9 min readConcept sheet2-mark and 3-mark questions; the √(a² − x²) result is used again for areas of circles and ellipses

Integrals of √(a² − x²), √(x² ± a²) and √(ax² + bx + c)

With a square root on top, each answer is (x/2) × root ± (a²/2) × a standard integral. For a quadratic, complete the square first.

√(a² − x²)∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C

√(x² + a²)∫ √(x² + a²) dx = (x/2)√(x² + a²) + (a²/2) log|x + √(x² + a²)| + C

√(x² − a²)∫ √(x² − a²) dx = (x/2)√(x² − a²) − (a²/2) log|x + √(x² − a²)| + C

How they are foundBy parts, taking the root as the first function and 1 as the second

Proof of the √(x² − a²) result

1
Let I = ∫ √(x² − a²) × 1 dx. Take √(x² − a²) as the first function and 1 as the second.
2
By parts: I = x√(x² − a²) − ∫ [x/√(x² − a²)] × x dx = x√(x² − a²) − ∫ x²/√(x² − a²) dx.
3
Write x² = (x² − a²) + a²: ∫ x²/√(x² − a²) dx = ∫ √(x² − a²) dx + a² ∫ dx/√(x² − a²).
4
So I = x√(x² − a²) − I − a² log|x + √(x² − a²)|.
5
2I = x√(x² − a²) − a² log|x + √(x² − a²)|.
6
So, I = (x/2)√(x² − a²) − (a²/2) log|x + √(x² − a²)| + C.

Compare the three

Integrand	First term	Second term
√(a² − x²)	(x/2)√(a² − x²)	+ (a²/2) sin⁻¹(x/a)
√(x² + a²)	(x/2)√(x² + a²)	+ (a²/2) log\|x + √(x² + a²)\|
√(x² − a²)	(x/2)√(x² − a²)	− (a²/2) log\|x + √(x² − a²)\|

Root at the bottom vs root on top

Root at the bottom: 1/√(a² − x²)One term only: sin⁻¹(x/a) + C.

Root on top: √(a² − x²)Two terms: (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C. Students often forget the a²/2.

The same pattern holds for the two log forms.

Example 1: a coefficient on x²

Problem

Find ∫ √(9 − 4x²) dx.

Working

Put t = 2x, so dx = dt/2 and 9 − 4x² = 3² − t².
The integral is (1/2) ∫ √(3² − t²) dt.
Use ∫ √(a² − t²) dt = (t/2)√(a² − t²) + (a²/2) sin⁻¹(t/a) + C with a = 3.
= (1/2)[(t/2)√(9 − t²) + (9/2) sin⁻¹(t/3)] + C.
Put back t = 2x: (1/2)[x√(9 − 4x²) + (9/2) sin⁻¹(2x/3)] + C.

Answer

(x/2)√(9 − 4x²) + (9/4) sin⁻¹(2x/3) + C

Example 2: a quadratic under the root

Problem

Find ∫ √(7 + 6x − x²) dx.

Working

Complete the square: 7 + 6x − x² = 7 − (x² − 6x) = 7 − [(x − 3)² − 9] = 4² − (x − 3)².
Put t = x − 3, dt = dx, a = 4.
∫ √(4² − t²) dt = (t/2)√(16 − t²) + (16/2) sin⁻¹(t/4) + C.
Put back t = x − 3, and note 16 − t² = 7 + 6x − x².

Answer

((x − 3)/2)√(7 + 6x − x²) + 8 sin⁻¹((x − 3)/4) + C

In real life

The √(a² − x²) result gives the area of a circle or an ellipse exactly. That is how the area of a round pond, a circular rangoli or an oval stadium ground is worked out.

Exam angle

2 marks: “Find ∫ √(x² + 9) dx.” Use the √(x² + a²) result with a = 3: (x/2)√(x² + 9) + (9/2) log|x + √(x² + 9)| + C.

3 marks: “Find ∫ √(x² + 6x + 10) dx.” x² + 6x + 10 = (x + 3)² + 1². Put t = x + 3, a = 1. Answer: ((x + 3)/2)√(x² + 6x + 10) + (1/2) log|x + 3 + √(x² + 6x + 10)| + C.

Remember it

Half x root, then half a-squared. Plus for sin⁻¹ and for x² + a²; minus only for x² − a².