CIE AS/A Level Maths-3.5 Integration- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-3.5 Integration- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-3.5 Integration- Study Notes
Key Concepts:
- Basic Integration
- Integration Using Trigonometric Relationships
- Integration of Rational Functions Using Partial Fractions
- Integration of Functions of the Form \(\dfrac{Kf'(x)}{f(x)}\)
- Integration by Parts
- Integration Using Substitution
Basic Integration
Basic Integration
| Type | Integral | Result |
|---|---|---|
| Exponential Functions | \(\displaystyle \int e^{ax+b}\, dx\) | \(\displaystyle \tfrac{1}{a} e^{ax+b} + C,\; a \neq 0\) |
| Reciprocal Linear | \(\displaystyle \int \tfrac{1}{ax+b}\, dx\) | \(\displaystyle \tfrac{1}{a} \ln|ax+b| + C,\; a \neq 0\) |
| Sine | \(\displaystyle \int \sin(ax+b)\, dx\) | \(\displaystyle -\tfrac{1}{a}\cos(ax+b) + C\) |
| Cosine | \(\displaystyle \int \cos(ax+b)\, dx\) | \(\displaystyle \tfrac{1}{a}\sin(ax+b) + C\) |
| Secant Squared | \(\displaystyle \int \sec^2(ax+b)\, dx\) | \(\displaystyle \tfrac{1}{a}\tan(ax+b) + C\) |
| Reciprocal Quadratic | \(\displaystyle \int \tfrac{1}{a^2+x^2}\, dx\) | \(\displaystyle \tfrac{1}{a}\tan^{-1}\!\Big(\tfrac{x}{a}\Big) + C\) |
Example:
Evaluate \(\displaystyle \int e^{3x-2} \, dx\)
▶️ Answer / Explanation
\(\displaystyle \int e^{3x-2} dx = \dfrac{1}{3} e^{3x-2} + C\)
Example:
Evaluate \(\displaystyle \int \dfrac{1}{2+3x^2} dx\)
▶️ Answer / Explanation
Step 1: Factor to match \(\int \dfrac{dx}{a^2 + x^2} = \dfrac{1}{a} \tan^{-1}\dfrac{x}{a} + C\)
\(\displaystyle \int \dfrac{1}{2+3x^2} dx = \int \dfrac{1}{\sqrt{2}^2 + (\sqrt{3} x)^2} dx\)
Step 2: Apply formula:
\(\displaystyle = \dfrac{1}{\sqrt{3}} \tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + C\)
Final Answer: \(\boxed{\dfrac{1}{\sqrt{3}} \tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + C}\)
Example:
Evaluate \(\displaystyle \int \cos(2x+1) dx\)
▶️ Answer / Explanation
\(\displaystyle \int \cos(2x+1) dx = \dfrac{1}{2} \sin(2x+1) + C\)
Example:
Evaluate \(\displaystyle \int \sec^2(3x-2) dx\)
▶️ Answer / Explanation
\(\displaystyle \int \sec^2(3x-2) dx = \dfrac{1}{3} \tan(3x-2) + C\)
Example:
Evaluate \(\displaystyle \int \dfrac{dx}{4 + x^2}\)
▶️ Answer / Explanation
\(\displaystyle \int \dfrac{dx}{4 + x^2} = \dfrac{1}{2} \tan^{-1}\left(\dfrac{x}{2}\right) + C\)
Integration Using Trigonometric Relationships
Integration Using Trigonometric Relationships
Using Double-Angle Formulae:
Double-angle identities useful for integration:
- \(\sin^2 x = \dfrac{1 – \cos 2x}{2}\)
- \(\cos^2 x = \dfrac{1 + \cos 2x}{2}\)
- \(\cos^2 2x = \dfrac{1 + \cos 4x}{2}\)
Example:
Evaluate \(\displaystyle \int \sin^2 x \, dx\)
▶️ Answer / Explanation
Step 1: Use double-angle formula: \(\sin^2 x = \dfrac{1 – \cos 2x}{2}\)
Step 2: Substitute into integral: \(\displaystyle \int \sin^2 x \, dx = \int \dfrac{1 – \cos 2x}{2} dx = \dfrac{1}{2} \int (1 – \cos 2x) dx\)
Step 3: Integrate term by term: \(\dfrac{1}{2} \left( \int 1 \, dx – \int \cos 2x \, dx \right) = \dfrac{1}{2} \left( x – \dfrac{1}{2} \sin 2x \right) + C\)
Final Answer: \(\boxed{\int \sin^2 x \, dx = \dfrac{x}{2} – \dfrac{\sin 2x}{4} + C}\)
Example:
