11 Integration Techniques
11.1 The Problem of Nonelementary Integrands
The Fundamental Theorem of Calculus reduces definite integration to antidifferentiation: if F' = f on [a,b], then \int_a^b f(x)\,dx = F(b) - F(a).
For elementary functions—polynomials, exponentials, basic trigonometric functions—reversing differentiation rules yields antiderivatives immediately. The derivative of x^{n+1}/(n+1) is x^n; the derivative of e^x is e^x; the derivative of -\cos x is \sin x. But compositions and products resist direct reversal.
Consider
\int xe^{x^2}\,dx \quad\text{or}\quad \int x\ln x\,dx.
These integrands are built from elementary functions via composition and multiplication, yet no obvious antiderivative presents itself. The chain rule and product rule, which generate such expressions through differentiation, suggest the structure we need: integration by substitution inverts the chain rule, and integration by parts inverts the product rule.
This chapter develops these two fundamental techniques systematically, proves their validity, and examines their range of applicability. We then address rational functions via partial fraction decomposition and algebraic expressions involving radicals via trigonometric substitution. Throughout, we emphasize structural understanding over mechanical application.
11.2 Change of Variables
11.2.1 The Chain Rule Inverted
The chain rule states that if F and g are differentiable, then \frac{d}{dx}F(g(x)) = F'(g(x)) \cdot g'(x).
Integrating both sides over [a,b] yields F(g(b)) - F(g(a)) = \int_a^b F'(g(x)) g'(x)\,dx.
If we set f = F', this becomes \int_a^b f(g(x)) g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du, where the substitution u = g(x) transforms the integral. This is the substitution theorem.
Theorem 11.1 (Change of Variables Theorem) Let g : [a,b] \to \mathbb{R} be continuously differentiable, and let f be continuous on the range of g. Then \int_a^b f(g(x)) g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.
Since f is continuous on a closed interval (the range of g, which is compact by continuity of g on the compact set [a,b]), it admits an antiderivative F by the Fundamental Theorem. Define \Phi(x) = F(g(x)).
By the chain rule, \Phi'(x) = F'(g(x)) \cdot g'(x) = f(g(x)) g'(x), so \Phi is an antiderivative of f(g(x))g'(x). Applying the Fundamental Theorem to \Phi: \int_a^b f(g(x))g'(x)\,dx = \Phi(b) - \Phi(a) = F(g(b)) - F(g(a)).
Since F is an antiderivative of f, F(g(b)) - F(g(a)) = \int_{g(a)}^{g(b)} f(u)\,du. \quad \square
Remark. The theorem requires g' to be continuous, not merely that g be differentiable. This ensures g' is Riemann integrable, which is essential for the integral \int_a^b f(g(x))g'(x)\,dx to exist as a Riemann integral. If g' had too many discontinuities, the composition f(g(x))g'(x) might fail to be integrable even when f and g are individually well-behaved.
11.2.2 The Indefinite Form
For indefinite integrals, the substitution rule takes the form \int f(g(x)) g'(x)\,dx = \int f(u)\,du \bigg|_{u=g(x)} = F(g(x)) + C,
where F is any antiderivative of f. The notation “\int f(u)\,du\big|_{u=g(x)}” means: find an antiderivative of f with respect to u, then substitute u = g(x).
Setting u = g(x) and du = g'(x)dx encodes the substitution compactly. The differential du is not rigorously defined as an independent quantity here—it’s shorthand for the relationship \frac{du}{dx} = g'(x). But the notation is useful: it makes the substitution visually clear and connects to the multivariable framework where differentials have geometric meaning.
11.2.3 Geometric Interpretation
Substitution is a change of coordinates on the domain of integration. The map x \mapsto u = g(x) reparametrizes [a,b] as [g(a), g(b)] (or [g(b), g(a)] if g is decreasing). The factor g'(x) appearing in the integrand is the Jacobian in one dimension—it measures how the transformation scales infinitesimal intervals.
