Math Analysis 1

Integrals in depth: fundamental theorem, integration methods proved and tables

Fundamental theorem of calculus proved, parts and substitution derived, cyclic integrals, Wallis formula, rational functions in every case, universal substitution, substitutions for roots, tables of basic and standard integrals, choosing the method.

ielettronica 25 October 2026 8 min read 0 comments

This page goes deeper into the lessons on indefinite and definite integrals. It proves the fundamental theorem of calculus and the integration rules, works out the method for rational functions in every case, derives the substitutions for trigonometric and irrational functions and the recurrence formula for powers of sine. At the end are the tables to remember: basic integrals, standard integrals and the “recipe” for choosing the method.

The fundamental theorem of calculus

Theorem. Let $f$ be continuous on $[a, b]$ and let $F(x) = \displaystyle\int_a^x f(t)\,dt$ be the integral function. Then $F$ is differentiable and $F'(x) = f(x)$ for every $x$ . Consequently, if $G$ is any antiderivative of $f$ :

$\int_a^b f(x)\,dx = G(b) - G(a)$

Proof. For $h \ne 0$ , by additivity of the integral,

$\frac{F(x + h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h}f(t)\,dt$

By the mean value theorem for integrals (lesson) there is $c_h$ between $x$ and $x + h$ with $\int_x^{x+h}f = h\,f(c_h)$ . So the difference quotient equals $f(c_h)$ . As $h \to 0$ , $c_h \to x$ , and by continuity of $f$ , $f(c_h) \to f(x)$ . Hence $F'(x) = f(x)$ .

For the second part: $F$ and $G$ are antiderivatives of the same function on an interval, so they differ by a constant (the derivative of $F - G$ is zero, and a function with zero derivative on an interval is constant, a consequence of the mean value theorem). $F(x) = G(x) + k$ ; with $x = a$ , $0 = G(a) + k$ . Then $\int_a^b f = F(b) = G(b) - G(a)$ . ∎

The fundamental theorem of calculus — The integral function F of x is the blue area under the graph of f from a to x. When x increases by h, the area grows by the red strip, which for small h is about f of x times h: this is why the derivative of F is f.

This is the theorem that links two apparently different problems, area and tangent. It has two consequences that are used often: every continuous function has an antiderivative (even if it cannot be written with elementary functions, like $e^{-x^2}$ ), and the derivative of $\int_{a}^{g(x)}f(t)\,dt$ is $f(g(x))\,g'(x)$ , by the chain rule.

The integration rules, derived

By parts. From the product rule $(fg)' = f'g + fg'$ , integrating both sides:

$\int f(x)\,g'(x)\,dx = f(x)\,g(x) - \int f'(x)\,g(x)\,dx$

Choose as $f$ the factor that simplifies when differentiated (polynomials, logarithms, inverse trigonometric functions) and as $g'$ the one you can integrate (exponentials, sine, cosine). The trick for $\int\ln x\,dx$ and $\int\arctan x\,dx$ is to take $g' = 1$ .

Cyclic integrals. For $\int e^x\sin x\,dx$ , integration by parts applied twice brings back the original integral with the opposite sign: $I = e^x\sin x - e^x\cos x - I$ , and so $I = \frac{1}{2}e^x(\sin x - \cos x) + C$ . One solves an equation in the unknown $I$ .

By substitution. From the chain rule, if $G' = g$ then $(G(\varphi(t)))' = g(\varphi(t))\,\varphi'(t)$ . So, with $x = \varphi(t)$ :

$\int g(x)\,dx = \int g(\varphi(t))\,\varphi'(t)\,dt,\qquad \int_a^b g(x)\,dx = \int_{\varphi^{-1}(a)}^{\varphi^{-1}(b)}g(\varphi(t))\,\varphi'(t)\,dt$

In a definite integral the limits change and there is no need to go back to the variable $x$ . The practical rule “ $dx = \varphi'(t)\,dt$ ” is just a mnemonic way of writing this formula.

Recurrence formula for powers of sine. With $I_n = \int_0^{\pi/2}\sin^n x\,dx$ , by parts with $f = \sin^{n-1}x$ and $g' = \sin x$ , and then $\cos^2 x = 1 - \sin^2 x$ :

$I_n = (n - 1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,dx = (n - 1)(I_{n-2} - I_n) \quad\Rightarrow\quad I_n = \frac{n - 1}{n}I_{n-2}$

with $I_0 = \frac{\pi}{2}$ and $I_1 = 1$ . For example $I_4 = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3\pi}{16}$ . These are the Wallis integrals; the same holds for cosine.

