CHAPTER 03 — INTEGRATION | PART 2: SUBSTITUTION

MATHEMATICS HIP-HOP ACADEMY · CHAPTER 03

INTEGRATION

Part 2 · Integration by Substitution and Recognising a Function with Its Derivative

A-LEVEL PURE MATHEMATICS

About this lesson

In Part 1, we integrated expressions directly using elementary rules. Some integrals, however, contain a function together with its derivative, or a complicated expression raised to a power. In such cases, substitution can turn the integral into a simpler one.

Central idea: Look for an inner function and its derivative. If both are present, a suitable substitution may reduce the integral to a standard form.

Learning outcomes

• Recognise a function and its derivative within an integrand.
• Choose a suitable substitution and calculate its differential.
• Integrate powers and reciprocal forms after substitution.
• Change the limits when evaluating definite integrals by substitution.
• Check an antiderivative by differentiating it.

1. Recognising a function and its derivative

Consider an integral containing an expression such as (3x² + 2)⁵. The expression inside the brackets is a function of x, and its derivative is 6x. If the integrand contains that derivative (or a constant multiple of it), substitution is likely to work.

If u = g(x), then du/dx = g′(x), so du = g′(x) dx.

This allows us to replace the combination g′(x) dx by du.

Recognition pattern: An integral of the form ∫g′(x)[g(x)]ⁿ dx can be simplified by putting u = g(x), provided the remaining factors match the derivative.

2. The substitution method

1. Identify a suitable inner function and let it equal u.
2. Differentiate: find du/dx.
3. Rearrange to express the required part of dx in terms of du.
4. Rewrite the entire integral using u. Do not leave stray x terms.
5. Integrate with respect to u.
6. For an indefinite integral, substitute back to x and include + C.
7. For a definite integral, either change the limits to u-values or substitute back before applying the original limits.

3. Worked examples

Example 1 — A simple power of a function

Evaluate ∫ 6x(3x² + 4)⁵ dx.

Step 1: Let u = 3x² + 4.

Step 2: du/dx = 6x, hence du = 6x dx.

Step 3: The integral becomes ∫u⁵ du.

Step 4: Integrate: u⁶/6 + C.

Step 5: Substitute back.

Answer: (3x² + 4)⁶/6 + C.

Example 2 — A constant multiple of the derivative

Evaluate ∫ x(2x² + 5)⁴ dx.

Let u = 2x² + 5. Then du = 4x dx, so x dx = du/4.

∫x(2x² + 5)⁴ dx = ¼∫u⁴ du.

= ¼ × u⁵/5 + C.

Answer: (2x² + 5)⁵/20 + C.

Example 3 — Recognising a reciprocal form

Evaluate ∫ [4x/(2x² + 7)] dx.

Let u = 2x² + 7. Then du = 4x dx.

The integral becomes ∫(1/u) du.

∫(1/u) du = ln|u| + C.

Answer: ln(2x² + 7) + C.

Since 2x² + 7 is always positive, the absolute value may be omitted.

Example 4 — Exponential function

Evaluate ∫ 3e^(3x − 1) dx.

Let u = 3x − 1. Then du = 3 dx.

The integral becomes ∫eᵘ du.

Answer: e^(3x − 1) + C.

Example 5 — A trigonometric function

Evaluate ∫ 2 sin(2x + 1) dx.

Let u = 2x + 1. Then du = 2 dx.

The integral becomes ∫sin u du = −cos u + C.

Answer: −cos(2x + 1) + C.

Example 6 — Definite integral with changed limits

Evaluate ∫₀¹ 2x( x² + 1)³ dx.

Let u = x² + 1, so du = 2x dx.

Change the limits: when x = 0, u = 1; when x = 1, u = 2.

The integral becomes ∫₁² u³ du.

= [u⁴/4]₁² = (16/4) − (1/4).

Answer: 15/4.

4. A useful recognition table

• Power pattern: ∫g′(x)[g(x)]ⁿ dx — let u = g(x), then use the power rule.
• Reciprocal pattern: ∫g′(x)/g(x) dx — the result is ln|g(x)| + C.
• Exponential pattern: ∫g′(x)eᵍ⁽ˣ⁾ dx — the result is eᵍ⁽ˣ⁾ + C.
• Sine pattern: ∫g′(x)sin(g(x)) dx — the result is −cos(g(x)) + C.
• Cosine pattern: ∫g′(x)cos(g(x)) dx — the result is sin(g(x)) + C.

These are recognition patterns, not permission to ignore missing factors. If the derivative is present only up to a constant multiple, adjust the integral by that constant.

5. Common mistakes to avoid

• Choosing u but forgetting to calculate du.
• Missing a numerical factor when rearranging du.
• Leaving x in the integral after changing to u.
• Forgetting to substitute back in an indefinite integral.
• Changing the integrand to u but forgetting to change the limits in a definite integral.
• Forgetting + C for an indefinite integral.

6. Practice exercises

Attempt each question before opening its answer.

1. Evaluate ∫ 8x(4x² + 3)³ dx.

Reveal answer

(4x² + 3)⁴/4 + C.

2. Evaluate ∫ 5x(5x² − 2)⁶ dx.

Reveal answer

(5x² − 2)⁷/14 + C.

3. Evaluate ∫ [6x/(3x² + 4)] dx.

Reveal answer

ln(3x² + 4) + C.

4. Evaluate ∫ 4e^(4x + 1) dx.

Reveal answer

e^(4x + 1) + C.

5. Evaluate ∫ 3cos(3x − 2) dx.

Reveal answer

sin(3x − 2) + C.

6. Evaluate ∫ 2x(x² + 5)⁴ dx.

Reveal answer

(x² + 5)⁵/5 + C.

7. Evaluate ∫₀² 2x(x² + 1)² dx.

Reveal answer

[(x² + 1)³/3]₀² = (125 − 1)/3 = 124/3.

8. Evaluate ∫ [2x/(x² + 9)] dx.

Reveal answer

ln(x² + 9) + C.

9. Evaluate ∫ 6(6x + 1)⁵ dx.

Reveal answer

(6x + 1)⁶/6 + C.

10. Evaluate ∫₁² 3x²( x³ + 2)⁴ dx.

Reveal answer

Let u = x³ + 2. Limits: 3 and 10. Answer = [u⁵/5]₃¹⁰ = (100000 − 243)/5 = 19951.

7. Lesson summary

• Substitution simplifies an integral by replacing an inner function with a new variable.
• Always differentiate the substitution to obtain du.
• Match all factors carefully, including constant multipliers.
• For definite integrals, change the limits whenever you integrate entirely in u.
• Differentiate your final answer to check an indefinite integral.

NEXT: PART 3 — TRIGONOMETRIC INTEGRATION

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