Show that

We have,

$f(x)=x^{2}-x \sin x$

$f^{\prime}(x)=2 x-\sin x-x \cos x$

Now,

$X \in\left(0, \frac{\pi}{2}\right)$

$\Rightarrow 0 \leq \sin x \leq 1,0 \leq \cos x \leq 1$,

$\Rightarrow 2 x-\sin x-x \cos x>0$

$\Rightarrow f^{\prime}(x) \geq 0$

Hence, $f(x)$ is an increasing function on $\left(0, \frac{\pi}{2}\right)$.