The value of int_0^pi {{e^{{{cos }^2}x}}{{cos | vedclass

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(C) Let $I = \int_0^\pi e^{\cos^2 x} \cos^5(3x) \,dx$.Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get:$I = \int_0^\pi e^{\cos^2(

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(C) Let $I = \int_0^\pi e^{\cos^2 x} \cos^5(3x) \,dx$.Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get:$I = \int_0^\pi e^{\cos^2(\pi-x)} \cos^5(3(\pi-x)) \,dx$Since $\cos(\pi-x) = -\cos x$,we have $\cos^2(\pi-x) = \cos^2 x$.Also,$\cos(3(\pi-x)) = \cos(3\pi - 3x) = \cos(3\pi) \cos(3x) + \sin(3\pi) \sin(3x) = -\cos(3x)$.Therefore,$\cos^5(3(\pi-x)) = (-\cos(3x))^5 = -\cos^5(3x)$.Substituting these into the integral:$I = \int_0^\pi e^{\cos^2 x} (-\cos^5(3x)) \,dx = -\int_0^\pi e^{\cos^2 x} \cos^5(3x) \,dx = -I$.Thus,$I = -I$,which implies $2I = 0$,so $I = 0$.

The value of $\int_0^\pi {{e^{{{\cos }^2}x}}{{\cos }^5}3x} \,dx$ is

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