Calculus

Furthermore

Official Guidance, clarification and syllabus links:

On examination papers, substitutions will be provided if the integral is not of the form

$$\int kg'(x)f(g(x)) dx.$$

Integration by parts examples:

$$\int x \sin{x} dx, \quad \int \ln{x} dx \quad \int \arcsin{x} dx.$$

Repeated integration by parts:

$$\int x^2e^x dx \text{ and } \int e^x \sin{x}dx.$$

Formula Booklet:

Integration by parts

\(\int u \frac{dv}{dx}dx = uv - \int v \frac{du}{dx}dx\) or \( \int u dv = uv - \int v du \)

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 $$ \frac{1}{3} \int e^u \, du = \frac{1}{3} e^u + C = \frac{1}{3} e^{3x} + C $$

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:

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

or

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

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.

$$\text{LIATE Rule:}$$

The LIATE rule is a mnemonic device used to determine the choice of u and dv when applying integration by parts. LIATE stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. When faced with an integral that involves a product of functions, LIATE helps prioritise which function should be chosen as u and which should be chosen as dv.

$$\text{Derivation of Integration by Parts:}$$

Integration by parts can be derived from the product rule for differentiation. Consider the product of two functions, \(u(x)\) and \(v(x)\). The product rule states that the derivative of their product is given by:

$$\frac{d}{dx}(uv) = u'v + uv'$$

Rearranging the equation:

$$uv' = \frac{d}{dx}(uv) - u'v$$

Integrate both sides with respect to \(x\):

$$\int uv' \, dx = \int \frac{d}{dx}(uv) \, dx - \int u'v \, dx$$

Applying the fundamental theorem of calculus, the right-hand side can be simplified:

$$\int uv' \, dx = uv - \int u'v \, dx$$

This is the formula for integration by parts.