If g is increasing, g' > 0 and the transformation preserves orientation. If g is decreasing, g' < 0 and orientation reverses, reflected in the fact that \int_a^b = -\int_b^a when limits are exchanged. The absolute value |g'(x)| measures pure scaling without regard to direction.
As we will see, this generalizes to the change of variables formula for multiple integrals, where the Jacobian determinant |\det Dg| plays the role of |g'(x)|. The one-dimensional case studied here is the prototype.
11.2.4 Examples
Example 11.1 (Substitution with Exponential) The integral \int xe^{x^2}\,dx fits the pattern \int f(g(x))g'(x)\,dx with f(u) = \frac{1}{2}e^u and g(x) = x^2. Since g'(x) = 2x, we have x\,dx = \frac{1}{2}g'(x)\,dx = \frac{1}{2}\,du. Thus \int xe^{x^2}\,dx = \frac{1}{2}\int e^u\,du = \frac{1}{2}e^u + C = \frac{1}{2}e^{x^2} + C.
Example 11.2 (Substitution with Logarithm) For \int \frac{\cos(\ln x)}{x}\,dx, set u = \ln x, du = \frac{1}{x}\,dx. Then \int \frac{\cos(\ln x)}{x}\,dx = \int \cos u\,du = \sin u + C = \sin(\ln x) + C.
Example 11.3 (Substitution with Tangent) For \int_0^{\pi/4} \tan x\,dx, write \tan x = \frac{\sin x}{\cos x} and substitute u = \cos x, du = -\sin x\,dx. The limits transform: x = 0 \Rightarrow u = 1, x = \pi/4 \Rightarrow u = 1/\sqrt{2}. Thus \int_0^{\pi/4} \tan x\,dx = \int_1^{1/\sqrt{2}} \frac{-du}{u} = -[\ln|u|]_1^{1/\sqrt{2}} = -\ln(2^{-1/2}) = \frac{1}{2}\ln 2.
11.3 Integration by Parts
11.3.1 The Product Rule Inverted
The product rule for differentiation states \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x).
Integrating both sides over [a,b]: f(b)g(b) - f(a)g(a) = \int_a^b f'(x)g(x)\,dx + \int_a^b f(x)g'(x)\,dx.
Rearranging yields the integration by parts formula.
Theorem 11.2 (Integration by Parts) Let f, g : [a,b] \to \mathbb{R} be continuously differentiable. Then \int_a^b f(x)g'(x)\,dx = [f(x)g(x)]_a^b - \int_a^b f'(x)g(x)\,dx.
The product rule gives (fg)' = f'g + fg'. Since f and g are continuously differentiable, fg is also continuously differentiable, and the Fundamental Theorem applies: \int_a^b (fg)'(x)\,dx = f(b)g(b) - f(a)g(a).
But (fg)' = f'g + fg', so \int_a^b f'(x)g(x)\,dx + \int_a^b f(x)g'(x)\,dx = f(b)g(b) - f(a)g(a).
Solving for \int_a^b f(x)g'(x)\,dx gives the stated formula. \square
Indefinite form. For antiderivatives, the formula becomes \int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx.
Strategy. The formula exchanges \int fg' for fg - \int f'g. For this to simplify the problem, we require:
g' integrates easily (yielding g)
f simplifies upon differentiation (yielding a simpler f')
Functions that simplify upon differentiation include logarithms, inverse trigonometric functions, and polynomials. Functions that integrate easily include exponentials, trigonometric functions, and powers.
11.3.2 Iterated Application
Integration by parts may be applied repeatedly when necessary. Each application reduces one factor (typically a polynomial degree or an inverse function) at the cost of potentially increasing another factor. The process terminates when one factor vanishes or when the integral becomes elementary.
Example 11.4 (Iterated Integration by Parts) For \int x^2 e^x\,dx, set f(x) = x^2, g'(x) = e^x. Then f'(x) = 2x, g(x) = e^x, giving \int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx = x^2 e^x - 2\int x e^x\,dx.
Apply parts again to \int xe^x\,dx with f(x) = x, g'(x) = e^x \int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x.
