If $\int\limits_e^x t f(t) d t=\sin x-x \cos x-\frac{x^2}{2}$ for all $x \in R-\{0\}$, then the value of $f\left(\frac{\pi}{6}\right)$ will be equal to

$-\frac{1}{2}$

We have,

$\int\limits_e^x t f(t) d t=\sin x-x \cos x-\frac{x^2}{2}$

Differentiating both sides w.r. to $x$, we get

$x f(x)=\cos x-\cos x+x \sin x-x$

$\Rightarrow f(x)=\sin x-1 \Rightarrow f\left(\frac{\pi}{6}\right)=\frac{1}{2}-1=-\frac{1}{2}$