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H2 Mathematics · JC 1 and 2

Integration Techniques

Integration techniques extend A-Math integration to a much wider range of functions. In H2 Mathematics you integrate using standard results from the formula list, recognise integrals of the forms f'(x)/f(x) and f'(x)[f(x)]^n, and use partial fractions, trigonometric identities, substitution and integration by parts. Choosing the right method is half the work. These techniques are then used for areas, volumes of revolution and differential equations, so speed and accuracy here pay off across the whole calculus section.

What the syllabus expects

  • Integrate standard functions, including results given in the formula list such as 1/(a² + x²) and 1/√(a² - x²)
  • Recognise and integrate f'(x)/f(x) and f'(x)[f(x)]^n
  • Integrate rational functions by first expressing them in partial fractions
  • Use trigonometric identities to integrate functions such as sin²x, cos²x, tan²x and sin 3x cos x
  • Integrate using a given substitution, changing the limits of a definite integral
  • Integrate by parts, including repeated use and integrals such as ∫ln x dx
  • Evaluate definite integrals exactly and check them with the graphing calculator

Standard forms and recognition

Start every integral by asking whether it matches a known form. Beyond powers, exponentials and basic trigonometric integrals, the formula list gives results such as ∫1/(a² + x²) dx = (1/a) tan⁻¹(x/a) + c and ∫1/√(a² - x²) dx = sin⁻¹(x/a) + c. Two patterns come from reversing the chain rule. If the numerator is the derivative of the denominator, ∫f'(x)/f(x) dx = ln|f(x)| + c, so ∫2x/(x² + 5) dx = ln(x² + 5) + c. If a bracket is raised to a power and its derivative is outside, ∫f'(x)[f(x)]^n dx = [f(x)]^(n+1)/(n + 1) + c, so ∫x(x² + 1)³ dx = (x² + 1)⁴/8 + c. Adjust constants as needed: ∫x/(x² + 5) dx = ½ ln(x² + 5) + c. For an integrand such as 1/(4 + 9x²), take out the 9 first to get (1/9) × 1/(4/9 + x²), then use a = 2/3.

Partial fractions and trigonometric identities

A rational function with a factorised denominator is split into partial fractions, then each part is integrated, often giving logarithms or inverse tangents. For example, 1/(x² - 1) = ½[1/(x - 1) - 1/(x + 1)], so ∫1/(x² - 1) dx = ½ ln|(x - 1)/(x + 1)| + c, which matches the formula list result. Remember the modulus inside ln. Trigonometric powers and products are rewritten using identities so that each term is a standard integral. Use cos²x = ½(1 + cos 2x) and sin²x = ½(1 - cos 2x), so ∫sin²x dx = x/2 - (sin 2x)/4 + c. Use tan²x = sec²x - 1, so ∫tan²x dx = tan x - x + c. For products, the factor formulae turn 2 sin 3x cos x into sin 4x + sin 2x, so ∫sin 3x cos x dx = -(cos 4x)/8 - (cos 2x)/4 + c.

Integration by substitution

A substitution changes the variable to simplify the integrand, and H2 questions usually give the substitution to use. With x = g(u), replace dx by g'(u) du, rewrite everything in terms of u, and for a definite integral change the limits to u-values so that you never need to substitute back. For example, to find the integral of x√(x - 1) from x = 1 to x = 2 using u = x - 1: x = u + 1, dx = du, and the limits become u = 0 and u = 1. The integrand becomes (u + 1)u^(1/2) = u^(3/2) + u^(1/2), whose integral from 0 to 1 is [2/5 u^(5/2) + 2/3 u^(3/2)] from 0 to 1 = 2/5 + 2/3 = 16/15. Trigonometric substitutions such as x = 2 sin θ are used for expressions like √(4 - x²), since 4 - 4sin²θ = 4cos²θ.

