Specialist Mathematics

Advanced Calculus: Techniques of Integration and Differential Equations

Year 12

✓Apply integration by substitution to evaluate definite and indefinite integrals.
✓Utilise integration by parts to solve integrals involving products of functions.
✓Solve first-order separable differential equations.
✓Solve first-order linear differential equations using an integrating factor.
✓Understand and apply differential equations to model real-world phenomena.

Key concepts

Integration by Substitution (u-substitution)

This technique simplifies integrals by changing the variable of integration. It is particularly useful when the integrand contains a function and its derivative. The core idea is to let 'u' be a function of 'x', then find 'du/dx' and rewrite the integral in terms of 'u' and 'du'.

∫ f(g(x))g'(x) dx = ∫ f(u) du, where u = g(x) and du = g'(x) dx

Integration by Parts

A technique for integrating products of functions, derived from the product rule for differentiation. It is effective when one part of the product becomes simpler when differentiated and the other part is easily integrated. The choice of 'u' and 'dv' is crucial; the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) can help in choosing 'u' (the function that comes first in LIATE is often chosen as 'u').

∫ u dv = uv - ∫ v du

Integration of Rational Functions (Partial Fractions)

This technique is used to integrate rational functions (polynomials divided by polynomials) by decomposing them into simpler fractions. This is possible when the degree of the numerator is less than the degree of the denominator. The denominators of the partial fractions are the factors of the original denominator.

First-Order Separable Differential Equations

A first-order differential equation is separable if it can be written in the form dy/dx = f(x)g(y). The method involves separating the variables such that all terms involving 'y' (and 'dy') are on one side, and all terms involving 'x' (and 'dx') are on the other side, then integrating both sides.

If dy/dx = f(x)g(y), then ∫ (1/g(y)) dy = ∫ f(x) dx

First-Order Linear Differential Equations

A first-order linear differential equation has the form dy/dx + P(x)y = Q(x). These are solved by multiplying the entire equation by an 'integrating factor', I(x), which makes the left-hand side the derivative of a product.

The integrating factor is I(x) = e^(∫ P(x) dx). Once multiplied, the equation becomes d/dx [y * I(x)] = Q(x) * I(x), which can then be integrated.

Key facts to remember

1The Fundamental Theorem of Calculus links differentiation and integration.
2Integration by substitution is effective when an integrand contains a function and its derivative.
3Integration by parts is used for products of functions, using the formula ∫ u dv = uv - ∫ v du.
4Partial fractions decompose rational functions into simpler terms for easier integration.
5A first-order differential equation is separable if it can be written as dy/dx = f(x)g(y).
6A first-order linear differential equation dy/dx + P(x)y = Q(x) is solved using an integrating factor I(x) = e^(∫ P(x) dx).
7General solutions to differential equations include an arbitrary constant C. Particular solutions require initial conditions to determine C.

Worked examples

Example 1

Evaluate ∫ x^2 * e^(x^3) dx.

ILet u = x^3.

IIThen du/dx = 3x^2, so du = 3x^2 dx.

IIIRewrite x^2 dx as (1/3) du.

IVSubstitute into the integral: ∫ e^u * (1/3) du.

VIntegrate with respect to u: (1/3) ∫ e^u du = (1/3) e^u + C.

VISubstitute back u = x^3: (1/3) e^(x^3) + C.

Answer

(1/3) e^(x^3) + C

This technique simplifies integrals where an inner function's derivative is present.

Example 2

Evaluate ∫ x * cos(x) dx.

IChoose u and dv. Let u = x (algebraic, simplifies on differentiation) and dv = cos(x) dx (easily integrated).

IIDifferentiate u to find du: du = dx.

IIIIntegrate dv to find v: v = ∫ cos(x) dx = sin(x).

IVApply the integration by parts formula: ∫ u dv = uv - ∫ v du.

VSubstitute: x * sin(x) - ∫ sin(x) dx.

VIEvaluate the remaining integral: ∫ sin(x) dx = -cos(x).

VIICombine: x * sin(x) - (-cos(x)) + C.

VIIISimplify: x * sin(x) + cos(x) + C.

Answer

x * sin(x) + cos(x) + C

Remember the LIATE rule for choosing 'u' to simplify the process.

Example 3

Find the general solution to dy/dx = (2x) / (y^2).

ISeparate the variables: y^2 dy = 2x dx.

IIIntegrate both sides: ∫ y^2 dy = ∫ 2x dx.

IIIEvaluate the integrals: (y^3)/3 = x^2 + C. (Only one constant of integration is needed).

IVSolve for y^3: y^3 = 3x^2 + 3C. Let K = 3C be a new arbitrary constant.

Vy^3 = 3x^2 + K.

VIy = (3x^2 + K)^(1/3).

Answer

y = (3x^2 + K)^(1/3)

Always include the constant of integration when finding general solutions.

Example 4

Find the general solution to dy/dx + (1/x)y = e^x.

IIdentify P(x) and Q(x). Here, P(x) = 1/x and Q(x) = e^x.

IICalculate the integrating factor I(x) = e^(∫ P(x) dx).

III∫ P(x) dx = ∫ (1/x) dx = ln|x|. For simplicity, assume x > 0 and use ln(x).

IVI(x) = e^(ln(x)) = x.

VMultiply the entire differential equation by the integrating factor x: x * (dy/dx) + x * (1/x)y = x * e^x.

VIThis simplifies to x * (dy/dx) + y = x * e^x.

VIIRecognise the left side as the derivative of a product: d/dx (xy) = x * e^x.

VIIIIntegrate both sides with respect to x: ∫ d/dx (xy) dx = ∫ x * e^x dx.

9xy = ∫ x * e^x dx.

10Evaluate ∫ x * e^x dx using integration by parts (let u=x, dv=e^x dx): x * e^x - ∫ e^x dx = x * e^x - e^x + C.

11Substitute back: xy = x * e^x - e^x + C.

12Solve for y: y = (x * e^x - e^x + C) / x.

13Simplify: y = e^x - (e^x)/x + C/x.

Answer

y = e^x - (e^x)/x + C/x

The integrating factor transforms the left side into the derivative of a product, making it integrable.

Common mistakes

✗Forgetting the constant of integration, + C, for indefinite integrals and general solutions of differential equations.
✗Incorrectly choosing 'u' and 'dv' in integration by parts, leading to a more complex integral.
✗Algebraic errors when decomposing rational functions into partial fractions.
✗Not correctly separating variables in separable differential equations (e.g., leaving 'y' terms on the 'x' side).
✗Errors in calculating or applying the integrating factor for linear differential equations.

Exam tips

★Always check your indefinite integrals by differentiating your answer to see if you get the original integrand.
★For integration by parts, carefully consider your choice of 'u' and 'dv' to simplify the process.
★When solving differential equations, clearly state whether you are finding a general or particular solution, and show all steps for finding C if applicable.
★Practise a wide variety of problems for each technique to build confidence and speed.

Ready to practise?

Try a problem on this topic

Snap a photo or type a question — get step-by-step working instantly.

Solve a problem →← Maths curriculum