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User Lucaboni
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Hey there :)

Please check the attached image for answer with explanation.


Benjemin360 :)

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User Shalni
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Integrate by parts to find some useful recurrences relating Iₙ and Jₙ.


\displaystyle I_n = \int_(-\pi)^\pi x^n \cos(x) \, dx = uv \bigg|_(x=-\pi)^(x=\pi) - \int_(-\pi)^\pi v\, du \\\\ \implies I_n = -n \int_(-\pi)^\pi x^(n-1) \sin(x) \, dx \\\\ \implies I_n = -n J_(n-1)

where u = xⁿ and dv = cos(x) dx.


\displaystyle J_n = \int_(-\pi)^\pi x^n \sin(x) \, dx = uv \bigg|_(x=-\pi)^(x=\pi) + \int_(-\pi)^\pi v\, du \\\\ \implies J_n = \pi^n - (-\pi)^n + n \int_(-\pi)^\pi x^(n-1) \cos(x) \, dx \\\\ \implies J_n = \pi^n - (-\pi)^n + n I_(n-1)

The integrals for n = 0 are trivial:


\displaystyle I_0 = \int_(-\pi)^\pi \cos(x) \, dx = \sin(\pi) - \sin(-\pi) = 0


\displaystyle J_0 = \int_(-\pi)^\pi \sin(x) \, dx = -\cos(\pi) - (-\cos(-\pi)) = 0

Now, the integral we're interested in is


\displaystyle \int_(-\pi)^\pi x^n f(x) \cos(x) \, dx

but we know f(x) is quadratic, and we want to find its coefficients a, b, and c such that


\displaystyle \int_(-\pi)^\pi x^n (ax^2+bx+c) \cos(x) \, dx

But this is simply


\displaystyle \int_(-\pi)^\pi (ax^(n+2)+bx^(n+1)+cx^n) \cos(x) \, dx = aI_(n+2) + bI_(n+1) + cI_n

Using the recurrences above and the initial values we've computed, we find

• I₁ = -J₀ = 0

• J₁ = π - (-π) + I₀ = 2π

• I₂ = -2 J₁ = -4π

• J₂ = π² - (-π)² + 2 I₁ = 0

• I₃ = -3 J₂ = 0

• J₃ = π³ - (-π)³ + 3 I₂ = 2π³ - 12π

• I₄ = -4 J₃ = -8π³ + 48π

When n = 0, the integral we care about is


aI_2 + bI_1 + cI_0 = -4\pi a + 0 + 0 = 4\pi \implies a = -1

When n = 1,


aI_3 + bI_2 + cI_1 = 0 - 4\pi b + 0 = 4\pi \implies b = -1

When n = 2,


aI_4 + bI_3 + cI_2 = (48\pi - 8\pi^3)a + 0 - 4\pi c = 4\pi \implies c = 2\pi^2 - 13

so that


f(x) = \boxed{-x^2 - x + 2\pi^2 - 13}

User Willert
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