If ∫_e^x t f(t) d t=sin x-x cos x- x^2 2, for all x ∈ R-0, then the value of f ( π 6 ) is

Let $\displaystyle \int_e^x t f(t) d t=\sin x-x \cos x-\frac{x^2}{2}$∫ex​tf(t)dt=sinx−xcosx−2x2​ By using Leibnitz rule, we get $$\begin{aligned} & \frac{d}{d x}\left[\int_e^x t f(t) d t\right]=\frac{d}{d x}\left[\sin x-x \cos x-\frac{x^2}{2}\right] \\ & \Rightarrow x f(x)-e f(e) \cdot 0=x \sin x-x \end{aligned}$$​dxd​[∫ex​tf(t)dt]=dxd​[sinx−xcosx−2x2​]⇒xf(x)−ef(e)⋅0=xsinx−x​ Now, put $x=\frac{\pi}{6}$x=6π​, we get $$\begin{aligned} & \frac{\pi}{6} \cdot f\left(\frac{\pi}{6}\right)=\frac{\pi}{6} \cdot \sin \frac{\pi}{6}-\frac{\pi}{6} \\ & \Rightarrow f\left(\frac{\pi}{6}\right)=\sin \frac{\pi}{6}-1=\frac{1}{2}-1=-\frac{1}{2} \end{aligned}$$​6π​⋅f(6π​)=6π​⋅sin6π​−6π​⇒f(6π​)=sin6π​−1=21​−1=−21​​