MCQ
If $f(x)=\int_{0}^{x} t \sin t \,d t,$ then $f^{\prime}(x)$ is
- A$\cos \,x+x \sin\, x$
- ✓$x\, \sin \,x$
- C$x\, \cos \,x$
- D$\sin\, x+x \,\cos \,x$
✓
Answer
Correct option: B.
$x\, \sin \,x$
b
$f(x)=\int_{0}^{x} t \sin t d t$
$f(x)=\int_{0}^{x} t \sin t d t$
Integrating by parts, we obtain
$f(x)=t \int_{0}^{x} \sin t d t-\int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \int \sin t \, d t\right\} d t$
$=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cot t) d t$
$=[-t \cos t+\sin t]_{0}^{x}$
$=-x \cos x+\sin x$
$\Rightarrow f^{\prime}(x)=-[\{x(-\sin x)\}+\cos x]+\cos x$
$=x \sin x-\cos x+\cos x$
$=x \sin x$
Hence, the correct Answer is $B$
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