Question

Evaluate the integral by reversing the order of integration.
∫(0 t0 3)∫(y^2 to 9) ycos(x^2)dxdy=

Answer 1

$$\int_0^3\int_{y^2}^9y\cos(x^2)dx\,dy=\int_{y^2}^9\cos(x^2)\int_0^3y\,dy\,dx\\ =\int_{y^2}^9\cos(x^2)[y^2/2]_0^3dx\\ =\frac{9}{2}\int_{y^2}^9\cos(x^2)dx$$

Substitute $u=\dfrac{\sqrt{2}x}{\sqrt{{\pi}}}\implies \mathrm{d}x=\dfrac{\sqrt{{\pi}}}{\sqrt{2}}\,\mathrm{d}u$ ,thus

$$\int_0^3\int_{y^2}^9y\cos(x^2)dx\,dy=\frac{9}{2}\dfrac{\sqrt{{\pi}}}{\sqrt{2}}{\displaystyle\int_{\tiny\dfrac{\sqrt{2}y^2}{\sqrt{{\pi}}}}^{\tiny\dfrac{9\sqrt{2}}{\sqrt{{\pi}}}}}\cos\left(\dfrac{{\pi}u^2}{2}\right)\,\mathrm{d}u\\ \implies\frac{9}{2}\dfrac{\sqrt{{\pi}}}{\sqrt{2}}C(u)=\dfrac{\sqrt{{\pi}}\operatorname{C}\left(\frac{\sqrt{2}x}{\sqrt{{\pi}}}\right)}{\sqrt{2}}\bigg|_{y^2}^9$$

Because, as

$${\displaystyle\int}\cos\left(\dfrac{{\pi}u^2}{2}\right)\,\mathrm{d}u=C(u)$$

This is a special integral (Fresnel integral):

$$\int_{y^2}^9 \int_0^3 y \cos(x^2) dy dx = 9/2 \sqrt(\pi/2) (C(9 \sqrt(2/\pi)) - C(\sqrt(2/\pi) y^2))$$