Question

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

• A. $\cos x+x\sin x.$
• B. $\sin x-x\cos x$
• C. $x\cos x.$
• D. None of these

Answer 1

The correct option is B $\sin x-x\cos x$

$f\left(x\right)={\int }_0^{x}t\sin tdt$

Now using integration by parts-

$f\left(x\right)={[t\int \sin t-\int \frac{d}{dt}t(\int \sin t)dt]}_0^x$

$f\left(x\right)={[-t\cos t-\int (-\cos t\left)dt\right]}_0^x$

$f\left(x\right)={[-t\cos t+\sin t]}_0^x$

$f\left(x\right)=\sin x-x\cos x$

Hence, answer is option-(B).