For any integer n,the value of the integral i | vedclass

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

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(C) Let $I = \int_0^\pi e^{\cos^2 x} \cos^3((2n+1)x) \, dx \quad \dots(1)$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,$\cos((2n+1)(\pi-x)) = \cos((2n+1)\pi - (2n+1)x)$.Since $(2n+1)$ is an odd integer,$\cos((2n+1)\pi - 	heta) = -\cos 	heta$.Therefore,$\cos^3((2n+1)\pi - (2n+1)x) = (-\cos((2n+1)x))^3 = -\cos^3((2n+1)x)$.Substituting these into the integral:$I = \int_0^\pi e^{\cos^2 x} (-\cos^3((2n+1)x)) \, dx$$I = -\int_0^\pi e^{\cos^2 x} \cos^3((2n+1)x) \, dx \quad \dots(2)$Adding $(1)$ and $(2)$,we get:$I + I = 0 \implies 2I = 0 \implies I = 0$.

For any integer $n$,the value of the integral $\int_0^\pi e^{\cos^2 x} \cos^3(2n+1)x \, dx$ is:

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