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Integration Flowchart

This flowchart will only be of use to those who have already learnt the techniques mentioned in the International Baccalaureate Mathematics Analysis and Approaches Higher Level course. It is designed to help you decide which of the techniques to use.


Monday, January 8, 2024

"Back in 1985, in Valley Park School, my A-Level group and I filled a large wall with a huge integration flowchart. It was magnificent. I wish I had taken a photograph of it because it no longer exists. In recent years I have had various attempts of recreating it digitally and the latest version can be seen above."

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Exercises on indefinite and definite integration of basic algebraic and trigonometric functions.

The short web address is:



Trapezium Rule

Trapezium Rule

Practise using the trapezium rule to find an approximate value for the area under a curve.

The short web address is:



Polynomial Division

Polynomial Division

Practise dividing one algebraic expression by another in this set of exercises.

The short web address is:


Standard Derivatives and Integrals

Integration by Parts is a technique used to integrate the product of two functions. It involves choosing one of the functions as the "\(u\)" term and the other as the "\(\frac{dv}{dx}\)" term, then applying the formula:

$$\int u \frac{dv}{dx} = uv - \int v \frac{du}{dx}dx$$


$$\int udv = uv - \int v du$$

where "\(u\)" and "\(v\)" are functions, and "\(du\)" and "\(dv\)" represent their differentials. This formula is derived from the product rule for differentiation.

For example, consider the integral:

$$\int x\sin x \quad dx$$

Here, we can choose "\(u\)" to be "\(x\)" and "\(dv\)" to be "\(\sin x\)". Then, we have:

$$v = -\cos x$$

Using the formula, we get:

\begin{align*} \int x\sin x \quad dx &= -x\cos x - \int (-\cos x)(1) \quad dx \\ &= -x\cos x + \sin x + c \end{align*}

where \(c\) is the constant of integration.

Integration by Parts can be used for more complicated integrals as well, such as those involving logarithmic, exponential, or trigonometric functions.

5. Integrating factor

Consider a first-order linear differential equation of the form \(y' + P(x)y = Q(x)\). To solve this type of equation, we often use a technique called the integrating factor method. The key to this method is to multiply the equation by an "integrating factor", which is the exponential of the integral of the function \(P(x)\), i.e., \(e^{\int P(x) dx}\). This transformation converts the left-hand side of the equation into the derivative of a product, making it possible to solve the equation.

The key formula for the integrating factor method is as follows:

$$ \begin{align*} \quad \text{The integrating factor is: } e^{\int P(x) dx} \\\ \end{align*} $$

Now, let's see an example:

Solve the following differential equation: \(y' + 2xy = x\)

The solution proceeds as follows:

$$ \begin{align*} 1. & \quad \text{Identify } P(x) = 2x \text{ and } Q(x) = x \\\ 2. & \quad \text{The integrating factor is } e^{\int 2x dx} = e^{x^2} \\\ 3. & \quad \text{Multiply the entire equation by the integrating factor: } e^{x^2}y' + 2x e^{x^2}y = xe^{x^2} \\\ 4. & \quad \text{The left-hand side is now the derivative of } (y e^{x^2}), \text{ so: } (y e^{x^2})' = xe^{x^2} \\\ 5. & \quad \text{Integrate both sides to get: } y e^{x^2} = \int xe^{x^2} dx + C \\\ 6. & \quad \text{This integral is a standard result: } \frac{1}{2}e^{x^2} + C \\\ 7. & \quad \text{Finally, solving for } y(x), \text{ we get: } y(x) = \frac{\frac{1}{2}e^{x^2} + C}{e^{x^2}} = \frac{1}{2} + Ce^{-x^2} \end{align*} $$

Here is an example of how to use partial fractions to decompose the rational function \( \dfrac{2x-5}{(x-2)(x+1)} \):

$$\begin{aligned} \frac{2x-5}{(x-2)(x+1)} &= \frac{A}{x-2} + \frac{B}{x+1} \\ &= \frac{A(x+1) + B(x-2)}{(x-2)(x+1)} \end{aligned} $$

Multiplying both sides by the common denominator gives:

$$2x-5 = A(x+1) + B(x-2)$$

Substituting \(x = 2\) gives:

$$-1 = 3A$$ $$A = -\dfrac{1}{3}$$

Substituting \(x = -1\) gives:

$$-7 = -3B$$ $$B = \dfrac{7}{3}$$

Therefore, we have:

$$\frac{2x-5}{(x-2)(x+1)} = -\frac{1}{3(x-2)} + \frac{7}{3(x+1)}$$
Formula Booklet:

