Given[tex]g(x)=sinx;\text{ }\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}g(x)dx[/tex]Solution[tex]\int sinx=-cosx+C[/tex][tex]\begin{gathered} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sinxdx=-cos(\frac{\pi}{2})--cos(-\frac{\pi}{2}) \\ =-cos(\frac{\pi}{2})+cos(-\frac{\pi}{2}) \\ \\ =0 \end{gathered}[/tex]The final answer[tex]\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}g(x)dx=0[/tex]
