${\int }_0^X{\int }_{\sin \left(x\right)}^{f\left(x\right)}1 \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_X^{\mathrm{\π}\/2}{\int }_{f\left(x\right)}^{\sin \left(x\right)}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$ ≐ $0.302$

Initialize

$f\=\arctan \left(x\+1\right)-1\/2$$\stackrel{\text{assign}}{\to }$

$g\=\sin \left(x\right)$$\stackrel{\text{assign}}{\to }$

Obtain the intersections of the curves bounding $R$ 

$f\=g$

$\arctan \left(x\+1\right)-\frac{1}{2}\=\sin \left(x\right)$

$\stackrel{\text{solve}}{\to }$

$0.5059163080$

$\stackrel{\text{assign to a name}}{\to }$

$X$

Iterate in the order $\mathrm{dy} \mathrm{dx}$ via the template in the Calculus palette

${\int }_0^X{\int }_g^{f}1 \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_X^{\mathrm{\π}\/2}{\int }_f^{g}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $0.3016523782$ 

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

$$\begin{gathered} f≔\arctan \left(x\+1\right)-1\/2\: \\ g≔\sin \left(x\right)\: \end{gathered}$$

Solve the equation $f\left(x\right)\=g\left(x\right)$ for $x$ 

$X≔\mathrm{fsolve}\left(f\=g\,x\right)$

$0.5059163080$

Top-level, using the Int and int commands

$$\begin{gathered} \mathrm{Int}\left(1\,\left[y\=g..f\,x\=0..X\right]\right)\+\mathrm{Int}\left(1\,\left[y\=f..g\,x\=X..\mathrm{\π}\/2\right]\right)\= \\ \mathrm{int}\left(1\,\left[y\=g..f\,x\=0..X\right]\right)\+\mathrm{int}\left(1\,\left[y\=f..g\,x\=X..\mathrm{\pi }\/2\right]\right) \end{gathered}$$

${\∫}_0^{0.5059163080}{\∫}_{\sin \left(x\right)}^{\arctan \left(x\+1\right)-\frac{1}{2}}1\ⅆy\ⅆx\+{\∫}_{0.5059163080}^{\frac{1}{2}\mathrm{\π}}{\∫}_{\arctan \left(x\+1\right)-\frac{1}{2}}^{\sin \left(x\right)}1\ⅆy\ⅆx\=0.3016523782$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(1\,y\=g..f\,x\=0..X\right)\+\mathrm{MultiInt}\left(1\,y\=f..g\,x\=X..\mathrm{\π}\/2\right)$ = $0.3016523782$$$

$$\begin{gathered} \mathrm{MultiInt}\left(1\,y\=g..f\,x\=0..X\,\mathrm{output}\=\mathrm{integral}\right)\+ \\ \mathrm{MultiInt}\left(1\,y\=f..g\,x\=X..\mathrm{\π}\/2\,\mathrm{output}\=\mathrm{integral}\right) \end{gathered}$$

${\∫}_0^{0.5059163080}{\∫}_{\sin \left(x\right)}^{\arctan \left(x\+1\right)-\frac{1}{2}}1\ⅆy\ⅆx\+{\∫}_{0.5059163080}^{\frac{1}{2}\mathrm{\π}}{\∫}_{\arctan \left(x\+1\right)-\frac{1}{2}}^{\sin \left(x\right)}1\ⅆy\ⅆx$

Use the MultiInt command with a pre-defined domain option

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(0..X\,g..f\right)\right)\+ \mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(X..\mathrm{\π}\/2\,f..g\right)\right)$

$0.3016523782$

$$\begin{gathered} \mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(0..X\,g..f\right)\,\mathrm{output}\=\mathrm{integral}\right)\+ \\ \mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(X..\mathrm{\pi }\/2\,f..g\right)\,\mathrm{output}\=\mathrm{integral}\right) \end{gathered}$$

${\∫}_0^{0.5059163080}{\∫}_{\sin \left(x\right)}^{\arctan \left(x\+1\right)-\frac{1}{2}}1\ⅆy\ⅆx\+{\∫}_{0.5059163080}^{\frac{1}{2}\mathrm{\π}}{\∫}_{\arctan \left(x\+1\right)-\frac{1}{2}}^{\sin \left(x\right)}1\ⅆy\ⅆx$