Techniques of Integration — Rules and Formulas
Based on Stewart, Calculus, Chapter 7: Techniques of Integration
Throughout, \(C\) is an arbitrary constant and \(a \neq 0\).
Table of Integration Formulas (Stewart, §7.5)
| # | Integral |
|---|---|
| 1 | \(\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\) |
| 2 | \(\displaystyle\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C\) |
| 3 | \(\displaystyle\int e^x\,dx = e^x + C\) |
| 4 | \(\displaystyle\int b^x\,dx = \frac{b^x}{\ln b} + C\) |
| 5 | \(\displaystyle\int \sin x\,dx = -\cos x + C\) |
| 6 | \(\displaystyle\int \cos x\,dx = \sin x + C\) |
| 7 | \(\displaystyle\int \sec^2 x\,dx = \tan x + C\) |
| 8 | \(\displaystyle\int \csc^2 x\,dx = -\cot x + C\) |
| 9 | \(\displaystyle\int \sec x\tan x\,dx = \sec x + C\) |
| 10 | \(\displaystyle\int \csc x\cot x\,dx = -\csc x + C\) |
| 11 | \(\displaystyle\int \sec x\,dx = \ln\lvert\sec x + \tan x\rvert + C\) |
| 12 | \(\displaystyle\int \csc x\,dx = \ln\lvert\csc x - \cot x\rvert + C\) |
| 13 | \(\displaystyle\int \tan x\,dx = \ln\lvert\sec x\rvert + C\) |
| 14 | \(\displaystyle\int \cot x\,dx = \ln\lvert\sin x\rvert + C\) |
| 15 | \(\displaystyle\int \sinh x\,dx = \cosh x + C\) |
| 16 | \(\displaystyle\int \cosh x\,dx = \sinh x + C\) |
| 17 | \(\displaystyle\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C\) |
| 18 | \(\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\!\left(\frac{x}{a}\right) + C,\ a>0\) |
| 19 | \(\displaystyle\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left\lvert\frac{x-a}{x+a}\right\rvert + C\) |
| 20 | \(\displaystyle\int \frac{dx}{\sqrt{x^2 \pm a^2}} = \ln\left\lvert x + \sqrt{x^2 \pm a^2}\right\rvert + C\) |
7.1 Integration by Parts
Derived from the Product Rule \(\dfrac{d}{dx}[f(x)g(x)] = f(x)g'(x) + g(x)f'(x)\).
Indefinite form \[\int f(x)\,g'(x)\,dx = f(x)\,g(x) - \int g(x)\,f'(x)\,dx\]
Shorthand (with \(u = f(x)\), \(v = g(x)\)): \[\boxed{\int u\,dv = uv - \int v\,du}\]
Definite form \[\int_a^b f(x)\,g'(x)\,dx = f(x)\,g(x)\Big|_a^b - \int_a^b g(x)\,f'(x)\,dx\]
Choosing \(u\) (LIATE heuristic): pick \(u\) as the first type in the list Logarithmic → Inverse trig → Algebraic → Trigonometric → Exponential, and let \(dv\) be the rest. Goal: \(\int v\,du\) should be simpler than the original.
Tips - Repeated parts: e.g. \(\int x^2 e^x\,dx\) — apply twice. - “Cyclic” integrals such as \(\int e^x \sin x\,dx\): apply parts twice, then solve algebraically for the original integral. - Reduction formulas, e.g. \[\int \sin^n x\,dx = -\frac{1}{n}\cos x\,\sin^{n-1}x + \frac{n-1}{n}\int \sin^{n-2}x\,dx\]
Example: \(\displaystyle\int x\sin x\,dx\), \(u=x\), \(dv=\sin x\,dx \Rightarrow du=dx\), \(v=-\cos x\): \[\int x\sin x\,dx = -x\cos x + \int\cos x\,dx = -x\cos x + \sin x + C\]
7.2 Trigonometric Integrals
Useful identities
\[\sin^2 x + \cos^2 x = 1,\qquad \tan^2 x + 1 = \sec^2 x\] \[\sin^2 x = \tfrac12(1 - \cos 2x),\qquad \cos^2 x = \tfrac12(1 + \cos 2x),\qquad \sin x\cos x = \tfrac12\sin 2x\]
Strategy for \(\displaystyle\int \sin^m x\,\cos^n x\,dx\)
| Case | Strategy |
|---|---|
| (a) power of cosine is odd (\(n = 2k+1\)) | Save one \(\cos x\), write \(\cos^{2k}x = (1-\sin^2 x)^k\), substitute \(u = \sin x\) |
| (b) power of sine is odd (\(m = 2k+1\)) | Save one \(\sin x\), write \(\sin^{2k}x = (1-\cos^2 x)^k\), substitute \(u = \cos x\) |
| (c) both even | Use half-angle identities \(\sin^2 x = \tfrac12(1-\cos 2x)\), \(\cos^2 x = \tfrac12(1+\cos 2x)\) |
(If both are odd, either (a) or (b) works.)