Standard derivatives

\( \frac{d}{dx} \tan(x) = \sec^2(x) \)

\( \frac{d}{dx} \sec(x) = \sec(x) \tan(x) \)

\( \frac{d}{dx} \cosec(x) = -\cosec(x) \cot(x) \)

\( \frac{d}{dx} \cot(x) = -\cosec^2(x) \)

\( \frac{d}{dx} a^x = a^x \ln(a) \)

\( \frac{d}{dx} \log_a(x) = \frac{1}{x \ln(a)} \)

\( \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}} \)

\( \frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1 - x^2}} \)

\( \frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2} \)

Standard Integrals

\( \int a^x dx = \dfrac{1}{\ln{a}}a^x + C\)

\( \int \dfrac{1}{a^2+x^2} dx = \dfrac{1}{a} \arctan{\left(\frac{x}{a}\right)}+C \)

\( \int \dfrac{1}{\sqrt{a^2-x^2}} dx = \arcsin{\left(\frac{x}{a}\right)}+C, \quad |x|\lt a \)

Formula Booklet:

Integral of \(x^n\)

\( \int x^n dx = \dfrac{x^{n+1}}{n+1} + C, n \neq -1\)

Area between a curve \(y=f(x)\) and the x-axis, where \( f(x) \gt 0 \)

\( A = \int^b_a y dx\)

Formula Booklet:

Standard integrals

\( \int \frac{1}{x} \; dx = \ln{|x|} + C\)

\( \int \sin x \; dx = - \cos x + C\)

\( \int \cos x \; dx = \sin x + C\)

\( \int e^x \; dx = e^x + C\)

Integration by inspection, also known as the reverse chain rule, is a technique used to simplify the process of integration for certain functions. When faced with an integrand of the form \( \int k g'(x)f(g(x)) \; dx \), one can recognise that the derivative of some function is present, allowing for a more straightforward integration.

To apply this method, we look for a function whose derivative matches a part of the integrand. If \( u = g(x) \), then \( du = g'(x) \; dx \). By substituting these values into the integral, the expression often simplifies, making it easier to evaluate.


Consider the integral:

$$ \int x \sin(x^2) \; dx $$

Here, we can let \( u = x^2 \). This gives \( \frac{du}{dx} = 2x \) or \( du = 2x \; dx \). Making the substitution, we get:

$$ \int \frac{1}{2} \sin(u) \; du $$

This integral is now straightforward to evaluate:

$$ -\frac{1}{2} \cos(u) + C $$

Re-substituting for \( u \), we obtain the final result:

$$ -\frac{1}{2} \cos(x^2) + C $$

Integration by substitution is a technique used to evaluate integrals by changing the variable of integration to simplify the integral. This method is particularly useful when dealing with integrals of composite functions. The basic idea is to substitute a part of the integral with a new variable, which makes the integral easier to solve. The substitution is often chosen so that the derivative of the new variable is present in the integral. Once the integral is evaluated with the new variable, we substitute back to the original variable.

Example 1:
Consider the integral $$ \int x \cos(x^2 + 1) \, dx $$ Let \( u = x^2 + 1 \). Then \( du = 2x \, dx \) and \( \frac{1}{2} du = x \, dx \). The integral becomes $$ \frac{1}{2} \int \cos(u) \, du = \frac{1}{2} \sin(u) + C = \frac{1}{2} \sin(x^2 + 1) + C $$

Example 2:
Evaluate $$ \int \frac{4x}{1 + x^2} \, dx $$ Let \( u = 1 + x^2 \), then \( du = 2x \, dx \) or \( \frac{1}{2} du = x \, dx \). The integral becomes $$ \frac{4}{2} \int \frac{1}{u} \, du = 2 \ln|u| + C = 2 \ln|1 + x^2| + C $$

Example 3:
Consider $$ \int e^{3x} \, dx $$ Let \( u = 3x \), hence \( du = 3 \, dx \) or \( \frac{1}{3} du = dx \). The integral becomes $$ \frac{1}{3} \int e^u \, du = \frac{1}{3} e^u + C = \frac{1}{3} e^{3x} + C $$

Algebraic division, often referred to as polynomial long division, is a method for dividing a polynomial by another polynomial of the same or lower degree. This technique is similar to long division used in arithmetic. The primary objective is to determine how many times the divisor fits into the dividend and to find the remainder if there is one. This method is particularly useful in simplifying rational expressions and solving polynomial equations.

The key formula for polynomial long division is as follows:

$$ \text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder} $$

Let's consider an example:

Algebraic long division example



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