Evaluate the double integral of sin(x²) + cos(y²) over the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

Step-by-step solution

1. 1

   Split the sum into two double integrals. Integration is linear, so

   $\iint \left(\sin(x^2) + \cos(y^2)\right)dA = \iint \sin(x^2)\,dA + \iint \cos(y^2)\,dA$

   This is the key move: each piece now depends on a single variable, even though neither can be antidifferentiated in closed form.

2. 2

   Collapse each piece using the fact that one variable is absent. In the first piece $\sin(x^2)$ does not involve $y$, so the inner $y$-integration just multiplies by the interval length $1$:

   $\int_0^1\!\!\int_0^1 \sin(x^2)\,dy\,dx = \int_0^1 \sin(x^2)\left(\int_0^1 dy\right)dx = \int_0^1 \sin(x^2)\,dx$

   The same argument (integrating in $x$ first) reduces the second piece to $\int_0^1 \cos(y^2)\,dy$.

3. 3

   Recognise the remaining integrals as Fresnel integrals. Renaming both dummy variables to $t$:

   $I = \int_0^1 \sin(t^2)\,dt + \int_0^1 \cos(t^2)\,dt$

   Neither $\sin(t^2)$ nor $\cos(t^2)$ has an elementary antiderivative — they define the Fresnel functions $S$ and $C$ (up to the usual $\pi/2$ scaling). So the exact answer is $\sqrt{\tfrac{\pi}{2}}\left[S\!\left(\sqrt{\tfrac{2}{\pi}}\right) + C\!\left(\sqrt{\tfrac{2}{\pi}}\right)\right]$, and a numerical value is the practical answer.

4. 4

   Evaluate both numerically. High-order Gauss-Legendre quadrature gives

   $\int_0^1 \sin(t^2)\,dt = 0.3102683, \qquad \int_0^1 \cos(t^2)\,dt = 0.9045242$

   The series expansions confirm the sizes: $\sin(t^2) \approx t^2$ integrates to about $\tfrac13$, and $\cos(t^2) \approx 1 - \tfrac{t^4}{2}$ integrates to about $1 - \tfrac{1}{10} = 0.9$.

5. 5

   Add and check against direct 2-D quadrature.

   $I \approx 0.3102683 + 0.9045242 = 1.2147925$

   Evaluating the original double integral numerically over the square, without splitting, gives $1.21479254$ — identical to eight decimals, confirming both the split and the two one-dimensional values.