Answer
2
Work Step by Step
$\int ^{\pi }_{0}\int ^{\sin x}_{0}\left( 1+\cos x\right) dydx=\int ^{\pi }_{0}y ] ^{\sin x}_{0}\left( 1+\cos x\right) dx=\int ^{\pi }_{0}\sin x\left( 1+\cos x\right) dx =\int ^{\pi }_{0}\left( \sin x+\cos x\sin x\right) dx=\int ^{\pi }_{0}\left( \sin x+\dfrac {\sin 2x}{2}\right) dx=(-\cos x-\dfrac {\cos 2x}{4})] ^{\pi }_{0}=\left( -\left( -1\right) -\dfrac {1}{4}\right) -\left( -1-\dfrac {1}{4}\right) =2 $
