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 $z=y^2$, $\mathrm dz=2y\,\mathrm dy$:

$\displaystyle\frac12\int_{z=0}^{z=1}\cos z\,\mathrm dz=\frac12\sin1$