If f(x) = int x^2 cos^2 x (2x tan^2 x - 2x - | vedclass

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(C) Given $f(x) = \int x^2 \cos^2 x (2x 	an^2 x - 2x - 6 	an x) dx$. Expanding the integrand: $f(x) = \int (2x^3 \sin^2 x - 2x^3 \cos^2 x - 6x^2 \si

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$-x^3 \sin 2x + \pi$

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(C) Given $f(x) = \int x^2 \cos^2 x (2x 	an^2 x - 2x - 6 	an x) dx$. Expanding the integrand: $f(x) = \int (2x^3 \sin^2 x - 2x^3 \cos^2 x - 6x^2 \sin x \cos x) dx$. Using trigonometric identities $\sin^2 x - \cos^2 x = -\cos 2x$ and $2 \sin x \cos x = \sin 2x$: $f(x) = \int (-2x^3 \cos 2x - 3x^2 \sin 2x) dx$. Let $I = \int (-2x^3 \cos 2x - 3x^2 \sin 2x) dx$. Using integration by parts on the first term $\int u dv = uv - \int v du$ where $u = -x^3$ and $dv = 2 \cos 2x dx$: $I = -x^3 \sin 2x - \int (-\sin 2x)(3x^2) dx - \int 3x^2 \sin 2x dx$. $I = -x^3 \sin 2x + \int 3x^2 \sin 2x dx - \int 3x^2 \sin 2x dx$. $I = -x^3 \sin 2x + C$. Given $f(0) = \pi$,we have $-(0)^3 \sin(0) + C = \pi$,which implies $C = \pi$. Therefore,$f(x) = -x^3 \sin 2x + \pi$.

If $f(x) = \int x^2 \cos^2 x (2x \tan^2 x - 2x - 6 \tan x) dx$ and $f(0) = \pi$,then $f(x) =$

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