Substituting back \int x^2 e^x\,dx = x^2 e^x - 2(xe^x - e^x) + C = e^x(x^2 - 2x + 2) + C.
11.3.3 Reduction Formulas
Repeated application of integration by parts yields reduction formulas that express an integral involving a parameter n in terms of a similar integral with parameter n-1. Induction then produces a closed form.
Theorem 11.3 (Reduction Formula for \int x^n e^x\,dx) For n \in \mathbb{N}, \int x^n e^x\,dx = x^n e^x - n\int x^{n-1} e^x\,dx.
Set f(x) = x^n, g'(x) = e^x. Then f'(x) = nx^{n-1}, g(x) = e^x. Applying Theorem 11.2 \int x^n e^x\,dx = x^n e^x - \int nx^{n-1} e^x\,dx = x^n e^x - n\int x^{n-1}e^x\,dx. \quad \square
Corollary 11.1 Applying the reduction formula repeatedly \int x^n e^x\,dx = e^x \sum_{k=0}^{n} (-1)^{n-k} \frac{n!}{k!} x^k + C.
Base case (n=0). \int e^x\,dx = e^x + C, matching the formula.
Inductive step. Assume the formula holds for n-1. Then \int x^{n-1}e^x\,dx = e^x \sum_{k=0}^{n-1} (-1)^{n-1-k}\frac{(n-1)!}{k!}x^k + C.
By Theorem 11.3, \begin{align*} \int x^n e^x\,dx &= x^n e^x - n\int x^{n-1}e^x\,dx\\ &= x^n e^x - ne^x\sum_{k=0}^{n-1}(-1)^{n-1-k}\frac{(n-1)!}{k!}x^k\\ &= e^x\left(x^n + \sum_{k=0}^{n-1}(-1)^{n-k}n\frac{(n-1)!}{k!}x^k\right)\\ &= e^x\sum_{k=0}^{n}(-1)^{n-k}\frac{n!}{k!}x^k + C, \end{align*} where the last step uses n \cdot \frac{(n-1)!}{k!} = \frac{n!}{k!} for k \leq n-1 and recognizes the k=n term as x^n. \square
Remark. Similar reduction formulas exist for \int x^n \sin x\,dx, \int x^n \cos x\,dx, \int (\ln x)^n\,dx, and others. Each arises from systematic application of Theorem 11.2 with appropriate choices of f and g'.
11.3.4 The Logarithm and Inverse Functions
For inverse functions like \ln x and \arctan x, integration by parts with f equal to the inverse function and g' = 1 produces the antiderivative.
Example 11.5 (Integrating the Logarithm) For \int \ln x\,dx (x > 0), set f(x) = \ln x, g'(x) = 1. Then f'(x) = 1/x, g(x) = x \begin{align*} \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{align*}
Example 11.6 (Integrating Arctangent) For \int \arctan x\,dx, set f(x) = \arctan x, g'(x) = 1. Then f'(x) = \frac{1}{1+x^2}, g(x) = x \int \arctan x\,dx = x\arctan x - \int \frac{x}{1+x^2}\,dx.
For the remaining integral, substitute u = 1+x^2, du = 2x\,dx: \int \frac{x}{1+x^2}\,dx = \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln(1+x^2) + C.
Thus \int \arctan x\,dx = x\arctan x - \frac{1}{2}\ln(1+x^2) + C.
11.4 Trigonometric Substitution
11.4.1 Algebraic Expressions and the Pythagorean Identity
Integrals involving \sqrt{a^2 - x^2}, \sqrt{a^2 + x^2}, or \sqrt{x^2 - a^2} arise naturally in geometry (arc length, surface area) and physics (potential fields). These expressions resist both substitution and integration by parts because they are neither compositions of simple functions nor products with obvious factors.
We use the Pythagorean identity \sin^2\theta + \cos^2\theta = 1 and its variants (1 + \tan^2\theta = \sec^2\theta, \sec^2\theta - 1 = \tan^2\theta) provide algebraic relationships between trigonometric functions that mirror the structure of these radicals. Substituting x with an appropriate trigonometric function converts the radical into a trigonometric expression that simplifies via the identity.
