Multivariable calculus

Double Integral of y sin(xy) Over [1,2] by [0, pi/2]

Q: Why does the leading y vanish from the answer?

Because the substitution u = xy in the x integral has du = y\,dx, so the y is absorbed into du. Equivalently, \int \sin(xy)\,dx = -\frac{\cos(xy)}{y}, and the y in front cancels the \frac{1}{y}.

Q: What changes if x runs from 1 to 3 instead?

The inner integral becomes \cos y - \cos 3y, and \int_{0}^{\pi/2}\left(\cos y - \cos 3y\right)\,dy = 1 + \frac{1}{3} = \frac{4}{3}, since \sin\frac{3\pi}{2} = -1.

The double integral of y times sin(xy), with x running from 1 to 2 and y from 0 to pi/2, equals 1. Integrate in x first: the factor y is exactly what the substitution u = xy needs, so the inner integral is cos y minus cos 2y, and integrating that from 0 to pi/2 gives 1.

f(x,y) = y\sin(xy)

\int_{0}^{\pi/2}\int_{1}^{2} y\sin(xy)\,dx\,dy = 1

Numerically 1.000000, confirmed by quadrature on every build.

Integrate in x first and the leading y disappears

Hold $y$ fixed and substitute $u = xy$ , so $du = y\,dx$ . The loose factor $y$ is consumed exactly, leaving $\int \sin u\,du$ .

\int_{1}^{2} y\sin(xy)\,dx = \left[-\cos(xy)\right]_{x=1}^{x=2} = \cos y - \cos 2y

No $x$ survives, so the inner stage is done. The outer integral is now an ordinary trigonometric one.

\int_{0}^{\pi/2} \left(\cos y - \cos 2y\right)\,dy = \left[\sin y - \frac{\sin 2y}{2}\right]_{0}^{\pi/2} = 1 - 0 = 1

The second term vanishes because $\sin \pi = 0$ , so the whole answer comes from $\sin\frac{\pi}{2} = 1$ .

The other order needs integration by parts

Integrating in $y$ first, the factor $y$ is no longer the derivative of anything convenient, since $\frac{\partial}{\partial y}(xy) = x$ . You would need parts with $u = y$ and $dv = \sin(xy)\,dy$ .

\int y\sin(xy)\,dy = \frac{\sin(xy)}{x^2} - \frac{y\cos(xy)}{x}

It gets to the same 1 in the end. The point is that the loose factor tells you which order is cheap: match it against $\frac{\partial}{\partial x}(xy) = y$ and $\frac{\partial}{\partial y}(xy) = x$ and take the one that fits.

The mistake: dividing by the wrong variable

The antiderivative of $\sin(xy)$ in $x$ is $-\frac{\cos(xy)}{y}$ , dividing by $\frac{\partial}{\partial x}(xy) = y$ . Dividing by $x$ instead leaves $-\frac{y\cos(xy)}{x}$ , and differentiating that in $x$ does not return the integrand.

Multiply the correct antiderivative back out: $y \cdot \left(-\frac{\cos(xy)}{y}\right) = -\cos(xy)$ , with the $y$ gone.
If the inner answer still contains the loose factor, you divided by the wrong variable.
Differentiating your antiderivative with respect to the integration variable settles it every time.

The edge $y = 0$ is a useful check. There the integrand is 0, and the inner result $\cos 0 - \cos 0 = 0$ agrees, so no stray factor of $\frac{1}{y}$ ever appears in the final expression.

Frequently asked questions

Why does the leading y vanish from the answer?

Because the substitution $u = xy$ in the $x$ integral has $du = y\,dx$ , so the $y$ is absorbed into $du$ . Equivalently, $\int \sin(xy)\,dx = -\frac{\cos(xy)}{y}$ , and the $y$ in front cancels the $\frac{1}{y}$ .

What changes if x runs from 1 to 3 instead?

The inner integral becomes $\cos y - \cos 3y$ , and $\int_{0}^{\pi/2}\left(\cos y - \cos 3y\right)\,dy = 1 + \frac{1}{3} = \frac{4}{3}$ , since $\sin\frac{3\pi}{2} = -1$ .