Lecture #3: Integration by Substitution & Integration by Parts
MATH 146 – Calculus II
December 1, 2025
The Fundamental Theorem of Calculus
Recall the Fundamental Theorem of Calculus from last lecture: Suppose \(f(t)\) is a continuous function on the interval \([a, b]\) and that \(F(t)\) is an antiderivative of \(f(t)\). Then
\[ \int_a^b f(t) \, dt = F(b) - F(a) \]
Rules for Integrals
The FTC says that you can evaluate \(\displaystyle\int_a^b f(x) \, dx\) if you can find the antiderivative of \(f(x)\).
But how do we find \(\displaystyle\int f(x) \, dx\)?
There is no “Product Rule” for integrals, so while it’s easy to find \(\dfrac{d}{dx} f(x) g(x)\), it may not be as easy to find \(\displaystyle\int f(x) g(x)\).
The CDF of the Standard Normal Distribution
Remember that not all functions have clear antiderivatives, or antiderivatives which can be expressed in terms of elementary functions.
For example, the cumulative density function of the standard Normal distribution cannot be expressed easily and can generally only be found using numerical approximation methods.
\[ \Phi(z) = \mathbb{P}[Z \le z] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^z e^{\frac{t^2}{2}} \, dt \]
Rules for Integrals
We can, however, use the Chain Rule to give us a trick to help solve stubborn integrals.
Recall the Chain Rule:
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \, g'(x) \]
Then we can have:
\[ \int f'(g(x)) \, g'(x) \, dx = \int \frac{d}{dx} f(g(x)) \, dx = f(g(x)) \]
Integration by Substitution
\[ \int f'(g(x)) \, g'(x) \, dx = \int f(u) \, du, \]
where
\[ u = g(x). \]
Example: Integration by Substitution
Examples:
Find \(\displaystyle\int 2xe^{x^2} \, dx\).
Find \(\displaystyle\int (x+1) \sqrt{3x+2} \, dx\).
Find \(\displaystyle\int \dfrac{1}{4x+3} \, dx\).
Integration by Substitution for Definite Integrals
\[ \int_a^b f'(g(x)) \, g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du, \]
Examples:
Find \(\displaystyle\int_0^{\frac{\pi}{2}} \sin^2(x) \cos(x) \, dx\).
Find \(\displaystyle\int_0^1 e^{-x}(2e^{-x} + 3)^9 \, dx\).
Integration by Parts
There is something similar to the Product Rule we can use, by reserse engineering the Product Rule. Recall the Chain Rule:
\[ \frac{d}{dx} f(x) g(x) = f(x)g'(x) + f'(x)g(x) \]
Integration by Parts
Then we can have:
\[ \begin{aligned} f(x) g(x) &= \int \left( f(x)g(x) \right)' \, dx \\ f(x) g(x) &= \int f(x)g'(x) + f'(x)g(x) \, dx \\ f(x) g(x) &= \int f(x)g'(x) \, dx + \int f'(x)g(x) \, dx \\ \int f(x)g'(x) \, dx &= f(x) g(x) - \int f'(x)g(x) \, dx \\ \int f(x) \left( \frac{dg}{\cancel{dx}} \right) \, \cancel{dx} &= f(x) g(x) - \int g(x) \left( \frac{df}{\cancel{dx}} \right) \, \cancel{dx} \\ \int f(x) \, dg &= f(x) g(x) - \int g(x) \, df \end{aligned} \]
Integration by Parts
\[ \int u \, dv = uv - \int v \, du \]
Examples:
\(\displaystyle\int x \sin(x) \, dx\).
\(\displaystyle\int x \ln\left(x\right) \, dx\).
Example: Integration by Parts
Example: \(\displaystyle\int \ln\left(x\right) \, dx\).
Example: Show that \[ \int \left( \ln\left(x\right) \right)^n \, dx = x \Big[ \ln\left(x\right) \Big]^n - n \int \left( \ln\left(x\right) \right)^{n-1} \, dx. \]
