For any integer n, the integral int_0^pi {{e^ | vedclass

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

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(B) Let $I = \int_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}(2n + 1)x\,dx}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \int_0^\pi {{e^{{{\sin }^2}(\pi-x)}}{{\cos }^3}(2n + 1)(\pi-x)\,dx}$.Since $\sin(\pi-x) = \sin x$,we have $\sin^2(\pi-x) = \sin^2 x$.Also,$\cos((2n+1)(\pi-x)) = \cos((2n+1)\pi - (2n+1)x) = -\cos((2n+1)x)$ because $(2n+1)$ is an odd integer.Therefore,$\cos^3((2n+1)(\pi-x)) = (-\cos((2n+1)x))^3 = -\cos^3((2n+1)x)$.Substituting these into the integral:$I = \int_0^\pi {{e^{{{\sin }^2}x}} \cdot [-\cos^3(2n + 1)x]\,dx} = -I$.Thus,$2I = 0$,which implies $I = 0$.

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

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