University of Toronto Scarborough
MATA36
Calculus II for Physical Sciences
Text Book: Calculus Early Transcendentals 9th Edition by James
Stewart, Daniel Clegg, Saleem Watson
In calculus, there are two widely used methods of integrating, integration by substitution and
integration by parts.
Integration by Substitution
It is the reverse or antiderivative version of the chain rule for calculating derivatives.
Here, we seek to substitute a function inside a function, within the function we are provided to
integrate, with the variable u, then we find du, i.e., find the derivative of u in terms of x, then
rewrite the equation in terms of dx, and substitute the value of dx in the original function, and
proceed onwards.
The method is represented as follows,
∫ 𝑓(𝑔(𝑥)𝑔′ (𝑥)𝑑𝑥
𝑢 = 𝑔(𝑥)
𝑑𝑢
= 𝑔′ (𝑥)
𝑑𝑥
𝑑𝑢 = 𝑔′ (𝑥)𝑑𝑥
𝑑𝑢
∫ 𝑓(𝑢)𝑔′ (𝑥)
𝑔′(𝑥)
∫ 𝑓(𝑢)𝑑𝑢
Example:
∫(1 + 𝑥 3 )10 3𝑥 2 𝑑𝑥
Let’s substitute 𝑢 = (1 + 𝑥 3 ). Then, let’s find the derivative of the substitution.
𝑑𝑢 = 3𝑥 2 𝑑𝑥. Let’s rearrange the equation with dx on the left-hand side.
MATA36
Calculus II for Physical Sciences
Text Book: Calculus Early Transcendentals 9th Edition by James
Stewart, Daniel Clegg, Saleem Watson
In calculus, there are two widely used methods of integrating, integration by substitution and
integration by parts.
Integration by Substitution
It is the reverse or antiderivative version of the chain rule for calculating derivatives.
Here, we seek to substitute a function inside a function, within the function we are provided to
integrate, with the variable u, then we find du, i.e., find the derivative of u in terms of x, then
rewrite the equation in terms of dx, and substitute the value of dx in the original function, and
proceed onwards.
The method is represented as follows,
∫ 𝑓(𝑔(𝑥)𝑔′ (𝑥)𝑑𝑥
𝑢 = 𝑔(𝑥)
𝑑𝑢
= 𝑔′ (𝑥)
𝑑𝑥
𝑑𝑢 = 𝑔′ (𝑥)𝑑𝑥
𝑑𝑢
∫ 𝑓(𝑢)𝑔′ (𝑥)
𝑔′(𝑥)
∫ 𝑓(𝑢)𝑑𝑢
Example:
∫(1 + 𝑥 3 )10 3𝑥 2 𝑑𝑥
Let’s substitute 𝑢 = (1 + 𝑥 3 ). Then, let’s find the derivative of the substitution.
𝑑𝑢 = 3𝑥 2 𝑑𝑥. Let’s rearrange the equation with dx on the left-hand side.