If int(3t^2 sin frac{1}{t} - t cos frac{1}{t} | vedclass

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(C) Let $I = \int(3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}) dt$. We use the substitution $u = \frac{1}{t}$,so $du = -\frac{1}{t^2} dt$,which implies

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$8$

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(C) Let $I = \int(3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}) dt$. We use the substitution $u = \frac{1}{t}$,so $du = -\frac{1}{t^2} dt$,which implies $dt = -t^2 du = -\frac{1}{u^2} du$. The integral becomes: $I = \int(3(\frac{1}{u})^2 \sin u - \frac{1}{u} \cos u) (- \frac{1}{u^2}) du$ $I = \int(-\frac{3}{u^4} \sin u + \frac{1}{u^3} \cos u) du$. This does not simplify easily,so let's try to express the integrand as a derivative. Consider $f(t) = t^3$. Then $\frac{d}{dt} (t^3 \sin \frac{1}{t}) = 3t^2 \sin \frac{1}{t} + t^3 \cos(\frac{1}{t}) \cdot (-\frac{1}{t^2}) = 3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}$. Thus,$\int(3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}) dt = t^3 \sin \frac{1}{t} + c$. Comparing this with $f(t) \sin(\frac{1}{t}) + c$,we get $f(t) = t^3$. Therefore,$f(2) = 2^3 = 8$.

If $\int(3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}) dt = f(t) \sin(\frac{1}{t}) + c$,then $f(2) =$

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