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For n > 0, ∫x sin^2n x/(sin^2n x + cos^2n x) (for x → 0, 2pi) dx ...
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For n > 0, ∫x sin^2n x/(sin^2n x + cos^2n x) (for x → 0, 2pi) dx ...
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calculate-A-n-0-pi-2-sin-n-x-sin-nx-dx- – Tinku Tara

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λ=nω05(sin2nπ−sinnπ+sin2nπ) =nπ20sin(2nπ) | Chegg.com
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