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

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

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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 evaluate $g(x + \pi) = \int_0^{x + \pi} \cos^4 t \, dt$. Using the property of definite integrals,we can split the interval: $g(x + \pi) = \int_0^{\pi} \cos^4 t \, dt + \int_{\pi}^{x + \pi} \cos^4 t \, dt$. By definition,the first integral is $g(\pi)$. For the second integral,let $t = u + \pi$,then $dt = du$. When $t = \pi$,$u = 0$,and when $t = x + \pi$,$u = x$. Since $\cos^4(u + \pi) = (-\cos u)^4 = \cos^4 u$,the second integral becomes $\int_0^x \cos^4 u \, du = g(x)$. Therefore,$g(x + \pi) = g(\pi) + g(x)$.

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

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