Show that


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. $\sin x+x \cos x$


$f(x)=\int_{0}^{x} t \sin t d t$

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.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now