If frac{d}{dx}f(x) = xcos x + sin x and f(0) | vedclass

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(C) Given $\frac{d}{dx}f(x) = x\cos x + \sin x$. Integrating both sides with respect to $x$: $f(x) = \int (x\cos x + \sin x) dx$. Using integration by

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$x\sin x + 2$

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(C) Given $\frac{d}{dx}f(x) = x\cos x + \sin x$. Integrating both sides with respect to $x$: $f(x) = \int (x\cos x + \sin x) dx$. Using integration by parts for $\int x\cos x dx$: $\int x\cos x dx = x\sin x - \int (1 \cdot \sin x) dx = x\sin x - (-\cos x) = x\sin x + \cos x$. Thus,$f(x) = (x\sin x + \cos x) + \int \sin x dx + C$. $f(x) = x\sin x + \cos x - \cos x + C = x\sin x + C$. Given $f(0) = 2$,we substitute $x = 0$: $f(0) = 0 \cdot \sin(0) + C = 2 \Rightarrow C = 2$. Therefore,$f(x) = x\sin x + 2$.

If $\frac{d}{dx}f(x) = x\cos x + \sin x$ and $f(0) = 2$,then $f(x) = $

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