Rational functions, case by case

To integrate $\frac{N(x)}{D(x)}$ , with $N$ and $D$ polynomials, proceed as follows.

1. Degree. If the degree of $N$ is greater than or equal to that of $D$ , divide: $\frac{N}{D} = Q + \frac{R}{D}$ , with the degree of $R$ less than that of $D$ . The quotient $Q$ is a polynomial and is integrated straight away.

2. Factorisation. Factor $D$ into first-degree factors $(x - a)^k$ and irreducible second-degree factors $(x^2 + px + q)^k$ with $p^2 - 4q \lt 0$ . Every real polynomial factors this way (a consequence of the fundamental theorem of algebra).

3. Partial fractions. $\frac{R}{D}$ is written as a sum of terms $\frac{A}{x - a}$ , $\frac{A}{(x - a)^2}$ , … for first-degree factors (one for each power up to the multiplicity) and $\frac{Bx + C}{x^2 + px + q}$ for second-degree ones. The constants are found with the identity principle for polynomials, or faster by substituting the roots.

4. Integrating the fractions. $\int\frac{dx}{x - a} = \ln|x - a|$ ; $\int\frac{dx}{(x - a)^k} = -\frac{1}{(k - 1)(x - a)^{k-1}}$ for $k \ge 2$ . For the second-degree term, make the derivative of the denominator appear in the numerator:

$\frac{Bx + C}{x^2 + px + q} = \frac{B}{2}\cdot\frac{2x + p}{x^2 + px + q} + \left(C - \frac{Bp}{2}\right)\frac{1}{x^2 + px + q}$

The first piece gives $\frac{B}{2}\ln(x^2 + px + q)$ . For the second, complete the square:

Theorem. If $\Delta = p^2 - 4q \lt 0$ , with $k = \frac{\sqrt{-\Delta}}{2} = \sqrt{q - \frac{p^2}{4}}$ :

$\int\frac{dx}{x^2 + px + q} = \frac{1}{k}\arctan\frac{x + \frac{p}{2}}{k} + C$

Proof. $x^2 + px + q = \left(x + \frac{p}{2}\right)^2 + k^2$ . With $x + \frac{p}{2} = k\,u$ , $dx = k\,du$ : $\int\frac{k\,du}{k^2(u^2 + 1)} = \frac{1}{k}\arctan u$ . ∎

If $\Delta \gt 0$ the denominator has two real roots and one goes back to first-degree fractions; if $\Delta = 0$ it is a perfect square, $(x - a)^2$ , and the integral of $1/(x - a)^2$ is $-1/(x - a)$ .

Trigonometric functions: the universal substitution

Theorem. With $t = \tan\frac{x}{2}$ , for $-\pi \lt x \lt \pi$ :

$\sin x = \frac{2t}{1 + t^2},\qquad \cos x = \frac{1 - t^2}{1 + t^2},\qquad dx = \frac{2\,dt}{1 + t^2}$

Proof. With the double-angle formulas and dividing by $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$ : $\sin x = \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2} + \sin^2\frac{x}{2}}$ ; dividing numerator and denominator by $\cos^2\frac{x}{2}$ gives $\frac{2t}{1 + t^2}$ . In the same way $\cos x = \frac{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}}{\cos^2\frac{x}{2} + \sin^2\frac{x}{2}} = \frac{1 - t^2}{1 + t^2}$ . Finally $x = 2\arctan t$ , so $dx = \frac{2\,dt}{1 + t^2}$ . ∎

Every rational function of $\sin x$ and $\cos x$ thus becomes a rational function of $t$ . It always works but often leads to long calculations: first try the shortcuts, i.e. $u = \sin x$ if the rest contains $\cos x\,dx$ , and the formulas $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$ for even powers.

Roots: trigonometric and hyperbolic substitutions

For $\sqrt{a^2 - x^2}$ set $x = a\sin t$ , and the root becomes $a\cos t$ . For $\sqrt{x^2 + a^2}$ set $x = a\sinh t$ , and the root becomes $a\cosh t$ (because $\cosh^2 - \sinh^2 = 1$ ). For $\sqrt{x^2 - a^2}$ , $x = a\cosh t$ .

