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If $f(x)=\int_0^x t \sin t, \operatorname{then} f^{\prime}(x)$ is
(a) $\cos x+x \sin x$
(b) $x \sin x$
(c) $x \cos x$
(d) $\sin x+x \cos x$
If $f(x)=\int_0^x t \sin t, \operatorname{then} f^{\prime}(x)$ is
(a) $\cos x+x \sin x$
(b) $x \sin x$
(c) $x \cos x$
(d) $\sin x+x \cos x$
Solution:
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Verified Answer
(b) $f(x)=\int_0^x t \sin t d t$
$\left.=t(-\cos t)-\int 1 \cdot(-\cos t) d t\right]_0^x$
$=-\mathrm{t} \cos \mathrm{t}+\sin \mathrm{t}]_0^x=-x \cos x+\sin x$
$\left.=t(-\cos t)-\int 1 \cdot(-\cos t) d t\right]_0^x$
$=-\mathrm{t} \cos \mathrm{t}+\sin \mathrm{t}]_0^x=-x \cos x+\sin x$
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