If g(x) = int_0^x cos^4 t ,dt, then g(x+pi) e | vedclass

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(A) Given $g(x) = \int_0^x \cos^4 t \,dt$. We need to find $g(x+\pi) = \int_0^{x+\pi} \cos^4 t \,dt$. Using the property of definite integrals, $\int_

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$g(x) + g(\pi)$

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(A) Given $g(x) = \int_0^x \cos^4 t \,dt$. We need to find $g(x+\pi) = \int_0^{x+\pi} \cos^4 t \,dt$. Using the property of definite integrals, $\int_0^{x+\pi} f(t) \,dt = \int_0^x f(t) \,dt + \int_x^{x+\pi} f(t) \,dt$. Thus, $g(x+\pi) = g(x) + \int_x^{x+\pi} \cos^4 t \,dt$. Since $\cos^4 t$ is a periodic function with period $\pi$, the integral over any interval of length $\pi$ is equal to the integral over $[0, \pi]$. Therefore, $\int_x^{x+\pi} \cos^4 t \,dt = \int_0^\pi \cos^4 t \,dt = g(\pi)$. Hence, $g(x+\pi) = g(x) + g(\pi)$.

If $g(x) = \int_0^x \cos^4 t \,dt$, then $g(x+\pi)$ equals

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