If a function is defined by $f(x)=\arcsin(\sin(\frac{x+\sin x}{2}))~\forall x\in[0,\pi]$ then

Question

I tried differentiating the function given but the problem is that it changes with intervals.

Answer 1

Taking the usual codomain of $\;\arcsin\;$ , we have that $\;\arcsin:[-1,1]\to\left[-\frac\pi2,\,\frac\pi2\right]\;$ . Now

$$x\in[0,\pi]\implies0\le\frac{x+\sin x}2\le\frac{\pi+\sin\pi}2=\frac\pi2\implies0\le\sin\left(\frac{x+\sin x}2\right)\le1$$

and thus

$$f(x):=\arcsin\sin\left(\frac{x+\sin x}2\right)=\frac{x+\sin x}2$$

Try now to take it from here.

Answer 2

Observe that for $x \in [0, \pi]$ we have that

$$f(x)=\frac{x+ \sin x}{2}.$$