If f(x) = intlimits_0^x {2(cos^2 3t + 3sin^2 | vedclass

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(A) Given $f(x) = \int_{0}^{x} (2\cos^2 3t + 3\sin^2 3t) dt$.We know that $f(x+\pi) = \int_{0}^{x+\pi} (2\cos^2 3t + 3\sin^2 3t) dt$.Using the propert

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$f(x) + f(\pi)$

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(A) Given $f(x) = \int_{0}^{x} (2\cos^2 3t + 3\sin^2 3t) dt$.We know that $f(x+\pi) = \int_{0}^{x+\pi} (2\cos^2 3t + 3\sin^2 3t) dt$.Using the property of definite integrals,$f(x+\pi) = \int_{0}^{x} (2\cos^2 3t + 3\sin^2 3t) dt + \int_{x}^{x+\pi} (2\cos^2 3t + 3\sin^2 3t) dt$.This simplifies to $f(x+\pi) = f(x) + \int_{x}^{x+\pi} (2\cos^2 3t + 3\sin^2 3t) dt$.The integrand $g(t) = 2\cos^2 3t + 3\sin^2 3t$ is periodic with period $T = \frac{\pi}{3}$.Since the interval of integration $[x, x+\pi]$ has length $\pi$,which is a multiple of the period $T = \frac{\pi}{3}$ (specifically $\pi = 3 	imes \frac{\pi}{3}$),the integral over any interval of length $\pi$ is equal to the integral over $[0, \pi]$.Thus,$\int_{x}^{x+\pi} g(t) dt = \int_{0}^{\pi} g(t) dt = f(\pi)$.Therefore,$f(x+\pi) = f(x) + f(\pi)$.

If $f(x) = \int\limits_0^x {2(\cos^2 3t + 3\sin^2 3t)dt}$,then $f(x + \pi)$ is equal to

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