11.4.2 The Three Cases
We present the three standard substitutions systematically.
Case 1: \sqrt{a^2 - x^2}
Set x = a\sin\theta with -\pi/2 \leq \theta \leq \pi/2. Then dx = a\cos\theta\,d\theta and \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\sqrt{1-\sin^2\theta} = a|\cos\theta| = a\cos\theta, since \cos\theta \geq 0 in the chosen range. The radical vanishes.
Case 2: \sqrt{a^2 + x^2}
Set x = a\tan\theta with -\pi/2 < \theta < \pi/2. Then dx = a\sec^2\theta\,d\theta and \sqrt{a^2 + x^2} = \sqrt{a^2 + a^2\tan^2\theta} = a\sqrt{1+\tan^2\theta} = a|\sec\theta| = a\sec\theta, since \sec\theta > 0 in the range.
Case 3: \sqrt{x^2 - a^2}
Set x = a\sec\theta with 0 \leq \theta < \pi/2 (for x > a) or \pi \leq \theta < 3\pi/2 (for x < -a). Then dx = a\sec\theta\tan\theta\,d\theta and \sqrt{x^2 - a^2} = \sqrt{a^2\sec^2\theta - a^2} = a\sqrt{\sec^2\theta - 1} = a|\tan\theta| = a\tan\theta, since \tan\theta \geq 0 in the chosen ranges.
11.4.3 Justification and Uniqueness
Why these particular substitutions? Each exploits a different form of the Pythagorean identity to eliminate the radical. The choice of domain for \theta ensures the transformation is bijective (or bijective on the relevant part of the domain), so that solving for \theta in terms of x is unambiguous.
The geometric interpretation: x = a\sin\theta parametrizes a semicircle of radius a; x = a\tan\theta parametrizes a branch of a hyperbola; x = a\sec\theta parametrizes another branch. The substitution converts a problem about algebraic curves to one about trigonometric functions, which are easier to integrate due to known antiderivatives.
Example 11.7 (Antiderivative of 1/\sqrt{1+x^2}) Consider \int \frac{1}{\sqrt{1+x^2}}\,dx. Set x = \tan\theta, dx = \sec^2\theta\,d\theta. Then \sqrt{1+x^2} = \sec\theta, giving \int \frac{1}{\sqrt{1+x^2}}\,dx = \int \frac{\sec^2\theta}{\sec\theta}\,d\theta = \int \sec\theta\,d\theta.
To integrate \sec\theta, multiply by \frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}: \int \sec\theta\,d\theta = \int \frac{\sec^2\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\,d\theta.
The numerator is the derivative of \sec\theta + \tan\theta, so \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C.
Substituting back (\sec\theta = \sqrt{1+x^2}, \tan\theta = x): \int \frac{1}{\sqrt{1+x^2}}\,dx = \ln|\sqrt{1+x^2} + x| + C.
This is the inverse hyperbolic sine, \sinh^{-1}(x) = \ln(x + \sqrt{x^2+1}).
11.5 Partial Fractions
11.5.1 Rational Functions and Polynomial Factorization
A rational function is a quotient P(x)/Q(x) where P, Q are polynomials. If \deg P < \deg Q, the function is proper; otherwise, polynomial long division reduces it to a polynomial plus a proper rational function. We focus on proper rational functions, as the polynomial part integrates trivially.
The Fundamental Theorem of Algebra states that every polynomial Q with real coefficients factors completely over \mathbb{C}. Grouping complex conjugate pairs yields a factorization over \mathbb{R}: Q(x) = c \prod_{i=1}^k (x - r_i)^{m_i} \prod_{j=1}^\ell (x^2 + p_j x + q_j)^{n_j}, where r_i \in \mathbb{R} are real roots with multiplicities m_i, and the quadratic factors (with p_j^2 - 4q_j < 0) correspond to complex conjugate pairs.
