If I_1 = int_0^{3 pi} f(cos^2 x) dx and I_2 = | vedclass

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(C) Let $g(x) = f(\cos^2 x)$. Since $\cos^2(x + \pi) = (-\cos x)^2 = \cos^2 x$, the function $g(x)$ is periodic with period $\pi$. Using the property

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$I_1 = 3 I_2$

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(C) Let $g(x) = f(\cos^2 x)$. Since $\cos^2(x + \pi) = (-\cos x)^2 = \cos^2 x$, the function $g(x)$ is periodic with period $\pi$. Using the property of definite integrals for periodic functions, $\int_0^{n T} g(x) dx = n \int_0^T g(x) dx$, where $T$ is the period. Here, $T = \pi$ and $n = 3$. Therefore, $I_1 = \int_0^{3 \pi} f(\cos^2 x) dx = 3 \int_0^\pi f(\cos^2 x) dx$. Since $I_2 = \int_0^\pi f(\cos^2 x) dx$, we have $I_1 = 3 I_2$.

If $I_1 = \int_0^{3 \pi} f(\cos^2 x) dx$ and $I_2 = \int_0^\pi f(\cos^2 x) dx$, then

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