Show that `f(x)=x^2-xsinx` is an increasing function on `(0,\ pi//2)` .

Explanation

We have given

`f(x)=x^2-xsinx`

Then

`f′(x)=2x−sinx−xcosx=x(2−cosx)−sinx`

But for all `x` in `(0, pi/2)`,`0`<`cosx`<`1` and `0`<`x`<`sinx`,

so

`f′(x)`>`x−sinx`>`0`

which implies that f is strictly increasing on `(0,pi/2)`