How is the integral 2π∫0πx2cos⁡(nx)dx=4(−1)nn2? I thought it would be this : 2π∫0πx2cos⁡(nx)dx=2π∫0πx2(−1)n=2π(−1)n∫0πx2=2π(−1)n[x33]0π=2(−1)n3π3 But it is actually 2π∫0πx2cos⁡(nx)dx=4(−1)nn2

Hint: Two times integrating by parts we get ∫0πx2cos⁡(nx)dx=π2sin⁡(πn)n2+2ncos⁡(πn)π−2sin⁡(πn)n3

Resolved: "How is the integral 2π∫0πx2cos⁡(nx)dx=4(−1)nn2? I thought it..."

elbluffz1

2022-01-25

How is the integral

\frac{2}{π} \int_{0}^{π} x^{2} \cos (n x) d x = \frac{4 {(- 1)}^{n}}{n^{2}}

?
I thought it would be this :

\frac{2}{π} \int_{0}^{π} x^{2} \cos (n x) d x = \frac{2}{π} \int_{0}^{π} x^{2} {(- 1)}^{n} = \frac{2}{π} {(- 1)}^{n} \int_{0}^{π} x^{2} = \frac{2}{π {(- 1)}^{n}} {[\frac{x^{3}}{3}]}_{0}^{π} = \frac{2 {(- 1)}^{n}}{3 π^{3}}

But it is actually

\frac{2}{π} \int_{0}^{π} x^{2} \cos (n x) d x = \frac{4 {(- 1)}^{n}}{n^{2}}

Georgia Ingram

Beginner2022-01-26Added 11 answers

You cannot say that

\cos n x = {(- 1)}^{n}

. This is valid for

x = π

. A counterexample would be

\cos n \cdot 0 = 1

, and this doesnt

Eliza Norris

Beginner2022-01-27Added 15 answers

Hint: Two times integrating by parts we get

\int_{0}^{π} x^{2} \cos (n x) d x = \frac{π^{2} \sin (π n) n^{2} + 2 n \cos (π n) π - 2 \sin (π n)}{n^{3}}

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