Question:
If $f(x)=\int_{0}^{x} t \sin t d t$, 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:
$f(x)=\int_{0}^{x} t \sin t d t$
Integrating by parts, we obtain
$f(x)=t \int_{0}^{x} \sin t d t-\int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \int \sin t d t\right\} d t$
$=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cos t) d t$
$=[-t \cos t+\sin t]_{0}^{x}$
$=-x \cos x+\sin x$
$\Rightarrow f^{\prime}(x)=-[\{x(-\sin x)\}+\cos x]+\cos x$
$=x \sin x-\cos x+\cos x$
$=x \sin x$
Hence, the correct answer is B.