Integration Techniques

This page reviews some common techniques for evaluating integrals that are useful in physics.

Basic Integrals

Before discussing integration techniques, it's useful to recall some basic integrals:

Basic Integral Formulas

\begin{aligned} \int x^n \, dx &= \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) \\ \int \frac{1}{x} \, dx &= \ln|x| + C \\ \int e^x \, dx &= e^x + C \\ \int \sin(x) \, dx &= -\cos(x) + C \\ \int \cos(x) \, dx &= \sin(x) + C \\ \int \sec^2(x) \, dx &= \tan(x) + C \end{aligned}

Substitution (u-Substitution)

Substitution is the integration counterpart to the chain rule in differentiation. It allows us to simplify integrals by making a change of variables. The idea is to identify a part of the integrand whose derivative also appears in the integrand.

Substitution Method

If we can write an integral in the form $\int f(g(x))g'(x) \, dx$, then we can make the substitution $u = g(x)$ to obtain:

\int f(g(x))g'(x) \, dx = \int f(u) \, du

After integrating with respect to $u$, we substitute back $u = g(x)$.

Example: Evaluate $\int 2x(x^2 + 1)^3 \, dx$

Notice that the derivative of $x^2 + 1$ is $2x$, which appears in the integrand. Let $u = x^2 + 1$, so that $du = 2x \, dx$:

\begin{aligned} \int 2x(x^2 + 1)^3 \, dx &= \int u^3 \, du \\ &= \frac{u^4}{4} + C \\ &= \frac{(x^2 + 1)^4}{4} + C \end{aligned}

Example: Evaluate $\int x e^{x^2} \, dx$

Notice that the derivative of $x^2$ is $2x$. Let $u = x^2$, so that $du = 2x \, dx$, or $x \, dx = \frac{du}{2}$:

\begin{aligned} \int x e^{x^2} \, dx &= \int e^u \cdot \frac{du}{2} \\ &= \frac{1}{2} \int e^u \, du \\ &= \frac{1}{2} e^u + C \\ &= \frac{1}{2} e^{x^2} + C \end{aligned}

Example: Evaluate $\int \frac{x}{\sqrt{1 - x^2}} \, dx$

Let $u = 1 - x^2$, so that $du = -2x \, dx$, or $x \, dx = -\frac{du}{2}$:

\begin{aligned} \int \frac{x}{\sqrt{1 - x^2}} \, dx &= \int \frac{1}{\sqrt{u}} \cdot \left(-\frac{du}{2}\right) \\ &= -\frac{1}{2} \int u^{-1/2} \, du \\ &= -\frac{1}{2} \cdot 2u^{1/2} + C \\ &= -\sqrt{u} + C \\ &= -\sqrt{1 - x^2} + C \end{aligned}

Integration by Parts

Integration by parts is the integration counterpart to the product rule in differentiation. It's particularly useful when the integrand is a product of two functions where one function becomes simpler when differentiated and the other becomes manageable when integrated.

Integration by Parts

\int u \, dv = uv - \int v \, du

where $u$ and $v$ are functions of the integration variable. The choice of $u$ and $dv$ is crucial for success. A common mnemonic is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to help choose $u$: the function that is earlier in this list should be chosen for $u$ while $v$ should be the function that is later.

Example: Evaluate $\int x e^x \, dx$

Let $u = x$ and $dv = e^x \, dx$. Then $du = dx$ and $v = e^x$:

\begin{aligned} \int x e^x \, dx &= xe^x - \int e^x \, dx \\ &= xe^x - e^x + C \\ &= e^x(x - 1) + C \end{aligned}

Example: Evaluate $\int x \sin(x) \, dx$

Let $u = x$ and $dv = \sin(x) \, dx$. Then $du = dx$ and $v = -\cos(x)$:

\begin{aligned} \int x \sin(x) \, dx &= x(-\cos(x)) - \int (-\cos(x)) \, dx \\ &= -x\cos(x) + \int \cos(x) \, dx \\ &= -x\cos(x) + \sin(x) + C \end{aligned}

