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Integration Techniques

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Integration Techniques

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Integration Techniques

🎯 Learning Objectives

By the end of this topic, you will be able to:

Apply substitution method for composite functions
Use integration by parts for products of functions
Decompose rational functions using partial fractions
Integrate piecewise-defined functions

📋 Prerequisites

Definite Integrals — FTC, properties
Algebra — polynomial division, factoring

📖 Core Content

17.1 Review of Substitution

\int f(g(x)) \cdot g'(x)\,dx = \int f(u)\,du \quad \text{where } u = g(x)

Example:

\int 2x \cos(x^2)\,dx

Let

u = x^2

du = 2x\,dx

\int \cos u\,du = \sin u + C = \sin(x^2) + C

17.2 Integration by Parts

\int u\,dv = uv - \int v\,du

Intuition: The reverse of the product rule. Choose

u

as the part that simplifies when differentiated. LIATE rule (order of choosing

u

): Logarithms, Inverse trig, Algebraic, Trigonometric, Exponential.

Worked Examples

Example 2.1:

\int x e^x\,dx

Let

u = x

dv = e^x\,dx

. Then

du = dx

v = e^x

\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1) + C

\boxed{e^x(x-1) + C}

Example 2.2:

\int \ln x\,dx

Let

u = \ln x

dv = dx

. Then

du = \frac{1}{x}dx

v = x

\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C

\boxed{x\ln x - x + C}

Example 2.3:

\int x^2 e^x\,dx

(repeated parts) Let

u = x^2

dv = e^x\,dx

du = 2x\,dx

v = e^x

= x^2 e^x - \int 2x e^x\,dx = x^2 e^x - 2(x e^x - e^x) + C = e^x(x^2 - 2x + 2) + C

\boxed{e^x(x^2 - 2x + 2) + C}

17.3 Partial Fractions

For

\int \frac{P(x)}{Q(x)}\,dx

where degree of

P

< degree of

Q

: Step 1: Factor

Q(x)

Step 2: Write partial fraction decomposition Step 3: Integrate each term

Factor in $Q(x)$	Term in Decomposition
$(ax + b)$	$\frac{A}{ax+b}$
$(ax+b)^k$	$\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \cdots + \frac{A_k}{(ax+b)^k}$
$(ax^2+bx+c)$ irreducible	$\frac{Ax+B}{ax^2+bx+c}$

Worked Example

Example 3.1:

\int \frac{2x+3}{x^2 + x - 2}\,dx

Step 1 — Factor:

x^2+x-2 = (x-1)(x+2)

Step 2 — Decompose:

\frac{2x+3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}

2x+3 = A(x+2) + B(x-1) = (A+B)x + (2A - B)

A+B = 2

2A-B = 3 \implies A = 5/3

B = 1/3

Step 3 — Integrate:

\int \frac{5/3}{x-1}\,dx + \int \frac{1/3}{x+2}\,dx = \frac{5}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C

\boxed{\frac{1}{3}\left(5\ln|x-1| + \ln|x+2|\right) + C}

17.4 Trigonometric Integrals

Type	Strategy
$\int \sin^m x \cos^n x\,dx$	If $m$ odd, save $\sin x$ , convert rest to $\cos$
	If $n$ odd, save $\cos x$ , convert rest to $\sin$
	If both even, use half-angle formulas
$\int \tan^m x \sec^n x\,dx$	If $n$ even, save $\sec^2 x$
	If $m$ odd, save $\sec x \tan x$

Example:

\int \sin^2 x\,dx = \int \frac{1-\cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C

17.5 Integration of Piecewise Functions

Split the integral at the piece boundaries and integrate each piece separately. Example:

\int_{-1}^2 f(x)\,dx

where

f(x) = \begin{cases} x & x < 0 \\ x^2 & x \geq 0 \end{cases}

\int_{-1}^0 x\,dx + \int_0^2 x^2\,dx = \left[\frac{x^2}{2}\right]_{-1}^0 + \left[\frac{x^3}{3}\right]_0^2 = (0 - \frac{1}{2}) + (\frac{8}{3} - 0) = -\frac{1}{2} + \frac{8}{3} = \frac{13}{6}

📐 Key Formulas — Summary Table

Technique	Formula	When to Use
Substitution	$\int f(g(x))g'(x)\,dx = \int f(u)\,du$	Composite functions
Parts	$\int u\,dv = uv - \int v\,du$	Product of functions
Partial fractions	$\frac{P(x)}{Q(x)} = \sum \frac{A_i}{\text{factor}}$	Rational functions
Trig identities	$\sin^2 x = \frac{1-\cos 2x}{2}$	Powers of trig

⚠️ Common Pitfalls

Pitfall 1: Wrong Choice of $u$ in Parts

Choose

u

that simplifies when differentiated. Use LIATE:

\ln x

before algebraic, exponential last.

Pitfall 2: Forgetting Absolute Values in Log Integrals

\int \frac{1}{x}\,dx = \ln|x| + C

, not

\ln x + C

(which is only valid for

x > 0

Pitfall 3: Partial Fractions Requires Proper Rational Functions

If degree of numerator ≥ degree of denominator, divide first.

📝 Practice Questions

Q1: Find $\int x \sin x\,dx$ .

Parts: $u=x$ , $dv=\sin x\,dx$ , $du=dx$ , $v=-\cos x$ . $= -x\cos x + \int \cos x\,dx = -x\cos x + \sin x + C$

$\boxed{-x\cos x + \sin x + C}$ Q2: Find $\int \frac{dx}{x^2 - 1}$ .

$\frac{1}{(x-1)(x+1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1}$ $\int = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$

$\boxed{\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C}$ Q3: Find $\int \ln(x^2)\,dx$ .

$\ln(x^2) = 2\ln x$ $2\int \ln x\,dx = 2(x\ln x - x) + C$

$\boxed{2x\ln x - 2x + C}$ Q4: Find $\int xe^{2x}\,dx$ .

Parts: $u=x$ , $dv=e^{2x}dx$ , $du=dx$ , $v=e^{2x}/2$ . $= \frac{xe^{2x}}{2} - \frac{1}{2}\int e^{2x}dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C$

$\boxed{\frac{e^{2x}}{4}(2x - 1) + C}$ Q5: Find $\int_0^1 x e^x\,dx$ .

$[xe^x - e^x]_0^1 = (e - e) - (0 - 1) = 1$

$\boxed{1}$ Q6: Find $\int \frac{dx}{x^2 + 4}$ .

$\frac{1}{2}\arctan\left(\frac{x}{2}\right) + C$

$\boxed{\frac{1}{2}\arctan\left(\frac{x}{2}\right) + C}$ **Q7: Find $\int_{-1}^1 f(x)\,dx$ where

> f(x) = \begin{cases} 1 & x<0 \\ 2 & x \geq 0 \end{cases} >

.**

$\int_{-1}^0 1\,dx + \int_0^1 2\,dx = 1 + 2 = 3$

$\boxed{3}$ Q8: Find $\int \cos^2 x\,dx$ .

$\frac{1+\cos 2x}{2}$ → $\frac{x}{2} + \frac{\sin 2x}{4} + C$

$\boxed{\frac{x}{2} + \frac{\sin 2x}{4} + C}$

🔗 Cross-References

Previous: Definite Integrals
Next: Graph Fundamentals
Across courses: BSMA1003 Maths 2 (advanced integration techniques, multiple integrals) Join Discord Previous9.2 Definite Integrals & FTCNext10.1 Graph Fundamentals