If linear functions f(x) and g(x) satisfy int | vedclass

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(D) Let $I = \int[(3x-1) \cos x + (1-2x) \sin x] dx$. Using integration by parts,$\int u dv = uv - \int v du$: For $\int (3x-1) \cos x dx$: Let $u = 3

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$g(x) = 3(x-1)$

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(D) Let $I = \int[(3x-1) \cos x + (1-2x) \sin x] dx$. Using integration by parts,$\int u dv = uv - \int v du$: For $\int (3x-1) \cos x dx$: Let $u = 3x-1, dv = \cos x dx \Rightarrow du = 3 dx, v = \sin x$. $\int (3x-1) \cos x dx = (3x-1) \sin x - \int 3 \sin x dx = (3x-1) \sin x + 3 \cos x$. For $\int (1-2x) \sin x dx$: Let $u = 1-2x, dv = \sin x dx \Rightarrow du = -2 dx, v = -\cos x$. $\int (1-2x) \sin x dx = (1-2x)(-\cos x) - \int (-\cos x)(-2) dx = (2x-1) \cos x - 2 \sin x$. Adding these: $I = (3x-1) \sin x + 3 \cos x + (2x-1) \cos x - 2 \sin x + C$. $I = (3x-1-2) \sin x + (3+2x-1) \cos x + C = (3x-3) \sin x + (2x+2) \cos x + C$. Comparing with $f(x) \cos x + g(x) \sin x + C$,we get $f(x) = 2x+2$ and $g(x) = 3x-3 = 3(x-1)$.

If linear functions $f(x)$ and $g(x)$ satisfy $\int[(3x-1) \cos x + (1-2x) \sin x] dx = f(x) \cos x + g(x) \sin x + C$,then:

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