Example: Evaluate $\int \ln(x) \, dx$

This requires a clever choice: let $u = \ln(x)$ and $dv = dx$. Then $du = \frac{1}{x} \, dx$ and $v = x$:

\begin{aligned} \int \ln(x) \, dx &= x\ln(x) - \int x \cdot \frac{1}{x} \, dx \\ &= x\ln(x) - \int 1 \, dx \\ &= x\ln(x) - x + C \\ &= x(\ln(x) - 1) + C \end{aligned}

Trigonometric Substitution

Trigonometric substitution is useful for integrals involving expressions like $\sqrt{a^2 - x^2}$, $\sqrt{x^2 - a^2}$, or $\sqrt{x^2 + a^2}$. The key is to use trigonometric identities to simplify the square root.

Common Trigonometric Substitutions

For $\sqrt{a^2 - x^2}$: use $x = a\sin(\theta)$
For $\sqrt{a^2 + x^2}$: use $x = a\tan(\theta)$
For $\sqrt{x^2 - a^2}$: use $x = a\sec(\theta)$

Example: Evaluate $\int \frac{1}{\sqrt{1 - x^2}} \, dx$

Let $x = \sin(\theta)$, so that $dx = \cos(\theta) \, d\theta$. Then $\sqrt{1 - x^2} = \sqrt{1 - \sin^2(\theta)} = \cos(\theta)$:

\begin{aligned} \int \frac{1}{\sqrt{1 - x^2}} \, dx &= \int \frac{1}{\cos(\theta)} \cdot \cos(\theta) \, d\theta \\ &= \int 1 \, d\theta \\ &= \theta + C \\ &= \arcsin(x) + C \end{aligned}

Example: Evaluate $\int \frac{1}{x^2 + 1} \, dx$

Let $x = \tan(\theta)$, so that $dx = \sec^2(\theta) \, d\theta$:

\begin{aligned} \int \frac{1}{x^2 + 1} \, dx &= \int \frac{1}{\tan^2(\theta) + 1} \cdot \sec^2(\theta) \, d\theta \\ &= \int \frac{1}{\sec^2(\theta)} \cdot \sec^2(\theta) \, d\theta \\ &= \int 1 \, d\theta \\ &= \theta + C \\ &= \arctan(x) + C \end{aligned}

Partial Fractions

Partial fractions is a technique for integrating rational functions (ratios of polynomials). The idea is to decompose a complex rational function into simpler fractions that can be integrated individually.

Partial Fraction Decomposition

For a rational function $\frac{P(x)}{Q(x)}$ where the degree of $P(x)$ is less than the degree of $Q(x)$, we decompose it into a sum of simpler fractions. The form depends on the factorization of $Q(x)$:

For a linear factor $(ax + b)$: $\frac{A}{ax + b}$
For a repeated linear factor $(ax + b)^n$: $\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$
For an irreducible quadratic factor $(ax^2 + bx + c)$: $\frac{Ax + B}{ax^2 + bx + c}$

Example: Evaluate $\int \frac{1}{x^2 - 1} \, dx$

Factor the denominator: $x^2 - 1 = (x - 1)(x + 1)$. Decompose:

\frac{1}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}

Multiply both sides by $(x - 1)(x + 1)$:

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

Solving: $A = \frac{1}{2}$, $B = -\frac{1}{2}$. Therefore:

\begin{aligned} \int \frac{1}{x^2 - 1} \, dx &= \int \left(\frac{1/2}{x - 1} - \frac{1/2}{x + 1}\right) \, dx \\ &= \frac{1}{2}\ln|x - 1| - \frac{1}{2}\ln|x + 1| + C \\ &= \frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right| + C \end{aligned}

Combining Techniques

Many integrals require the application of multiple techniques. The key is to recognize which technique to apply first and work systematically through the problem.