Theorem 11.4 (Partial Fraction Decomposition) Let P(x)/Q(x) be a proper rational function with Q factored as above. Then there exist unique real numbers A_{ij}, B_{jk}, C_{jk} such that \frac{P(x)}{Q(x)} = \sum_{i=1}^k \sum_{j=1}^{m_i} \frac{A_{ij}}{(x-r_i)^j} + \sum_{j=1}^\ell \sum_{k=1}^{n_j} \frac{B_{jk}x + C_{jk}}{(x^2 + p_jx + q_j)^k}.
The theorem’s proof uses linear algebra: multiplying both sides by Q(x) yields a polynomial identity, and equating coefficients produces a system of linear equations for the unknowns A_{ij}, B_{jk}, C_{jk}. Uniqueness follows from the fact that the system has a unique solution (the coefficient matrix is invertible by properties of Vandermonde-type matrices).
We omit the full proof, which is standard in abstract algebra. For practical computation, the method of undetermined coefficients suffices: write the decomposition with unknown constants, clear denominators, and solve by substitution or coefficient comparison.
11.5.2 Integration of Decomposed Terms
Once decomposed, each term integrates via known formulas:
Linear factors: \int \frac{A}{(x-r)^n}\,dx = \begin{cases} A\ln|x-r| + C & n = 1,\\ \frac{A(x-r)^{1-n}}{1-n} + C & n \geq 2. \end{cases}
Quadratic factors (degree 1): \int \frac{Bx + C}{x^2 + px + q}\,dx splits into two parts. Complete the square, x^2 + px + q = (x + p/2)^2 + (q - p^2/4). Set u = x + p/2, a^2 = q - p^2/4 > 0. Then \int \frac{Bx + C}{x^2 + px + q}\,dx = \int \frac{B(u - p/2) + C}{u^2 + a^2}\,du = \frac{B}{2}\ln(u^2 + a^2) + \frac{C - Bp/2}{a}\arctan(u/a) + C.
Quadratic factors (higher degree): For \int \frac{Bx + C}{(x^2+px+q)^n}\,dx with n \geq 2, similar completion of the square reduces to integrals of \frac{u}{(u^2+a^2)^n} (which substitute further with v = u^2 + a^2) and \frac{1}{(u^2+a^2)^n} (which require reduction formulas or trigonometric substitution).
The key point: every proper rational function integrates to a combination of logarithms, arctangents, and rational functions. This is a remarkable closure property—rational functions remain in the realm of elementary functions upon integration.
Example 11.8 (Partial Fractions with Linear and Quadratic Factors) Consider \int \frac{x^2}{(x-1)(x^2+1)}\,dx. Decompose: \frac{x^2}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}.
Multiply by (x-1)(x^2+1): x^2 = A(x^2+1) + (Bx+C)(x-1).
Setting x = 1: 1 = 2A, so A = 1/2.
Then
x^2 = Ax^2 + A + Bx^2 - Bx + Cx - C.
Equating coefficients:
x^2: 1 = A + B \Rightarrow B = 1/2
x^1: 0 = -B + C \Rightarrow C = 1/2
x^0: 0 = A - C (consistent)
Thus \frac{x^2}{(x-1)(x^2+1)} = \frac{1/2}{x-1} + \frac{(x+1)/2}{x^2+1}.
Integrate: \int \frac{x^2}{(x-1)(x^2+1)}\,dx = \frac{1}{2}\ln|x-1| + \frac{1}{2}\int \frac{x}{x^2+1}\,dx + \frac{1}{2}\int \frac{1}{x^2+1}\,dx.
The first remaining integral is \frac{1}{4}\ln(x^2+1) (via u = x^2+1), the second is \frac{1}{2}\arctan x. Combining: \int \frac{x^2}{(x-1)(x^2+1)}\,dx = \frac{1}{2}\ln|x-1| + \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\arctan x + C.
11.6 Summary and Table of Integrals
We have developed four fundamental techniques:
Substitution (u = g(x)): Inverts the chain rule, transforming integrals via change of variables
Integration by parts: Inverts the product rule, trading \int fg' for fg - \int f'g
Trigonometric substitution: Exploits the Pythagorean identity to eliminate radicals
Partial fractions: Decomposes rational functions into integrable pieces
These techniques suffice for a large class of integrands.
