8√2 sin((x/2)+(1/2)) * 3x^2 dx

Evaluating the limits of integration gives:

(-2)^3 * 8√2 sin((-2/2)+(1/2)) * √(4-(-2)^2) – (2)^3 * 8√2 sin((2/2)+(1/2)) * √(4-(2)^2)

= -16√2 sin(3/2) + 16√2 sin(5/2)

2nd integral can be evaluated by substitution of u = (x/2)+(1/2):

∫[-2,2] 8√2 sin((x/2)+(1/2)) * 3x^2 dx
= 24√2 ∫[-2,2] sin(u) * (2u-1)^2 du
= 24√2 ∫[-3/2,5/2] sin(u) * (2u-1)^2 du (since…since (x/2)+(1/2) goes from -3/2 to 5/2 when x goes from -2 to 2.

Now expand (2u-1)^2 to get:

∫[-3/2,5/2] sin(u) * (2u-1)^2 du
= ∫[-3/2,5/2] sin(u) * (4u^2 – 4u + 1) du
= 4 ∫[-3/2,5/2] sin(u) * u^2 du – 4 ∫[-3/2,5/2] sin(u) * u du + ∫[-3/2,5/2] sin(u) du

The third integral is simply -cos(u) evaluated at the limits of integration:

cos(5/2) + cos(3/2)
The first two integrals can be evaluated using integration by parts with:

u = u^2 dv = sin(u) du
du/dx = 1/2 v = -cos(u)

Then:

∫ sin(u) * u^2 du
= -cos(u) * u^2|[-3/2,5/2] + 2 ∫ cos(u) * u du
= cos(5/2) * (25/4) – cos(3/2) * (9/4) + 2 [sin(u) * u – ∫sin(u) du]|[-3/2,5/2]
= (25/4) cos(5/2) – (9/4) cos(3/2) + 2 [sin(u) * u + cos(u)]|[-3/2,5/2]

Finally, plugging in all the evaluated integrals and simplifying gives:

∫[-2,2] (x^3 cos((x/2)+(1/2))) * √(4-x^2) dx
= -16√2 sin(3/2) + 16√2 sin(5/2) + (25/4) cos(5/2) – (9/4) cos(3/2) + 4√2

Therefore, the value of the integral is approximately 8.564

Cheers!
Shiva

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