Answer
$F(x)=\sin x-\sin 1$
$F(2)\approx 0.0678$
$F(5)\approx-1.8004$
$F(8)\approx 0.1479$
Work Step by Step
Apply The Second Fundamental Theorem of Calculus (Th.4.11 )
If $f$ is continuous on an open interval I containing $a$, then, for every $x$ in the
interval,
$\displaystyle \frac{d}{dx}[\int_{a}^{x}f(t)dt]=f(x)$.
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$F(x)=\displaystyle \int_{1}^{x}\cos\theta d\theta=$
$=[\sin\theta]_{1}^{x}=\sin x-\sin 1$
$F(x)=\sin x-\sin 1$
$F(2)=\sin 2-\sin 1\approx 0.0678$
$F(5)=\sin 5-\sin 1\approx-1.8004$
$F(8)=\sin 8-\sin 1\approx 0.1479$
