- Chapter 7
- Calculus
- 24 questions
Integrals
Integration as the inverse of differentiation, the standard methods, definite integrals and their properties.
Integration is differentiation run backwards, and that is exactly why it is harder. Differentiating is a procedure: apply the rules and an answer comes out. Integrating is a matter of recognition — deciding which method will turn an unfamiliar integrand into a familiar one.
This chapter is therefore mostly about choosing well. Substitution, partial fractions and integration by parts each suit a particular shape of integrand, and building the judgement to see which applies is worth more than memorising any individual result.
What you should be able to do
- Integrate standard functions and recognise integration as the inverse of differentiation.
- Apply integration by substitution, including trigonometric substitutions.
- Integrate rational functions using partial fractions.
- Apply integration by parts, choosing the first function using the ILATE guideline.
- Evaluate integrals of the standard quadratic forms.
- Evaluate definite integrals using the Fundamental Theorem of Calculus and simplify them using their properties.
Topics
7 topics
- Integration as an Inverse Process of Differentiation
Indefinite integrals, the constant of integration, and the standard results.
- Integration by Substitution
Changing the variable to reduce an integral to a standard form.
- Integration by Partial Fractions
Decomposing a rational function before integrating it, for each standard denominator type.
- Integration by Parts
The product rule reversed, the ILATE guideline, and integrals requiring repeated application.
- Integrals of Special Forms
The standard quadratic-denominator and square-root forms, reached by completing the square.
- Forms with $x^2 \pm a^2$ and $a^2 - x^2$
- Forms with $ax^2 + bx + c$
- Linear numerator over a quadratic
- Definite Integrals
The Fundamental Theorem of Calculus and evaluation of definite integrals.
- Properties of Definite Integrals
The standard properties, including symmetry about the interval, and their use in simplifying integrals.
Formulas
15 items · All formulas
Power rule
\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + C- Valid when
- n \neq -1. For n = -1 the integral is \ln|x| + C.
Reciprocal
\int \dfrac{1}{x}\,dx = \ln|x| + CThe modulus matters — the integrand is defined for negative x too.
Standard trigonometric integrals
\int \sin x\,dx = -\cos x + C, \quad \int \cos x\,dx = \sin x + C, \quad \int \sec^{2}x\,dx = \tan x + CIntegrals giving logarithms of trigonometric functions
\int \tan x\,dx = \ln|\sec x| + C, \qquad \int \cot x\,dx = \ln|\sin x| + CIntegration by parts
\int u\,v'\,dx = uv - \int u'\,v\,dxChoose u by ILATE: Inverse, Logarithmic, Algebraic, Trigonometric, Exponential.
The exponential shortcut
\int e^{x}\left[f(x) + f'(x)\right]dx = e^{x}f(x) + CLook for an e^x multiplying a bracket where one term is the derivative of the other.
Standard form: difference of squares in the denominator
\int \dfrac{dx}{x^{2} - a^{2}} = \dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right| + CStandard form: sum of squares in the denominator
\int \dfrac{dx}{x^{2} + a^{2}} = \dfrac{1}{a}\tan^{-1}\!\left(\dfrac{x}{a}\right) + CStandard form: square root of a difference
\int \dfrac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{-1}\!\left(\dfrac{x}{a}\right) + C- Valid when
- |x| < a.
Standard form: square root of a sum
\int \dfrac{dx}{\sqrt{x^{2} + a^{2}}} = \ln\left|x + \sqrt{x^{2}+a^{2}}\right| + CIntegral of a square root
\int \sqrt{a^{2} - x^{2}}\,dx = \dfrac{x}{2}\sqrt{a^{2}-x^{2}} + \dfrac{a^{2}}{2}\sin^{-1}\!\left(\dfrac{x}{a}\right) + CFundamental Theorem of Calculus
\int_{a}^{b} f(x)\,dx = F(b) - F(a) \quad \text{where } F' = fReversal property
\int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dxThe reflection property
\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dxThe single most useful property in examinations — it often turns an intractable integrand into a tractable sum.
Even and odd functions
\int_{-a}^{a} f(x)\,dx = \begin{cases} 2\displaystyle\int_{0}^{a} f(x)\,dx, & f \text{ even} \\[6pt] 0, & f \text{ odd} \end{cases}Check the parity before integrating — an odd integrand vanishes immediately.
Key points
8 items · All key points
Never omit the constant of integration C in an indefinite integral.
\displaystyle\int \frac{1}{x}\,dx = \ln|x| + C — the modulus is part of the answer.
When substituting in a definite integral, change the limits to match the new variable.
Either convert the limits, or convert back to the original variable before substituting. Doing neither is a common and costly error.
Choose u for integration by parts using ILATE.
Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential — earliest in the list becomes u.
Check whether \int_{-a}^{a} has an odd integrand — if so the answer is 0 with no work at all.
Before using partial fractions, confirm the degree of the numerator is less than that of the denominator; otherwise divide first.
For \int \frac{dx}{ax^2+bx+c}, complete the square to reach one of the standard forms.
\displaystyle\int e^{x}[f(x) + f'(x)]\,dx = e^{x}f(x) + C — recognising this shape saves a full round of integration by parts.
Common mistakes
6 items · All common mistakes
- Mistake
Omitting the constant of integration.
InsteadEvery indefinite integral carries +\,C.
WhyIt contributes nothing to the visible working, so it is dropped when copying the final line.
- Mistake
Substituting in a definite integral but leaving the original limits in place.
InsteadIf u = g(x), the limits become g(a) and g(b). Alternatively substitute back to x before applying the original limits.
WhyThe substitution step feels complete once the integrand is transformed, so the limits are overlooked.
- Mistake
Writing \int \frac{1}{x}\,dx = \ln x + C without the modulus.
Instead\ln|x| + C.
- Mistake
Choosing u and v' for integration by parts in the order they happen to be written.
InsteadApply ILATE. Choosing badly usually produces an integral harder than the original.
- Mistake
Applying partial fractions when the numerator has degree greater than or equal to the denominator.
InsteadDivide first to get a polynomial plus a proper fraction, then decompose the proper part.
WhyThe decomposition set-up looks applicable regardless of degree, and produces an inconsistent system if used improperly.
- Mistake
Treating \int \frac{dx}{x^2 - a^2} and \int \frac{dx}{x^2 + a^2} as the same standard form.
InsteadThe first gives a logarithm; the second gives an inverse tangent. The sign changes the answer completely.