Solve Integral: \int_{-n \pi /2}^{n \pi / 2} y^2[1 + \cos(2y)] dy

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SUMMARY

The integral \(\int_{-n \pi /2}^{n \pi / 2} y^2[1 + \cos(2y)] dy\) can be solved by distributing \(y^2\) into two separate integrals: \(\int_{-n \pi /2}^{n \pi / 2} y^2 dy\) and \(\int_{-n \pi /2}^{n \pi / 2} y^2 \cos(2y) dy\). The first integral evaluates to \(\frac{(n \pi)^3}{12}\) for odd \(n\). The second integral requires integration by parts applied twice, leading to a solvable expression involving sine and cosine functions.

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SoggyBottoms
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I have the following integral, but I don't know how to solve it: \int_{-n \pi /2}^{n \pi / 2} y^2[1 + \cos(2y)] dy, with n = 1, 3, 5... Any ideas?
 
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Could you distribute the y^2? The first term would just be integral of y^2 and the second you could use Int by parts twice.
 

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