If n is any integer,then int_0^pi {e^{cos^2 x | vedclass

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(C) Let $I = \int_0^\pi {e^{\cos^2 x} \cos^3((2n + 1)x)} \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have: $I = \int

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(C) Let $I = \int_0^\pi {e^{\cos^2 x} \cos^3((2n + 1)x)} \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have: $I = \int_0^\pi {e^{\cos^2(\pi - x)} \cos^3((2n + 1)(\pi - x))} \, dx$. Since $\cos(\pi - x) = -\cos x$,we have $\cos^2(\pi - x) = \cos^2 x$. Also,$\cos((2n + 1)(\pi - x)) = \cos((2n + 1)\pi - (2n + 1)x) = \cos((2n + 1)\pi) \cos((2n + 1)x) + \sin((2n + 1)\pi) \sin((2n + 1)x)$. Since $(2n + 1)$ is an odd integer,$\cos((2n + 1)\pi) = -1$ and $\sin((2n + 1)\pi) = 0$. Thus,$\cos((2n + 1)(\pi - x)) = -\cos((2n + 1)x)$. Therefore,$\cos^3((2n + 1)(\pi - x)) = (-\cos((2n + 1)x))^3 = -\cos^3((2n + 1)x)$. Substituting this back into the integral,we get $I = \int_0^\pi e^{\cos^2 x} (-\cos^3((2n + 1)x)) \, dx = -I$. $2I = 0 \implies I = 0$.

If $n$ is any integer,then $\int_0^\pi {e^{\cos^2 x} \cos^3((2n + 1)x)} \, dx = $

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