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Ultimate Guide to Further Integration in HSC Mathematics Extension 2

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    Vu Hung
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Introduction

Integration in HSC Mathematics Extension 2 builds significantly on the foundations laid in Advanced and Extension 1. Further Integration requires a deep toolkit of algebraic and trigonometric manipulations to transform complex integrands into integrable forms. This guide covers the essential techniques you need to master for the HSC exams, from t-formulas to integration by parts.

Executive Summary

This comprehensive guide walks you through the core components of the Further Integration syllabus:

  • Trigonometric Products and Sums: Converting products of sines and cosines into sums or differences.
  • The t-Formulas: Using the universal substitution t=tan(x2)t = \tan(\frac{x}{2}) to solve trigonometric equations and integrals.
  • Advanced Substitution: Evaluating complex integrals using provided or derived substitutions.
  • Partial Fractions: Decomposing rational functions to integrate expressions with distinct linear, irreducible quadratic, or perfect square factors.
  • Completing the Square: A vital algebraic step for integrating quadratic denominators.
  • Integration by Parts: The product rule for integration, applied once or multiple times, and used to derive recurrence relations.

What is this about?

Further integration is about expanding your mathematical arsenal. In Extension 1, you learned the basics of substitution and integrating inverse trigonometric functions. Extension 2 demands that you recognise patterns and apply the correct technique—whether it's an algebraic decomposition using partial fractions or a trigonometric identity substitution—to solve highly complex problems.

Main Content

1. Trigonometric Products as Sums and Differences

The products to sums identities are crucial for integrating expressions like sin(mx)cos(nx)\sin(mx)\cos(nx). We can derive these from the compound angle formulas:

  • sin(A+B)+sin(AB)=2sinAcosB\sin(A+B) + \sin(A-B) = 2\sin A \cos B
  • sin(A+B)sin(AB)=2cosAsinB\sin(A+B) - \sin(A-B) = 2\cos A \sin B
  • cos(AB)cos(A+B)=2sinAsinB\cos(A-B) - \cos(A+B) = 2\sin A \sin B
  • cos(A+B)+cos(AB)=2cosAcosB\cos(A+B) + \cos(A-B) = 2\cos A \cos B

Example: Evaluate sin(3x)cos(5x)dx\int \sin(3x) \cos(5x) \, dx. Using the identity 2sinAcosB=sin(A+B)+sin(AB)2\sin A \cos B = \sin(A+B) + \sin(A-B): sin(3x)cos(5x)=12[sin(8x)+sin(2x)]=12[sin(8x)sin(2x)]\sin(3x) \cos(5x) = \frac{1}{2} [\sin(8x) + \sin(-2x)] = \frac{1}{2} [\sin(8x) - \sin(2x)] sin(3x)cos(5x)dx=12(sin(8x)sin(2x))dx=12[cos(8x)8+cos(2x)2]+C\int \sin(3x) \cos(5x) \, dx = \frac{1}{2} \int (\sin(8x) - \sin(2x)) \, dx = \frac{1}{2} \left[ -\frac{\cos(8x)}{8} + \frac{\cos(2x)}{2} \right] + C

2. The t-Formulas

The tt-formulas provide a powerful method for solving trigonometric equations and integrals by converting them into algebraic expressions. Let t=tan(x2)t = \tan(\frac{x}{2}). Then:

  • sinx=2t1+t2\sin x = \frac{2t}{1+t^2}
  • cosx=1t21+t2\cos x = \frac{1-t^2}{1+t^2}
  • tanx=2t1t2\tan x = \frac{2t}{1-t^2}
  • dx=21+t2dtdx = \frac{2}{1+t^2} dt

Example: Evaluate 15+4cosxdx\int \frac{1}{5 + 4\cos x} \, dx. Substitute t=tan(x2)t = \tan(\frac{x}{2}), dx=21+t2dtdx = \frac{2}{1+t^2} dt: 15+4(1t21+t2)21+t2dt=25(1+t2)+4(1t2)dt=29+t2dt\int \frac{1}{5 + 4(\frac{1-t^2}{1+t^2})} \frac{2}{1+t^2} \, dt = \int \frac{2}{5(1+t^2) + 4(1-t^2)} \, dt = \int \frac{2}{9 + t^2} \, dt =23tan1(t3)+C=23tan1(tan(x/2)3)+C= \frac{2}{3} \tan^{-1}\left(\frac{t}{3}\right) + C = \frac{2}{3} \tan^{-1}\left(\frac{\tan(x/2)}{3}\right) + C

3. Partial Fractions

When integrating a rational function P(x)Q(x)\frac{P(x)}{Q(x)} where the degree of the numerator P(x)P(x) is less than the denominator Q(x)Q(x), we decompose it into partial fractions.

Types of Denominators:

  • Distinct Linear Factors: 1(xa)(xb)=Axa+Bxb\frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}
  • Perfect Square Factors: 1(xa)2(xb)=Axa+B(xa)2+Cxb\frac{1}{(x-a)^2(x-b)} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b}
  • Irreducible Quadratic Factors: 1(x2+a2)(xb)=Ax+Bx2+a2+Cxb\frac{1}{(x^2+a^2)(x-b)} = \frac{Ax+B}{x^2+a^2} + \frac{C}{x-b}

Note: If the degree of the numerator is \ge the denominator, you must use polynomial long division first.

4. Completing the Square

For integrals with a quadratic denominator ax2+bx+cax^2 + bx + c that cannot be factorised into real linear factors, complete the square to form a(xh)2+ka(x-h)^2 + k. This usually leads to an inverse tangent function or requires a trigonometric substitution.

Example: 1x2+4x+13dx=1(x+2)2+9dx=13tan1(x+23)+C\int \frac{1}{x^2 + 4x + 13} \, dx = \int \frac{1}{(x+2)^2 + 9} \, dx = \frac{1}{3}\tan^{-1}\left(\frac{x+2}{3}\right) + C

5. Integration by Parts

Derived from the product rule for differentiation, integration by parts is essential for integrating products of different types of functions (e.g., algebraic and transcendental). Formula: udv=uvvdu\int u \, dv = uv - \int v \, du

Example: Recurrence Relations Let In=xnexdxI_n = \int x^n e^x \, dx. Let u=xn    du=nxn1dxu = x^n \implies du = n x^{n-1} dx Let dv=exdx    v=exdv = e^x dx \implies v = e^x In=xnexnxn1exdx=xnexnIn1I_n = x^n e^x - \int n x^{n-1} e^x \, dx = x^n e^x - n I_{n-1}

mini-FAQ page

Q: Do I always have to use t=tan(x/2)t = \tan(x/2) for trigonometric integrals? A: Not always. If the integrand has an odd power of sine or cosine, a standard substitution like u=cosxu = \cos x or u=sinxu = \sin x is usually faster. Use tt-formulas when the denominator contains a linear combination of sinx\sin x and cosx\cos x.

Q: How do I know when to use integration by parts? A: Use it when you have a product of two unrelated functions, like a polynomial and an exponential (xexx e^x), or a polynomial and a logarithmic function (xlnxx \ln x). The acronym LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) can help you choose uu.

Common mistakes to avoid

  • Forgetting the +C+ C: Always remember the constant of integration for indefinite integrals.
  • Incorrect Partial Fraction Setup: Ensure you use the correct numerator forms (Ax+BAx+B for irreducible quadratics, AA for linear factors).
  • Ignoring the Numerator Degree: If integrating x3x2+1\frac{x^3}{x^2+1}, you must perform long division first because the numerator's degree is greater than the denominator's.

Practice on Vu's Maths Hub

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