Evaluate \(\displaystyle \int \cos^2 (2x) \, dx\)
▶️ Answer / Explanation
Step 1: Use double-angle formula: \(\cos^2 2x = \dfrac{1 + \cos 4x}{2}\)
Step 2: Substitute into integral: \(\displaystyle \int \cos^2 2x \, dx = \int \dfrac{1 + \cos 4x}{2} dx = \dfrac{1}{2} \int (1 + \cos 4x) dx\)
Step 3: Integrate term by term: \(\dfrac{1}{2} \left( x + \dfrac{1}{4} \sin 4x \right) + C = \dfrac{x}{2} + \dfrac{\sin 4x}{8} + C\)
Final Answer: \(\boxed{\int \cos^2 2x \, dx = \dfrac{x}{2} + \dfrac{\sin 4x}{8} + C}\)
Integration of Rational Functions Using Partial Fractions
Integration of Rational Functions Using Partial Fractions
If a rational function \(\dfrac{P(x)}{Q(x)}\) can be decomposed into partial fractions, then
\(\displaystyle \frac{P(x)}{Q(x)} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2} + \frac{Dx + E}{cx^2 + d} + \dots \)
Then integrate each term separately using basic formulas:
- \(\displaystyle \int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C\)
- \(\displaystyle \int \frac{1}{(ax+b)^2} dx = -\frac{1}{a(ax+b)} + C\)
- \(\displaystyle \int \frac{Dx+E}{cx^2 + d} dx = \frac{D}{2c} \ln|cx^2 + d| + \frac{E}{\sqrt{d}} \tan^{-1}\frac{\sqrt{c}x}{\sqrt{d}} + C\) (if needed)
Example:
Evaluate \(\displaystyle \int \frac{7x+3}{(x+1)(2x-1)} dx\)
▶️ Answer / Explanation
Step 1: Decompose into partial fractions:
\(\frac{7x+3}{(x+1)(2x-1)} = \frac{A}{x+1} + \frac{B}{2x-1}\)
Step 2: Multiply both sides by \((x+1)(2x-1)\) and equate coefficients:
\(7x + 3 = A(2x-1) + B(x+1)\)
\(7x + 3 = (2A + B)x + (-A + B)\)
Equating coefficients: 2A + B = 7, -A + B = 3 → Solve: A = 2, B = 3
Step 3: Integrate each term:
\(\int \frac{2}{x+1} dx + \int \frac{3}{2x-1} dx = 2 \ln|x+1| + \frac{3}{2} \ln|2x-1| + C\)
Final Answer: \(\boxed{2 \ln|x+1| + \frac{3}{2} \ln|2x-1| + C}\)
Example:
Evaluate \(\displaystyle \int \frac{3x+5}{(x+1)(x+2)^2} dx\)
▶️ Answer / Explanation
Step 1: Decompose into partial fractions:
\(\frac{3x+5}{(x+1)(x+2)^2} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}\)
Step 2: Multiply both sides by \((x+1)(x+2)^2\) and equate coefficients:
\(3x+5 = A(x+2)^2 + B(x+1)(x+2) + C(x+1)\)
Solve for A, B, C → A = 1, B = 2, C = 2
Step 3: Integrate each term:
\(\int \frac{1}{x+1} dx + \int \frac{2}{x+2} dx + \int \frac{2}{(x+2)^2} dx = \ln|x+1| + 2 \ln|x+2| – \frac{2}{x+2} + C\)
Final Answer: \(\boxed{\ln|x+1| + 2 \ln|x+2| – \frac{2}{x+2} + C}\)
Integration of Functions of the Form \(\dfrac{Kf'(x)}{f(x)}\)
Integration of Functions of the Form \(\dfrac{Kf'(x)}{f(x)}\)
If an integrand can be written in the form \(\dfrac{f'(x)}{f(x)}\), the integral is given by:
\(\displaystyle \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C\)
- This works because the derivative of \(\ln|f(x)|\) is \(\dfrac{f'(x)}{f(x)}\).
Steps:
- Step 1: Identify the denominator or inner function \(f(x)\).
- Step 2: Check if the numerator is a constant multiple of \(f'(x)\).
- Step 3: If needed, factor out constants to match the formula.
- Step 4: Apply \(\int f’/f = \ln|f| + C\).
Example:
Evaluate \(\displaystyle \int \frac{x}{x^2 + 1} dx\)
▶️ Answer / Explanation
Step 1: Identify the denominator as \(f(x) = x^2 + 1\) with derivative \(f'(x) = 2x\).
