an iterated integral question This iterated integral is proving harder than I thought. Evaluate by reversing the order of integration: 
$$
\int_{0}^{1}\left(\int_{y=x}^{\sqrt{x}}\frac{\sin y}{y}dy\right)dx
$$ 
 A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#00f}{\large%
\int_{0}^{1}\bracks{\int_{x}^{\root{x}}{\sin\pars{y} \over y}\,dy}\,\dd x}
=\int_{0}^{1}\bracks{\int_{0}^{1}\Theta\pars{y - x}\Theta\pars{\root{x} - y}{\sin\pars{y} \over y}\,dy}\,\dd x
\\[3mm]&=\int_{0}^{1}\bracks{\int_{0}^{1}\Theta\pars{y - x}\Theta\pars{x - y^{2}}{\sin\pars{y} \over y}\,dy}\,\dd x=
\int_{0}^{1}{\sin\pars{y} \over y}\bracks{\int_{y^{2}}^{y}\dd x}\,\dd y
\\[3mm]&=
\int_{0}^{1}{\sin\pars{y} \over y}\pars{y - y^{2}}\,\dd y
=
\int_{0}^{1}\bracks{\sin\pars{y} - y\sin\pars{y}}\,\dd y
=
\cos\pars{1} + \int_{0}^{1}\bracks{\sin\pars{y} - \cos\pars{y}}\,\dd y
\\[3mm]&=
\cos\pars{1} + \bracks{-\cos\pars{y} - \sin\pars{y}}_{0}^{1}
=
\cos\pars{1} + \bracks{-\cos\pars{1} - \sin\pars{1} + 1}
=\color{#00f}{\large 1 - \sin\pars{1}}
\end{align}
