Question

Prove that \( \displaystyle \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \).

Hint:

For integrals of the form \( \sqrt{a^2 - x^2} \), always try substitution \( x = a \sin \theta \).

Answer 1

Concept: Use trigonometric substitution: \[ x = a \sin \theta \] This simplifies expressions of the form \( \sqrt{a^2 - x^2} \).
Step 1: Substitute Let: \[ x = a \sin \theta \quad \Rightarrow \quad dx = a \cos \theta \, d\theta \] Then, \[ \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sin^2 \theta} = a \cos \theta \]
Step 2: Transform the integral \[ \int \sqrt{a^2 - x^2} \, dx = \int (a \cos \theta)(a \cos \theta \, d\theta) \] \[ = a^2 \int \cos^2 \theta \, d\theta \]
Step 3: Use identity \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \] \[ = \frac{a^2}{2} \int (1 + \cos 2\theta) d\theta \] \[ = \frac{a^2}{2} \left( \theta + \frac{\sin 2\theta}{2} \right) + C \]
Step 4: Simplify \[ = \frac{a^2}{2}\theta + \frac{a^2}{4} \sin 2\theta + C \] Using: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] \[ = \frac{a^2}{2}\theta + \frac{a^2}{2} \sin \theta \cos \theta + C \]

Step 5: Back-substitute Since: \[ \sin \theta = \frac{x}{a}, \quad \cos \theta = \frac{\sqrt{a^2 - x^2}}{a}, \quad \theta = \sin^{-1}\left(\frac{x}{a}\right) \] So, \[ \sin \theta \cos \theta = \frac{x}{a} \cdot \frac{\sqrt{a^2 - x^2}}{a} = \frac{x\sqrt{a^2 - x^2}}{a^2} \] Substitute back: \[ \frac{a^2}{2}\theta + \frac{a^2}{2} \cdot \frac{x\sqrt{a^2 - x^2}}{a^2} \] \[ = \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + \frac{x}{2}\sqrt{a^2 - x^2} + C \] Final Result: \[ \boxed{\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C} \] Explanation: Trigonometric substitution converts the radical into a simple trigonometric integral, which is then transformed back into algebraic form.