Strategy for \(\displaystyle\int \tan^m x\,\sec^n x\,dx\)
| Case | Strategy |
|---|---|
| (a) power of secant is even (\(n = 2k,\ k\ge2\)) | Save \(\sec^2 x\), write the rest via \(\sec^2 x = 1+\tan^2 x\), substitute \(u = \tan x\) |
| (b) power of tangent is odd (\(m = 2k+1\)) | Save \(\sec x\tan x\), write the rest via \(\tan^2 x = \sec^2 x - 1\), substitute \(u = \sec x\) |
| Other cases | Use identities, integration by parts, and \(\int\tan x\,dx\), \(\int\sec x\,dx\) |
Useful results: \[\int \tan x\,dx = \ln\lvert\sec x\rvert + C,\qquad \int \sec x\,dx = \ln\lvert\sec x + \tan x\rvert + C\] \[\int \sec^3 x\,dx = \tfrac12\left(\sec x\tan x + \ln\lvert\sec x + \tan x\rvert\right) + C\]
(Integrals of \(\cot^m x\,\csc^n x\) are handled the same way using \(1 + \cot^2 x = \csc^2 x\).)
Products of sines and cosines with different angles
For \(\int \sin mx\cos nx\,dx\), \(\int \sin mx\sin nx\,dx\), \(\int \cos mx\cos nx\,dx\) use: \[\sin A\cos B = \tfrac12\left[\sin(A-B) + \sin(A+B)\right]\] \[\sin A\sin B = \tfrac12\left[\cos(A-B) - \cos(A+B)\right]\] \[\cos A\cos B = \tfrac12\left[\cos(A-B) + \cos(A+B)\right]\]
7.3 Trigonometric Substitution
Used for integrands containing \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\) (with \(a > 0\)).
| Expression | Substitution | Identity used | Restriction on \(\theta\) |
|---|---|---|---|
| \(\sqrt{a^2 - x^2}\) | \(x = a\sin\theta\) | \(1 - \sin^2\theta = \cos^2\theta\) | \(-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}\) |
| \(\sqrt{a^2 + x^2}\) | \(x = a\tan\theta\) | \(1 + \tan^2\theta = \sec^2\theta\) | \(-\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2}\) |
| \(\sqrt{x^2 - a^2}\) | \(x = a\sec\theta\) | \(\sec^2\theta - 1 = \tan^2\theta\) | \(0 \le \theta < \tfrac{\pi}{2}\) or \(\pi \le \theta < \tfrac{3\pi}{2}\) |
Steps 1. Substitute \(x\) and \(dx\) (e.g. \(x = a\sin\theta \Rightarrow dx = a\cos\theta\,d\theta\)). 2. Simplify the radical with the identity, e.g. \(\sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta\). 3. Integrate in \(\theta\). 4. Return to \(x\) using a reference right triangle (e.g. \(\sin\theta = x/a\) gives opposite \(x\), hypotenuse \(a\), adjacent \(\sqrt{a^2-x^2}\)). 5. For definite integrals, change the limits to \(\theta\)-values instead.
Tip: for \(\sqrt{ax^2 + bx + c}\), first complete the square, then use a trig substitution.
7.4 Integration of Rational Functions by Partial Fractions
To integrate \(f(x) = \dfrac{P(x)}{Q(x)}\):
Step 0 — Improper fraction: if \(\deg P \ge \deg Q\), use long division: \[\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)},\qquad \deg R < \deg Q\]
Step 1 — Factor \(Q(x)\) into linear factors \((ax+b)\) and irreducible quadratic factors \((ax^2+bx+c)\) with \(b^2 - 4ac < 0\).
Step 2 — Decompose according to the cases:
| Case | Factor in \(Q(x)\) | Partial fraction terms |
|---|---|---|
| I | distinct linear factors \((a_1x+b_1)\cdots(a_kx+b_k)\) | \(\dfrac{A_1}{a_1x+b_1} + \cdots + \dfrac{A_k}{a_kx+b_k}\) |
| II | repeated linear factor \((ax+b)^r\) | \(\dfrac{A_1}{ax+b} + \dfrac{A_2}{(ax+b)^2} + \cdots + \dfrac{A_r}{(ax+b)^r}\) |
| III | distinct irreducible quadratic \(ax^2+bx+c\) | \(\dfrac{Ax+B}{ax^2+bx+c}\) |
| IV | repeated irreducible quadratic \((ax^2+bx+c)^r\) | \(\dfrac{A_1x+B_1}{ax^2+bx+c} + \cdots + \dfrac{A_rx+B_r}{(ax^2+bx+c)^r}\) |
Step 3 — Solve for the constants (multiply through by \(Q(x)\); equate coefficients or plug in convenient \(x\)-values).
Step 4 — Integrate each term. Key formulas: \[\int \frac{dx}{ax+b} = \frac{1}{a}\ln\lvert ax+b\rvert + C,\qquad \int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C\]
Rationalizing substitutions: if the integrand contains \(\sqrt[n]{g(x)}\), try \(u = \sqrt[n]{g(x)}\) to obtain a rational function.
