If f(x)=∫0x tsin⁡ tdt, then f′(x) is

f(x)=∫0x tsin tdtIntegrating by parts, we obtainf(x)=t∫0x sin tdt−∫0x ddtt∫sin⁡ tdtdt=[t(−cos⁡t)]0x−∫0x (−cos⁡t)dt=[−tcos⁡t+sin⁡t]0x=−xcos x+sin x⇒f′(x)=−[{x(−sin⁡x)}+cos⁡x]+cos⁡x =xsin⁡x−cos⁡x+cos⁡x =xsin⁡x

If f(x)=∫0x tsin⁡ tdt, then f′(x) is

f(x)=∫0x tsin tdtIntegrating by parts, we obtainf(x)=t∫0x sin tdt−∫0x ddtt∫sin⁡ tdtdt=[t(−cos⁡t)]0x−∫0x (−cos⁡t)dt=[−tcos⁡t+sin⁡t]0x=−xcos x+sin x⇒f′(x)=−[{x(−sin⁡x)}+cos⁡x]+cos⁡x =xsin⁡x−cos⁡x+cos⁡x =xsin⁡x

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If $f (x) = \int_{0}^{x} t \sin t d t$ , then $f^{'} (x)$ is

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$\sin x + x \cos x$

$x \sin x$

$\cos x + x \sin x$

$x \cos x$

answer is B.

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Detailed Solution

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

Integrating by parts, we obtain

$\begin{array}{l} f (x) = t \int_{0}^{x} \sin t d t - \int_{0}^{x} \{(\frac{d}{d t} t) \int \sin t d t\} d t \\ = [t (- \cos t)]_{0}^{x} - \int_{0}^{x} (- \cos t) d t \\ = [- t \cos t + \sin t]_{0}^{x} \\ = - x \cos x + \sin x \end{array}$