If f x = ∫ 0 x t sin x - sin t d t , then

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

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

On differentiating, we get

$\Rightarrow f^{\text{'}}\left(x\right)=\left(\sin x\right)x+\cos x{\int }_0^{x}tdt-x\sin x$

$\Rightarrow f^{\text{'}}\left(x\right)=\cos x{\int }_0^{x}tdt$

Differentiating the above equation again, we get

$\Rightarrow f^{\text{'}\text{'}}\left(x\right)=\left(\cos x\right)x-\left(\sin x\right){\int }_0^{x}tdt$

$\Rightarrow f^{\text{'}\text{'}\text{'}}\left(x\right)=x\left(-\sin x\right)+\cos x-\left(\sin x\right)x-\left(\cos x\right){\int }_0^{x}tdt$

$\Rightarrow f^{\text{'}\text{'}\text{'}}\left(x\right)+f^{\text{'}}\left(x\right)=\cos x-2x\sin x$.