7.5 Strategy for Integration
- Simplify the integrand if possible (algebra or trig identities). e.g. \(\int \sqrt{x}(1+\sqrt{x})\,dx = \int (\sqrt{x} + x)\,dx\), \(\ \int \frac{\tan\theta}{\sec^2\theta}\,d\theta = \int \sin\theta\cos\theta\,d\theta\).
- Look for an obvious substitution \(u = g(x)\) where \(du = g'(x)\,dx\) also appears (up to a constant).
- Classify the integrand by form:
- Trigonometric functions → §7.2 strategies
- Rational functions → partial fractions (§7.4)
- Products of different kinds of functions → integration by parts (§7.1)
- Radicals:
- \(\sqrt{\pm x^2 \pm a^2}\) → trig substitution (§7.3)
- \(\sqrt[n]{ax+b}\) → rationalizing substitution \(u = \sqrt[n]{ax+b}\)
- Try again:
- Try a different substitution.
- Try integration by parts again (e.g. single functions like \(\tan^{-1}x\), \(\ln x\)).
- Manipulate the integrand: rationalize, multiply by a clever form of 1, use identities.
- Relate to previous problems.
- Use several methods together.
Note: some continuous functions have no elementary antiderivative, e.g. \[\int e^{x^2}dx,\quad \int \frac{e^x}{x}dx,\quad \int \sin(x^2)\,dx,\quad \int \frac{1}{\ln x}dx,\quad \int \sqrt{x^3+1}\,dx,\quad \int \frac{\sin x}{x}dx\]
7.6 Integration Using Tables and Computer Algebra Systems
- Match the integrand to an entry in a table of integrals (Stewart’s Reference Pages), possibly after a substitution or completing the square.
- A CAS (e.g. SymPy, Mathematica) may give an answer in a different but equivalent form; it may omit absolute values or the constant \(C\).
7.7 Approximate Integration
Divide \([a, b]\) into \(n\) subintervals of width \(\Delta x = \dfrac{b-a}{n}\), with \(x_i = a + i\,\Delta x\) and midpoints \(\bar{x}_i = \tfrac12(x_{i-1} + x_i)\).
Left / Right endpoint rules \[L_n = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x,\qquad R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x\]
Midpoint Rule \[\int_a^b f(x)\,dx \approx M_n = \Delta x\left[f(\bar{x}_1) + f(\bar{x}_2) + \cdots + f(\bar{x}_n)\right]\]
Trapezoidal Rule \[\int_a^b f(x)\,dx \approx T_n = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right]\]
Simpson’s Rule (\(n\) even) \[\int_a^b f(x)\,dx \approx S_n = \frac{\Delta x}{3}\left[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)\right]\]
Also: \(S_{2n} = \tfrac13 T_n + \tfrac23 M_n\).
Error bounds If \(\lvert f''(x)\rvert \le K\) on \([a,b]\): \[\lvert E_T\rvert \le \frac{K(b-a)^3}{12n^2},\qquad \lvert E_M\rvert \le \frac{K(b-a)^3}{24n^2}\] If \(\lvert f^{(4)}(x)\rvert \le K\) on \([a,b]\): \[\lvert E_S\rvert \le \frac{K(b-a)^5}{180n^4}\]
7.8 Improper Integrals
Type 1: Infinite intervals
If \(\int_a^t f(x)\,dx\) exists for every \(t \ge a\): \[\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx\]
If \(\int_t^b f(x)\,dx\) exists for every \(t \le b\): \[\int_{-\infty}^b f(x)\,dx = \lim_{t\to-\infty}\int_t^b f(x)\,dx\] The integral is convergent if the limit exists (finite) and divergent otherwise.
If both \(\int_{-\infty}^a f\,dx\) and \(\int_a^\infty f\,dx\) converge (for any \(a\)): \[\int_{-\infty}^\infty f(x)\,dx = \int_{-\infty}^a f(x)\,dx + \int_a^\infty f(x)\,dx\]
\(p\)-integral: \[\int_1^\infty \frac{1}{x^p}\,dx \text{ is convergent if } p > 1 \text{ and divergent if } p \le 1\]
Type 2: Discontinuous integrands
- If \(f\) is continuous on \([a, b)\) and discontinuous at \(b\): \[\int_a^b f(x)\,dx = \lim_{t\to b^-}\int_a^t f(x)\,dx\]
- If \(f\) is continuous on \((a, b]\) and discontinuous at \(a\): \[\int_a^b f(x)\,dx = \lim_{t\to a^+}\int_t^b f(x)\,dx\]
- If \(f\) has a discontinuity at \(c\), \(a < c < b\), and both pieces converge: \[\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx\]
Also: \(\displaystyle\int_0^1 \frac{1}{x^p}\,dx\) converges if \(p < 1\) and diverges if \(p \ge 1\).
Comparison Test
Suppose \(f\) and \(g\) are continuous with \(f(x) \ge g(x) \ge 0\) for \(x \ge a\). - (a) If \(\int_a^\infty f(x)\,dx\) is convergent, then \(\int_a^\infty g(x)\,dx\) is convergent. - (b) If \(\int_a^\infty g(x)\,dx\) is divergent, then \(\int_a^\infty f(x)\,dx\) is divergent.