Step 2: Adjust numerator to match derivative: \(\frac{x}{x^2+1} = \frac{1}{2} \cdot \frac{2x}{x^2+1}\)
Step 3: Apply formula: \(\int \frac{2x}{x^2+1} dx = \ln|x^2+1| + C\)
Include the factor \(\frac{1}{2}\): \(\int \frac{x}{x^2+1} dx = \frac{1}{2} \ln|x^2+1| + C\)
Final Answer: \(\boxed{\frac{1}{2} \ln|x^2+1| + C}\)
Example:
Evaluate \(\displaystyle \int \tan x \, dx\)
▶️ Answer / Explanation
Step 1: Rewrite \(\tan x\) as \(\frac{\sin x}{\cos x} = – \frac{d}{dx}(\cos x)/\cos x\)
Step 2: Recognize \(-\frac{d}{dx}(\cos x)/\cos x\) is in the form \(f’/f\).
Step 3: Apply formula: \(\int \tan x dx = \int -\frac{d}{dx}(\cos x)/\cos x dx = -\ln|\cos x| + C\)
Final Answer: \(\boxed{-\ln|\cos x| + C}\)
Example:
Evaluate \(\displaystyle \int \frac{3x+1}{3x^2 + 2x + 5} dx\)
▶️ Answer / Explanation
Step 1: Identify \(f(x) = 3x^2 + 2x + 5\) with derivative \(f'(x) = 6x + 2\)
Step 2: Factor numerator to match derivative: \(3x+1 = \frac{1}{2} (6x+2) – 0\)
Step 3: Apply formula: \(\int \frac{6x+2}{3x^2+2x+5} dx = \ln|3x^2+2x+5| + C\)
Include factor \(\frac{1}{2}\): \(\int \frac{3x+1}{3x^2+2x+5} dx = \frac{1}{2} \ln|3x^2+2x+5| + C\)
Final Answer: \(\boxed{\frac{1}{2} \ln|3x^2+2x+5| + C}\)
Integration by Parts
Integration by Parts
Integration by parts is based on the formula derived from the product rule:
\(\displaystyle \int u \, dv = uv – \int v \, du\)
- Where \(u\) is a function chosen to simplify upon differentiation, and \(dv\) is the remaining part of the integrand.
Steps:
- Step 1: Identify parts of the integrand: choose \(u\) and \(dv\).
- Step 2: Compute \(du = \frac{du}{dx} dx\) and \(v = \int dv\).
- Step 3: Substitute into \(\int u\, dv = uv – \int v\, du\).
- Step 4: Simplify and integrate the remaining integral if possible.
Guidelines for Choosing \(u\) and \(dv\):
- Use the ILATE rule (Inverse trig, Logarithmic, Algebraic, Trig, Exponential) to choose \(u\).
- The function that simplifies upon differentiation is usually chosen as \(u\).
Example:
Evaluate \(\displaystyle \int x \sin 2x \, dx\)
▶️ Answer / Explanation
Step 1: Choose \(u = x \implies du = dx\), and \(dv = \sin 2x dx \implies v = -\frac{1}{2} \cos 2x\)
Step 2: Apply formula: \(\int u\, dv = uv – \int v\, du\)
\(\displaystyle \int x \sin 2x dx = x \cdot \left(-\frac{1}{2} \cos 2x \right) – \int \left(-\frac{1}{2} \cos 2x\right) dx\)
\(\displaystyle = -\frac{x}{2} \cos 2x + \frac{1}{2} \int \cos 2x dx\)
\(\displaystyle = -\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x + C\)
Final Answer: \(\boxed{-\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x + C}\)
Example:
Evaluate \(\displaystyle \int x^2 e^{-x} dx\)
▶️ Answer / Explanation
Step 1: Choose \(u = x^2 \implies du = 2x dx\), and \(dv = e^{-x} dx \implies v = – e^{-x}\)
Step 2: Apply formula: \(\int u\, dv = uv – \int v\, du\)
\(\displaystyle \int x^2 e^{-x} dx = -x^2 e^{-x} + \int 2x e^{-x} dx\)
Step 3: Use integration by parts again on \(\int 2x e^{-x} dx\):
Let \(u = 2x \implies du = 2 dx\), \(dv = e^{-x} dx \implies v = -e^{-x}\)
\(\int 2x e^{-x} dx = -2x e^{-x} + \int 2 e^{-x} dx = -2x e^{-x} – 2 e^{-x}\)
Step 4: Combine results:
\(\displaystyle \int x^2 e^{-x} dx = -x^2 e^{-x} – 2x e^{-x} – 2 e^{-x} + C = -(x^2 + 2x + 2) e^{-x} + C\)
