"\\begin{aligned}\n&\\iint y \\sin (xy)dA\\textcolor{blue}{=\\int_{0}^{\\pi}\\textcolor{red}{ \\int_{1}^{2} y \\sin (xy) dx}dy }\\\\[1 em]\n\n\\textcolor{red}{ \\int_{1}^{2} y \\sin (xy)dx}& =\\frac{-y}{y}\\cos (xy)\\mid _{1}^{2} =-\\cos (xy)\\mid _{1}^{2}\\\\[1 em]\n &=-(\\cos 2y -\\cos y) =-\\cos 2y +\\cos y\\\\[1 em]\n\\textcolor{blue}{ \\int_{0}^{\\pi} -\\cos 2y +\\cos y dy} &=\\frac{-1}{2}\\sin 2y+\\sin y\\mid _{0}^{\\pi}\\\\[1 em]\n &=(\\frac{-1}{2}\\sin 2\\pi+\\sin \\pi)-(\\frac{-1}{2}\\sin 0+\\sin 0)=\\fcolorbox{red}{aqua}{0}\\\\[1 em]\n\\text{The answer is }\\fcolorbox{red}{aqua}{a}\n\\end{aligned}"
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