Example: the area of the circle. With $x = \sin t$ , $dx = \cos t\,dt$ :

$\int\sqrt{1 - x^2}\,dx = \int\cos^2 t\,dt = \frac{t}{2} + \frac{\sin 2t}{4} = \frac{\arcsin x}{2} + \frac{x\sqrt{1 - x^2}}{2} + C$

Between 0 and 1 it gives $\frac{\pi}{4}$ , a quarter of the circle of radius 1. The two terms have a precise geometric meaning, shown in the figure.

Geometric meaning of the integral of the square root of 1 minus x squared — The area under the quarter circle between 0 and x is the sum of the circular sector S, with angle t equal to arcsine of x and area t over 2, and of the triangle T, with base x and height the square root of 1 minus x squared.

Tables to remember

Basic integrals (the constant $C$ is understood)

$f(x)$	$\int f(x)\,dx$
$x^\alpha$ , $\alpha \ne -1$	$\dfrac{x^{\alpha+1}}{\alpha + 1}$
$\dfrac{1}{x}$	$\ln\|x\|$
$e^x$ , $a^x$	$e^x$ , $\dfrac{a^x}{\ln a}$
$\sin x$ , $\cos x$	$-\cos x$ , $\sin x$
$\dfrac{1}{\cos^2 x}$ , $\dfrac{1}{\sin^2 x}$	$\tan x$ , $-\cot x$
$\dfrac{1}{1 + x^2}$	$\arctan x$
$\dfrac{1}{\sqrt{1 - x^2}}$	$\arcsin x$
$\sinh x$ , $\cosh x$	$\cosh x$ , $\sinh x$

Composite forms (with $g$ differentiable): $\int g'(x)\,g(x)^\alpha\,dx = \frac{g^{\alpha+1}}{\alpha + 1}$ ; $\int\frac{g'}{g}dx = \ln|g|$ ; $\int g'\,e^{g}dx = e^{g}$ ; $\int\frac{g'}{1 + g^2}dx = \arctan g$ . Recognising them at a glance avoids many substitutions.

Standard integrals ( $a \gt 0$ )

$f(x)$	$\int f(x)\,dx$
$\ln x$	$x\ln x - x$
$\arctan x$	$x\arctan x - \frac{1}{2}\ln(1 + x^2)$
$\tan x$	$-\ln\|\cos x\|$
$\dfrac{1}{\cos x}$	$\ln\left\|\dfrac{1 + \sin x}{\cos x}\right\|$
$\sin^2 x$ , $\cos^2 x$	$\dfrac{x}{2} - \dfrac{\sin 2x}{4}$ , $\dfrac{x}{2} + \dfrac{\sin 2x}{4}$
$\dfrac{1}{x^2 + a^2}$	$\dfrac{1}{a}\arctan\dfrac{x}{a}$
$\dfrac{1}{x^2 - a^2}$	$\dfrac{1}{2a}\ln\left\|\dfrac{x - a}{x + a}\right\|$
$\dfrac{1}{\sqrt{a^2 - x^2}}$	$\arcsin\dfrac{x}{a}$
$\dfrac{1}{\sqrt{x^2 + a^2}}$	$\ln\left(x + \sqrt{x^2 + a^2}\right)$
$\sqrt{a^2 - x^2}$	$\dfrac{a^2}{2}\arcsin\dfrac{x}{a} + \dfrac{x}{2}\sqrt{a^2 - x^2}$
$e^{ax}\sin bx$	$\dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2}$

Which method to use

If the integrand is…	Method
a composite form $g'\cdot h(g)$	substitution $u = g(x)$
polynomial times exponential, sine or cosine	by parts, differentiating the polynomial
polynomial times logarithm or arctangent	by parts, differentiating the logarithm or arctangent
exponential times sine or cosine	by parts twice, cyclic integral
rational function	division and partial fractions
rational in $\sin x$ and $\cos x$	shortcuts, otherwise $t = \tan\frac{x}{2}$
with $\sqrt{a^2 - x^2}$ , $\sqrt{x^2 \pm a^2}$	$x = a\sin t$ , $x = a\sinh t$ , $x = a\cosh t$
with $\sqrt[n]{ax + b}$	$t = \sqrt[n]{ax + b}$

Exercises from basic to exam level, with areas, volumes and integral functions, are on the exercise page linked at the top.

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The fundamental theorem of calculus

The integration rules, derived

Rational functions, case by case

Trigonometric functions: the universal substitution

Roots: trigonometric and hyperbolic substitutions

Tables to remember

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