Answer
$$\frac{1}{3}\left( {2\sqrt 2 - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^{\arccos y} {\sin x\sqrt {1 + {{\sin }^2}x} } dxdy} \cr
& {\text{Switching the order of integration, the new region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant \cos x,{\text{ }}0 \leqslant x \leqslant \frac{\pi }{2}} \right\} \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\int_0^{\arccos y} {\sin x\sqrt {1 + {{\sin }^2}x} } dxdy} = \int_0^{\pi /2} {\int_0^{\cos x} {\sin x\sqrt {1 + {{\sin }^2}x} } dydx} \cr
& = \int_0^{\pi /2} {\left[ {\int_0^{\cos x} {\sin x\sqrt {1 + {{\sin }^2}x} dy} } \right]} dx \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^{\pi /2} {\left[ {\sin x\sqrt {1 + {{\sin }^2}x} y} \right]} _0^{\cos x}dx \cr
& = \int_0^{\pi /2} {\sin x\sqrt {1 + {{\sin }^2}x} \left( {\cos x} \right)} dx \cr
& = \int_0^{\pi /2} {\sqrt {1 + {{\sin }^2}x} \left( {\sin x\cos x} \right)} dx \cr
& {\text{Let }}u = 1 + {\sin ^2}x,\,{\text{ }}du = 2\sin x\cos xdx \cr
& = \frac{1}{2}\int_1^2 {\sqrt u } du \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\left[ {\frac{2}{3}{u^{3/2}}} \right]_1^2 \cr
& = \frac{1}{3}\left[ {{u^{3/2}}} \right]_1^2 \cr
& = \frac{1}{3}\left( {2\sqrt 2 - 1} \right) \cr} $$
