Dealing with Radicals
When you see expressions like \(\sqrt{a^2 - x^2}\), ordinary U-substitution fails. Instead, we substitute \(x = a \sin \theta\). This uses the Pythagorean identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to "cancel" the square root.
Worked Examples
Example 1: The Area of a Quarter Circle
Evaluate \(\int_0^1 \sqrt{1 - x^2} dx\).
- Let \(x = \sin \theta\). Then \(dx = \cos \theta d\theta\).
- Limits: If \(x=0, \theta=0\). If \(x=1, \theta=\pi/2\).
- Integral: \(\int_0^{\pi/2} \sqrt{1 - \sin^2 \theta} \cos \theta d\theta = \int_0^{\pi/2} \cos^2 \theta d\theta\).
- Use identity: \(\int_0^{\pi/2} \frac{1 + \cos(2\theta)}{2} d\theta = [\frac{1}{2}\theta + \frac{\sin(2\theta)}{4}]_0^{\pi/2} = \frac{\pi}{4}\).
- Result: \(\pi/4\). (Which is correct: the area of a full unit circle is \(\pi\), so a quarter is \(\pi/4\)).
The Bridge to Quantum Mechanics
Trigonometric substitution is used in Semiclassical Quantum Mechanics (the WKB approximation). We often have to integrate expressions like \(\sqrt{E - V(x)}\) to find the "phase" of the wavefunction. If the potential \(V(x)\) is parabolic or circular, this method is the only way to calculate the tunneling probability or the energy levels. It bridges the gap between the geometry of classical orbits and the wave-nature of quantum particles.