Example: Evaluate $\int x^3 e^{x^2} \, dx$

First, use substitution: let $u = x^2$, so that $du = 2x \, dx$, or $x \, dx = \frac{du}{2}$. Notice that $x^3 = x^2 \cdot x$, so $x^3 \, dx = x^2 \cdot x \, dx = u \cdot \frac{du}{2}$:

\begin{aligned} \int x^3 e^{x^2} \, dx &= \int u e^u \cdot \frac{du}{2} \\ &= \frac{1}{2} \int u e^u \, du \end{aligned}

Now use integration by parts with $v = u$ and $dw = e^u \, du$, so $dv = du$ and $w = e^u$:

\begin{aligned} \frac{1}{2} \int u e^u \, du &= \frac{1}{2}\left(ue^u - \int e^u \, du\right) \\ &= \frac{1}{2}(ue^u - e^u) + C \\ &= \frac{1}{2}e^u(u - 1) + C \\ &= \frac{1}{2}e^{x^2}(x^2 - 1) + C \end{aligned}

Example: Evaluate $\int e^x \sin(x) \, dx$

Use integration by parts twice. Let $u = e^x$ and $dv = \sin(x) \, dx$, so $du = e^x \, dx$ and $v = -\cos(x)$:

\begin{aligned} \int e^x \sin(x) \, dx &= -e^x\cos(x) + \int e^x \cos(x) \, dx \end{aligned}

Apply integration by parts again to $\int e^x \cos(x) \, dx$ with $u = e^x$ and $dv = \cos(x) \, dx$, so $du = e^x \, dx$ and $v = \sin(x)$:

\begin{aligned} \int e^x \cos(x) \, dx &= e^x\sin(x) - \int e^x \sin(x) \, dx \end{aligned}

Substituting back:

\begin{aligned} \int e^x \sin(x) \, dx &= -e^x\cos(x) + e^x\sin(x) - \int e^x \sin(x) \, dx \\ 2\int e^x \sin(x) \, dx &= e^x(\sin(x) - \cos(x)) \\ \int e^x \sin(x) \, dx &= \frac{e^x(\sin(x) - \cos(x))}{2} + C \end{aligned}

Definite Integrals

When evaluating definite integrals using substitution or other techniques, we can either:

Evaluate the indefinite integral and then apply the limits, or
Change the limits of integration when making the substitution and evaluate the new integral directly.

Example: Evaluate $\int_0^2 x\sqrt{x^2 + 1} \, dx$

Use substitution: let $u = x^2 + 1$, so that $du = 2x \, dx$, or $x \, dx = \frac{du}{2}$. When $x = 0$, $u = 1$; when $x = 2$, $u = 5$:

\begin{aligned} \int_0^2 x\sqrt{x^2 + 1} \, dx &= \int_1^5 \sqrt{u} \cdot \frac{du}{2} \\ &= \frac{1}{2} \int_1^5 u^{1/2} \, du \\ &= \frac{1}{2} \cdot \frac{2}{3} u^{3/2}\bigg|_1^5 \\ &= \frac{1}{3}(5^{3/2} - 1^{3/2}) \\ &= \frac{1}{3}(5\sqrt{5} - 1) \end{aligned}

Summary

Substitution: Use when you can identify $u = g(x)$ such that $g'(x)$ appears in the integrand: $\int f(g(x))g'(x) \, dx = \int f(u) \, du$
Integration by Parts: $\int u \, dv = uv - \int v \, du$; choose $u$ using LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential)
Trigonometric Substitution: Use for integrals involving $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$
Partial Fractions: Decompose rational functions into simpler fractions that can be integrated individually
Many integrals require combining multiple techniques—start with the technique that simplifies the integral most
For definite integrals, either evaluate the indefinite integral first or change the limits with the substitution