So you want the Gaussian integral trick again?

Overview

blackpenredpen revisits the Gaussian integral, whose square becomes a double integral of e^{-(x²+y²)} over the plane and evaluates to √π using polar coordinates and the Jacobian factor r. He then constructs a related unit-disk integral of sin(x²)cos(y²): swapping variables and adding the equal integrals combines the integrand into sin(x²+y²), yielding I = (π/2)(1 − cos 1).

Key takeaways

Squaring ∫ e^(−x²) dx converts a one-dimensional integral into a two-dimensional radial integral, which polar coordinates evaluate as √π.
For the unit disk, swapping x and y preserves the region and makes the integral of sin(x²)cos(y²) equal to that of sin(y²)cos(x²).
Adding those equal integrals applies the sine angle-sum identity and reduces the two-variable integrand to sin(x²+y²).
The polar-coordinate area element includes a Jacobian factor r; omitting it would change the result.
The constructed unit-disk integral evaluates to (π/2)(1−cos 1), using ∫₀¹ r sin(r²) dr and a full angular range of 2π.

Chapters

0:00 The Gaussian Integral: Squaring the Integral and Using Polar Coordinates

Set I = ∫ from −∞ to ∞ e^(−x²) dx, then square it by introducing a second variable y.
The product becomes a plane integral of e^(−(x²+y²)); polar coordinates turn x²+y² into r².
Including the area element r dr dθ and integrating over the full plane gives I = √π.

3:10 Building a Symmetric Sine–Cosine Integral Over the Unit Disk

Define I as the integral of sin(x²)cos(y²) over the unit disk, with y ranging from −1 to 1 and x from −√(1−y²) to √(1−y²).
The unit-circle bounds provide the rotationally symmetric region needed for a polar-coordinate evaluation.
Swapping x and y leaves the disk unchanged and produces an equal integral with sin(y²)cos(x²).

9:08 Combining the Integrals and Evaluating in Polar Coordinates

Add the two equal integrals and use sin A cos B + sin B cos A = sin(A+B) to obtain 2I with integrand sin(x²+y²).
Convert the unit disk to polar bounds 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π; the area element is r dr dθ.
Integrating gives 2I = π(1−cos 1), so the original integral is I = (π/2)(1−cos 1).

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, blackpenredpen.