The given integral can be solved using numerical calculation methods. The integral by part formula also would be applicable in the given integral.
The value of the given integral is [tex]3.14\rm\: or\: \pi[/tex].
Given:
Write the given integral.
[tex]\int_{-2}^{2}\sqrt{4-x^2}\left(\cos\left(\dfrac{x}{2}\right)x^3+\dfrac{1}{2}\right)\\[/tex]
Write the formula for the integration by part formula.
[tex]\int u.v=u\int vdx-\int u'(\int vdx)dx[/tex]
Let,
[tex]u=\sqrt{4-x^2}[/tex]
[tex]v=\left (x^3\cos\dfrac{x}{2} +\dfrac{1}{2} \right)[/tex]
Now put the value in above formula and integrate.
[tex]\begin{aligned}\int_{-2}^{2}\sqrt{4-x^2}\left(\cos\left(\dfrac{x}{2}\right)x^3+\dfrac{1}{2}\right)&=\left [\dfrac{\frac{x\sqrt{4-x^2}}{2}+2\arcsin\left(\frac{x}{2}\right)}{2}\right]_{-2}^{2}\\&=\left [\dfrac{x\sqrt{4-x^2}}{4}+\arcsin\left(\dfrac{x}{2}\right)\right]_{-2}^{2}\\\end[/tex]
Now, further simplify the expression.
[tex]\int_{-2}^{2}\sqrt{4-x^2}\left(\cos\left(\dfrac{x}{2}\right)x^3+\dfrac{1}{2}\right)\\=\left [\dfrac{2\sqrt{4-2^2}}{4}+\arcsin\left(\dfrac{2}{2}\right)\right]-\left [\dfrac{2\sqrt{4-(-2)^2}}{4}+\arcsin\left(\dfrac{-2}{2}\right)\right]\\=1.57-(-1.57)\\=3.14[/tex]
The value of the given integral is [tex]3.14\rm\: or\: \pi[/tex].
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