Final Answer: \(\boxed{-(x^2 + 2x + 2) e^{-x} + C}\)
Example:
Evaluate \(\displaystyle \int \ln x \, dx\)
▶️ Answer / Explanation
Step 1: Choose \(u = \ln x \implies du = \frac{1}{x} dx\), and \(dv = dx \implies v = x\)
\(\int \ln x \, dx = x \ln x – \int x \cdot \frac{1}{x} dx = x \ln x – \int 1 dx = x \ln x – x + C\)
Final Answer: \(\boxed{x \ln x – x + C}\)
Example:
Evaluate \(\displaystyle \int x \tan^{-1} x \, dx\)
▶️ Answer / Explanation
Step 1: Choose \(u = \tan^{-1} x \implies du = \frac{1}{1+x^2} dx\), and \(dv = x dx \implies v = \frac{x^2}{2}\)
\(\int x \tan^{-1} x dx = \frac{x^2}{2} \tan^{-1} x – \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} dx = \frac{x^2}{2} \tan^{-1} x – \frac{1}{2} \int \frac{x^2}{1+x^2} dx\)
Step 2: Simplify \(\frac{x^2}{1+x^2} = 1 – \frac{1}{1+x^2}\), so integral becomes:
\(-\frac{1}{2} \int \left(1 – \frac{1}{1+x^2}\right) dx = -\frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{1}{1+x^2} dx = -\frac{x}{2} + \frac{1}{2} \tan^{-1} x\)
Step 3: Combine results:
\(\int x \tan^{-1} x dx = \frac{x^2}{2} \tan^{-1} x – \frac{x}{2} + \frac{1}{2} \tan^{-1} x + C = \frac{x^2 + 1}{2} \tan^{-1} x – \frac{x}{2} + C\)
Final Answer: \(\boxed{\frac{x^2 + 1}{2} \tan^{-1} x – \frac{x}{2} + C}\)
Integration Using Substitution
Integration Using Substitution
Substitution is used to simplify an integral by changing variables.
If \(u = g(x)\), then \(du = g'(x) dx\) and the integral can be rewritten in terms of \(u\):
\(\displaystyle \int f(g(x)) g'(x) dx = \int f(u) du\)
Steps:
- Step 1: Identify a part of the integrand to substitute (usually inner function).
- Step 2: Set \(u = g(x)\) and compute \(du = g'(x) dx\).
- Step 3: Replace all instances of \(x\) in the integral using \(u\) and \(du\).
- Step 4: Integrate with respect to \(u\).
- Step 5: Substitute back \(x\) to get the final answer.
Example:
Evaluate \(\displaystyle \int \sin^2 2x \cos x \, dx\) using \(u = \sin x\)
▶️ Answer / Explanation
Step 1: Substitute \(u = \sin x \implies du = \cos x dx\)
Step 2: Rewrite \(\sin^2 2x \cos x dx\) in terms of \(u\). First, \(\sin 2x = 2 \sin x \cos x = 2 u \cos x\)
So \(\sin^2 2x \cos x dx = (2 u \cos x)^2 \cos x dx = 4 u^2 (\cos x)^2 \cos x dx\)
Notice \(cos x dx = du\), so one \(\cos x\) is replaced:
\(\displaystyle \int \sin^2 2x \cos x dx = \int 4 u^2 (\cos x)^2 du\)
Also, \(\cos^2 x = 1 – u^2\), so integral becomes:
\(\int 4 u^2 (1-u^2) du = \int 4(u^2 – u^4) du = 4 \int u^2 du – 4 \int u^4 du\)
\(\displaystyle = 4 \cdot \frac{u^3}{3} – 4 \cdot \frac{u^5}{5} + C = \frac{4}{3} u^3 – \frac{4}{5} u^5 + C\)
Step 3: Substitute back \(u = \sin x\):
Final Answer: \(\boxed{\frac{4}{3} \sin^3 x – \frac{4}{5} \sin^5 x + C}\)
Example :
Evaluate \(\displaystyle \int x \sqrt{x^2 + 1} \, dx\) using \(u = x^2 + 1\)
▶️ Answer / Explanation
Step 1: Substitute \(u = x^2 + 1 \implies du = 2x dx \implies x dx = \frac{1}{2} du\)
Step 2: Replace in integral: \(\int x \sqrt{x^2 + 1} dx = \int \sqrt{u} \cdot \frac{1}{2} du = \frac{1}{2} \int u^{1/2} du\)
Step 3: Integrate: \(\frac{1}{2} \cdot \frac{u^{3/2}}{3/2} = \frac{1}{3} u^{3/2}\)
Step 4: Substitute back \(u = x^2 + 1\):
Final Answer: \(\boxed{\frac{1}{3} (x^2 + 1)^{3/2} + C}\)