Integration by parts

Integration by parts reverses the product rule: ∫u (dv/dx) dx = uv - ∫v (du/dx) dx. Choose u to be the factor that becomes simpler when differentiated. A common order of preference for u is logarithms first, then inverse trigonometric functions, then powers of x, then trigonometric and exponential functions. For ∫xe^(2x) dx, let u = x and dv/dx = e^(2x), so v = ½e^(2x) and the integral is ½xe^(2x) - ∫½e^(2x) dx = ½xe^(2x) - ¼e^(2x) + c. For ∫ln x dx, write it as ∫1 × ln x dx with u = ln x and dv/dx = 1, which gives x ln x - x + c. Some integrals, such as ∫x² sin x dx, need parts twice. Others, such as ∫e^x cos x dx, return to the original integral after two applications, so you then solve for it as an unknown.

Worked example

Find the exact value of the integral of x² ln x from x = 1 to x = e.

  1. Use integration by parts with u = ln x and dv/dx = x², since ln x becomes simpler when differentiated.
  2. Then du/dx = 1/x and v = x³/3.
  3. The integral equals [(x³/3) ln x] from 1 to e minus the integral from 1 to e of (x³/3)(1/x) dx.
  4. Simplify the remaining integrand: (x³/3)(1/x) = x²/3, which integrates to x³/9.
  5. So the integral equals [(x³/3) ln x - x³/9] from 1 to e.
  6. At x = e, ln e = 1, giving e³/3 - e³/9 = 2e³/9. At x = 1, ln 1 = 0, giving 0 - 1/9 = -1/9.
  7. Subtract: 2e³/9 - (-1/9) = (2e³ + 1)/9. The graphing calculator gives 4.57 for the integral, which agrees.

Answer: (2e³ + 1)/9, which is about 4.57

Common mistakes

  • Choosing u = x² and dv/dx = ln x in ∫x² ln x dx. You would then need to integrate ln x as part of the first step, which makes the problem harder. Let u = ln x.
  • Forgetting to change the limits after a substitution, or changing them and then also substituting back to x. Do one or the other, not both.
  • Leaving out the modulus in ∫1/x dx = ln|x| + c. It matters when x is negative, as in the integral of 1/(x - 3) from 0 to 2, which is ln|-1| - ln|-3| = -ln 3.
  • Integrating sin²x as sin³x/3 or as -cos²x. Rewrite it with a double angle formula first.
  • Forgetting to replace dx when making a substitution. Every dx must be replaced, using dx = g'(u) du.
  • Missing the factor 1/a in standard results, such as writing ∫1/(9 + x²) dx as tan⁻¹(x/3) + c instead of (1/3) tan⁻¹(x/3) + c.

Exam tips

  • Before integrating, spend a few seconds choosing the method: standard form, f'(x)/f(x), partial fractions, identity, substitution or parts. Starting with the wrong method wastes the most time.
  • Check any definite integral with the graphing calculator's numerical integration. If the decimal does not match your exact answer, look for a slip before moving on.
  • Copy standard results from the formula list exactly, then adapt the constant a to your question in a separate line.

Practice questions

1. Find ∫(3x² + 2)/(x³ + 2x + 1) dx.

Show answer

ln|x³ + 2x + 1| + c. The numerator 3x² + 2 is exactly the derivative of the denominator, so this is of the form f'(x)/f(x).

2. Find ∫cos²3x dx.

Show answer

x/2 + (sin 6x)/12 + c. Using cos²3x = ½(1 + cos 6x), the integral is ½x + ½ × (sin 6x)/6 + c. Check: differentiating gives ½ + ½cos 6x = cos²3x.

3. Using the substitution x = 2 sin θ, find the exact value of the integral of √(4 - x²) from x = 0 to x = 1.

Show answer

π/3 + √3/2. With x = 2 sin θ, dx = 2 cos θ dθ, √(4 - x²) = 2 cos θ, and the limits become θ = 0 and θ = π/6. The integral becomes the integral of 4cos²θ = 2(1 + cos 2θ) from 0 to π/6, which is [2θ + sin 2θ] from 0 to π/6 = π/3 + sin(π/3) = π/3 + √3/2, about 1.91.

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