Evaluate the double integral of xy times sin(xy²) over the region 0 ≤ x ≤ π, 0 ≤ y ≤ π.

Step-by-step solution

1. 1

   Spot the substitution hiding in the integrand. The sine argument is $xy^2$ and the factor outside is $xy$. Since $\frac{d}{dy}(xy^2) = 2xy$, the outside factor is half the derivative of the inside — the signature of a clean u-substitution in $y$. Integrating in $x$ first would not work nearly as neatly, so keep the given order $dy\,dx$.

2. 2

   Substitute in the inner integral with x held fixed. Let

   $u = xy^2, \qquad du = 2xy\,dy \ \Longrightarrow \ xy\,dy = \tfrac12\,du$

   The limits move with the variable: $y = 0$ gives $u = 0$, and $y = \pi$ gives $u = x\pi^2$. So

   $\int_0^{\pi} xy\sin(xy^2)\,dy = \frac{1}{2}\int_0^{x\pi^2} \sin u\,du$

3. 3

   Evaluate the inner integral.

   $\frac{1}{2}\bigl[-\cos u\bigr]_0^{x\pi^2} = \frac{1}{2}\left(1 - \cos(x\pi^2)\right)$

   The $x$ that was outside has been absorbed by the substitution, which is why no stray factor of $x$ survives.

4. 4

   Integrate the result in x.

   $I = \frac{1}{2}\int_0^{\pi}\left(1 - \cos(x\pi^2)\right) dx = \frac{1}{2}\left(\pi - \left[\frac{\sin(x\pi^2)}{\pi^2}\right]_0^{\pi}\right)$

   The $\frac{1}{\pi^2}$ comes from the chain rule: $\pi^2$ here is a constant multiplying $x$ inside the cosine.

5. 5

   Assemble and evaluate.

   $I = \frac{\pi}{2} - \frac{\sin(\pi^3)}{2\pi^2}$

   With $\pi^3 \approx 31.00628$ and $\sin(\pi^3) \approx -0.39829$:

   $I \approx 1.570796 + 0.020178 = 1.590974$

   Direct two-dimensional quadrature gives $1.59097384$, matching. (The stated region had $y < \pi$ rather than $y \le \pi$; a single boundary line has zero area, so the value is unchanged.)