Multivariable calculus

Double Integral of cos(x^2) Over a Triangle

Q: Why is the upper limit sqrt(pi/2) rather than something rounder?

It is chosen so that x^{2} reaches exactly \pi/2, where \sin\!\left(x^{2}\right) = 1. That makes the final answer the clean 1/2 instead of a decimal with no closed form.

Q: Is cos(x^2) positive over the whole region?

Yes. Here x^{2} runs from 0 to \pi/2, and cosine is non-negative on that interval. So 1/2 is an honest volume, not a signed cancellation.

The double integral of cos of x squared over the triangle where y runs from 0 to x and x runs from 0 to the square root of pi/2 equals exactly 1/2. Integrating in y first produces a factor of x, which turns the outer integral into the substitution u = x squared.

f(x,y) = \cos\!\left(x^{2}\right)

\dfrac{1}{2}

Numerically 0.500000, confirmed by quadrature on every build.

The inner integral supplies the factor the substitution needs

The region is the triangle below the line $y = x$ , with $x$ running from $0$ to $\sqrt{\pi/2} \approx 1.2533$ . The integrand has no $y$ in it, so the inner integral just measures the height of the slice.

\int_{0}^{x} \cos\!\left(x^{2}\right)\,dy = \cos\!\left(x^{2}\right)\Big[y\Big]_{0}^{x} = x\cos\!\left(x^{2}\right)

That $x$ out front is exactly half of $\frac{d}{dx}\left(x^{2}\right) = 2x$ , so the outer integral is a substitution with $u = x^{2}$ , $du = 2x\,dx$ .

\int_{0}^{\sqrt{\pi/2}} x\cos\!\left(x^{2}\right)\,dx = \left[\frac{\sin\!\left(x^{2}\right)}{2}\right]_{0}^{\sqrt{\pi/2}} = \frac{\sin(\pi/2)}{2} - 0 = \frac{1}{2}

Why the reversed order is a dead end

Describing the same triangle by horizontal slices puts $y$ outside and $x$ from $y$ to $\sqrt{\pi/2}$ inside. The value is unchanged, but the inner integral becomes impossible by hand.

\int_{0}^{\sqrt{\pi/2}}\!\!\int_{y}^{\sqrt{\pi/2}} \cos\!\left(x^{2}\right)\,dx\,dy = \frac{1}{2}

The function $\cos\!\left(x^{2}\right)$ has no elementary antiderivative; the integral $\int \cos(x^2)\,dx$ is the Fresnel integral, a named special function. Without the $x$ that the $dy$ integration hands you, there is nothing to substitute with.

If a problem hands you this order, reversing it is the intended move, not an optional shortcut.
Sketch the triangle, read it by vertical slices instead, and the $x$ appears on its own.
The same pattern rescues $e^{x^{2}}$ , $\sin\!\left(x^{2}\right)$ and $\frac{\sin x}{x}$ style integrands.

The mistake: dropping the chain-rule half

After reaching $\int x\cos\!\left(x^{2}\right)\,dx$ , a common slip is to write the antiderivative as $\sin\!\left(x^{2}\right)$ with no factor of $\frac{1}{2}$ , which doubles the answer to $1$ .

\frac{d}{dx}\sin\!\left(x^{2}\right) = 2x\cos\!\left(x^{2}\right), \qquad \frac{d}{dx}\left[\frac{\sin\!\left(x^{2}\right)}{2}\right] = x\cos\!\left(x^{2}\right)

Differentiating your antiderivative back before you evaluate it catches this every time, and costs about five seconds. The integrand has one $x$ , not two, so the $\frac{1}{2}$ has to be there.

Frequently asked questions

Why is the upper limit sqrt(pi/2) rather than something rounder?

It is chosen so that $x^{2}$ reaches exactly $\pi/2$ , where $\sin\!\left(x^{2}\right) = 1$ . That makes the final answer the clean $1/2$ instead of a decimal with no closed form.

Is cos(x^2) positive over the whole region?

Yes. Here $x^{2}$ runs from $0$ to $\pi/2$ , and cosine is non-negative on that interval. So $1/2$ is an honest volume, not a signed cancellation.