# Solving $\int_0^1\int_{y}^{\sqrt{y}}\frac{\sin x}{x} dxdy$

I am stuck at this question:

$$\int_0^1\int_{y}^{\sqrt{y}}\frac{\sin x}{x} dxdy$$

It seems maybe I can use change of variable theorem to solve this but I don't know how... Can someone show me how it can be done?

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Change order of integration:

$$\int\limits_0^1\int\limits_{x^2}^x\frac{\sin x}{x}dydx=\int_0^1(1-x)\sin xdx=\ldots$$

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Hint: If you draw a picture of the region of integration, you can see that we have $$\int_0^1\int_y^{\sqrt{y}}\!\frac{\sin x}{x}\,dxdy = \int_0^1\int_{x^2}^x\!\frac{\sin x}{x}\,dydx,$$ and the second integral should be easier for you to solve (the $x$ in the denominator will conveniently vanish).

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$$\int_0^1 \int_{x^2}^x \dfrac{\sin x}{x} dy dx$$
$$\int_0^1 (\sin x - x\sin x) dx$$