If P = int_0^{3pi} f(cos^2 x) dx and Q = int_ | vedclass

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(C) Given $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$. Using the property of definite integrals,$\int_0^{na} f(x) dx = n

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$P - 3Q = 0$

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(C) Given $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$. Using the property of definite integrals,$\int_0^{na} f(x) dx = n \int_0^a f(x) dx$ if $f(x+a) = f(x)$. Here,$f(\cos^2(x+\pi)) = f(\cos^2 x)$ because $\cos^2(x+\pi) = (-\cos x)^2 = \cos^2 x$. Thus,the function $f(\cos^2 x)$ is periodic with period $\pi$. Therefore,$P = \int_0^{3\pi} f(\cos^2 x) dx = 3 \int_0^{\pi} f(\cos^2 x) dx = 3Q$. This implies $P = 3Q$,or $P - 3Q = 0$.

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

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