11.6.1 Closing Remarks
Integration techniques are tools for rewriting problems into forms with known antiderivatives. The best strategy is conceptual: detect structure and choose the transformation that exposes a standard antiderivative. Practice with a variety of integrals deepens pattern recognition and reduces dependence on rote memorization.
Below is a compact reference of common antiderivatives. Use it as a checklist while you look for an underlying technique; whenever possible derive the entry from differentiation rules rather than memorize it in isolation.
11.6.2 Basic (Algebraic, Exponential, and Rational)
| Integrand f(x) | Antiderivative F(x) + C | Domain / Notes |
|---|---|---|
| x^n (n \neq -1) | \dfrac{x^{n+1}}{n+1} | Valid where x^n is defined (for non-integer n, typically x>0); n\neq-1 |
| \dfrac{1}{x} | \ln|x| | x\neq 0 |
| e^{ax} | \dfrac{1}{a}e^{ax} | All real x, a\neq 0 (for a=0 the integrand is constant) |
| a^x (a>0, a\neq1) | \dfrac{a^x}{\ln a} | a>0, a\neq1; all real x |
| \dfrac{1}{a^2 + x^2} | \dfrac{1}{a}\arctan\!\left(\dfrac{x}{a}\right) | All real x, a\neq 0 |
| \dfrac{1}{x^2 - a^2} | \dfrac{1}{2a}\ln\left|\dfrac{x - a}{x + a}\right| | x\neq \pm a |
| \dfrac{1}{\sqrt{a^2 - x^2}} | \arcsin\!\left(\dfrac{x}{a}\right) | |x| < a (principal branch) |
| \dfrac{1}{\sqrt{x^2 + a^2}} | \ln\left|x + \sqrt{x^2 + a^2}\right| | All real x, a\neq 0 |
11.6.3 Trigonometric & Hyperbolic
| Integrand f(x) | Antiderivative F(x) + C | Domain / Notes |
|---|---|---|
| \sin(ax) | -\dfrac{1}{a}\cos(ax) | All real x, a\neq 0 |
| \cos(ax) | \dfrac{1}{a}\sin(ax) | All real x, a\neq 0 |
| \sec^2(ax) | \dfrac{1}{a}\tan(ax) | Defined where \cos(ax)\neq 0 |
| \csc^2(ax) | -\dfrac{1}{a}\cot(ax) | Defined where \sin(ax)\neq 0 |
| \sec(ax)\tan(ax) | \dfrac{1}{a}\sec(ax) | Defined where \cos(ax)\neq 0 |
| \csc(ax)\cot(ax) | -\dfrac{1}{a}\csc(ax) | Defined where \sin(ax)\neq 0 |
| \tan(ax) | -\dfrac{1}{a}\ln|\cos(ax)| | Defined where \cos(ax)\neq 0; antiderivative uses principal branch of log |
| \cot(ax) | \dfrac{1}{a}\ln|\sin(ax)| | Defined where \sin(ax)\neq 0 |
| \sinh(ax) | \dfrac{1}{a}\cosh(ax) | All real x, a\neq 0 |
| \cosh(ax) | \dfrac{1}{a}\sinh(ax) | All real x, a\neq 0 |
11.6.4 Integration by Parts / Inverse-function Antiderivatives
| Integrand f(x) | Antiderivative F(x) + C | Domain / Notes |
|---|---|---|
| \ln x (x>0) | x\ln x - x | x>0; standard by parts with u=\ln x, dv=dx |
| \arctan x | x\arctan x - \dfrac{1}{2}\ln(1+x^2) | All real x; often derived via parts |
| \arcsin x | x\arcsin x + \sqrt{1-x^2} | |x|\le 1; can be derived via substitution or parts |
| \arccos x | x\arccos x - \sqrt{1-x^2} | |x|\le 1; standard ny parts with u=\arccos x, dv=dx |