The value of int_{0}^{2pi} [sin 2x(1 + cos 3x | vedclass

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(D) Let $I = \int_{0}^{2\pi} [\sin 2x(1 + \cos 3x)] \,dx$.Using the property $\int_{0}^{a} f(x) \,dx = \int_{0}^{a} f(a-x) \,dx$,we observe that the i

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$-\pi$

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(D) Let $I = \int_{0}^{2\pi} [\sin 2x(1 + \cos 3x)] \,dx$.Using the property $\int_{0}^{a} f(x) \,dx = \int_{0}^{a} f(a-x) \,dx$,we observe that the integrand $f(x) = [\sin 2x(1 + \cos 3x)]$ satisfies $f(2\pi - x) = [\sin(4\pi - 2x)(1 + \cos(6\pi - 3x))] = [-\sin 2x(1 + \cos 3x)]$.Using the property $\int_{0}^{a} [f(x) + f(a-x)] \,dx$,we know that for any $x$ such that $\sin 2x(1 + \cos 3x)$ is not an integer,$[t] + [-t] = -1$.Since the function is not an integer almost everywhere in the interval $[0, 2\pi]$,we have:$2I = \int_{0}^{2\pi} ( [\sin 2x(1 + \cos 3x)] + [-\sin 2x(1 + \cos 3x)] ) \,dx$$2I = \int_{0}^{2\pi} -1 \,dx = -2\pi$.Therefore,$I = -\pi$.

The value of $\int_{0}^{2\pi} [\sin 2x(1 + \cos 3x)] \,dx$,where $[t]$ denotes the greatest integer function,is

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