Question

Double integral of cos (y^2) dy dx, Where y=1 , y=x, x=0, x=1 are the bounds

Answer 1

`\displaystyle\int_(x=0)^(x=1)\int_(y=1)^(y=x)\cos y^2\,\mathrm dy\,\mathrm dx`

Change the order of integration. The region over which you're integrating can be equivalently described by the set of points in the plane,
`\{(x,y)\mid0\le x\le y,0\le y\le1\}`.

Then the integral becomes

`\displaystyle\int_(y=0)^(y=1)\int_(x=0)^(x=y)\cos y^2\,\mathrm dx\,\mathrm dy=\int_(y=0)^(y=1)y\cos y^2\,\mathrm dy`

Substitute
`z=y^2`,
`\mathrm dz=2y\,\mathrm dy`:

`\displaystyle\frac12\int_(z=0)^(z=1)\cos z\,\mathrm dz=\frac12\sin1`