Question

If $f\left(x\right)={\int }_0^{x}t\sin tdt$, then $f^{\prime }\left(x\right)$ is

• A. $\cos x+x\sin x$
• B. $x\sin x$
• C. $x\cos x$
• D. $\sin x+x\cos x$

Answer 1

Answer: (B)

$x\sin x$

$f\left(x\right)={\int }_0^{x}t\sin tdt$
Integrating by parts, we obtain
$f\left(x\right)=t{\int }_0^x\sin tdt-{\int }_0^x\{\left(\frac{d}{dt}t\right)\int \sin tdt\}dt$
$=\left[t\right(-\cos t){]}_0^x-{\int }_0^x(-\cos t)dt$
$=[-t\cos t+\sin t{]}_0^x$
$=-x\cos x+\sin x$
$\Rightarrow f^{\prime }\left(x\right)=-[\left\{x\right(-\sin x\left)\right\}+\cos x]+\cos x$
$=x\sin x-\cos x+\cos x$
$=x\sin x$
Hence, the correct Answer is B.