Carry out simple cases of integration by substitution and integration by parts; understand these methods as the inverse processes of the chain and product rules respectively (integration by substitution includes finding a suitable substitution and is limited to cases where one substitution will lead to a function which can be integrated; integration by parts includes more than one application of the method but excludes reduction formulae)

Integration by substitution and integration by parts are two essential techniques for finding antiderivatives of functions that are not immediately integrable. These methods are the inverse processes of the chain rule and product rule for differentiation, respectively. In A-Level Mathematics (Edexcel), you will learn to apply these techniques to a variety of functions, including those involving trigonometric, exponential, and algebraic expressions. Mastery of these methods is crucial for solving more complex integrals and for success in calculus-based problems in further study.

Integration by substitution involves replacing a part of the integrand with a new variable to simplify the integral. The key is to choose a suitable substitution that transforms the integral into a standard form. For example, substituting u = g(x) allows you to rewrite the integral in terms of u, often leading to a simpler expression. This method is particularly useful for integrals of composite functions, such as ∫ f(g(x)) g'(x) dx. You are expected to handle cases where one substitution suffices, and the resulting integral can be evaluated directly.

Integration by parts is based on the product rule for differentiation and is used to integrate products of two functions. The formula ∫ u dv = uv - ∫ v du is applied, where you choose u and dv strategically. Typically, you select u as a function that simplifies when differentiated (e.g., polynomials, logarithms) and dv as a function that is easy to integrate (e.g., exponentials, trigonometric functions). In some cases, you may need to apply integration by parts more than once, but you are not required to handle reduction formulae. Understanding when and how to apply these methods is key to solving a wide range of integrals efficiently.