Standard Integrals and Fundamental Theorems

Having understood the concept of an antiderivative and the geometry of the indefinite integral, we now begin building our mathematical toolkit.

To evaluate integrals efficiently, we must have a reliable dictionary of known antiderivatives and clear algebraic rules that govern how integrals behave when functions are multiplied by constants, added together, or subjected to linear transformations.

In this tutorial, we establish the standard table of elementary antiderivatives, verify the non-trivial logarithmic and trigonometric forms, and prove the three foundational theorems that unlock the integration of linear combinations and linear composite arguments.

The Verification Principle

Because indefinite integration is the inverse of differentiation, discovering an antiderivative does not require a lengthy algebraic proof from scratch.

To prove that:

\[\int f(x)\,dx = F(x) + C\]

we simply need to differentiate the proposed answer $F(x) + C$. If:

\[\frac{d}{dx}[F(x) + C] = f(x)\]

then by definition, the formula is 100% correct and verified.

Every standard derivative formula learned in differential calculus therefore gives us an immediate integration formula in reverse.

The Basic Table of Standard Antiderivatives

The table below summarizes the core elementary antiderivatives.

Integrand $f(x)$	Indefinite Integral $\int f(x)\,dx$	Domain / Restriction
$x^n$	$\frac{x^{n+1}}{n+1} + C$	$n \neq -1$
$1$	$x + C$	Constant function ($x^0$)
$\frac{1}{x}$	$\ln	x
$e^x$	$e^x + C$	Natural exponential
$a^x$	$\frac{a^x}{\ln a} + C$	$a > 0, a \neq 1$
$\cos x$	$\sin x + C$	Standard trigonometric
$\sin x$	$-\cos x + C$	Standard trigonometric
$\sec^2 x$	$\tan x + C$	Standard trigonometric
$\csc^2 x$	$-\cot x + C$	Standard trigonometric
$\sec x \tan x$	$\sec x + C$	Standard trigonometric
$\csc x \cot x$	$-\csc x + C$	Standard trigonometric
$\tan x$	$\ln	\sec x
$\cot x$	$\ln	\sin x
$\sec x$	$\ln	\sec x + \tan x
$\csc x$	$\ln	\csc x - \cot x
$\frac{1}{\sqrt{1 - x^2}}$	$\arcsin x + C = -\arccos x + C'$	$
$\frac{1}{1 + x^2}$	$\arctan x + C = -\operatorname{arccot} x + C'$	$x \in \mathbb{R}$
**$\frac{1}{	x	\sqrt{x^2 - 1}}$**

For the standard algebraic, exponential, and inverse trigonometric integrals, the results are immediate reversals of their derivative formulas:

$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}} \implies \int \frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C$
$\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2} \implies \int \frac{dx}{1 + x^2} = \arctan x + C$
$\frac{d}{dx}[\operatorname{arcsec} x] = \frac{1}{|x|\sqrt{x^2 - 1}} \implies \int \frac{dx}{|x|\sqrt{x^2 - 1}} = \operatorname{arcsec} x + C$

No separate derivations are needed for these. However, the logarithmic and trigonometric integrals involve crucial subtleties that deserve direct verification.

Verification of Logarithmic and Special Trigonometric Integrals

The Logarithmic Integral

For the function $f(x) = \frac{1}{x}$, the antiderivative is $\ln|x| + C$. The absolute value $|x|$ is necessary because $\frac{1}{x}$ is defined for both positive and negative values of $x$, whereas the real logarithm $\ln u$ requires a positive argument $u > 0$.

We verify this by checking both branches:

Branch 1 ($x > 0$):

Here $|x| = x$. Differentiating:

\[\frac{d}{dx}[\ln x] = \frac{1}{x}\]

Branch 2 ($x < 0$):

Here $|x| = -x > 0$. Using the Chain Rule:

\[\frac{d}{dx}[\ln(-x)] = \frac{1}{-x} \cdot \frac{d}{dx}[-x] = \frac{1}{-x} \cdot (-1) = \frac{1}{x}\]

In both branches, the derivative is $\frac{1}{x}$. Hence, for all $x \neq 0$:

\[\int \frac{1}{x}\,dx = \ln|x| + C\]

The Tangent and Cotangent Integrals

Differentiating $\ln|\sec x|$:

\[\frac{d}{dx}[\ln|\sec x|] = \frac{1}{\sec x} \cdot (\sec x \tan x) = \tan x \implies \int \tan x\,dx = \ln|\sec x| + C\]

Because $\ln|\sec x| = \ln\left|\frac{1}{\cos x}\right| = -\ln|\cos x|$, this is equivalently written as:

\[\int \tan x\,dx = -\ln|\cos x| + C\]

Similarly, differentiating $\ln|\sin x|$:

\[\frac{d}{dx}[\ln|\sin x|] = \frac{1}{\sin x} \cdot \cos x = \cot x \implies \int \cot x\,dx = \ln|\sin x| + C\]

The Secant Integral and its Half-Angle Form

1. Verification by Differentiation:

Let us differentiate the expression $\ln|\sec x + \tan x|$:

\[\frac{d}{dx}[\ln|\sec x + \tan x|] = \frac{\frac{d}{dx}(\sec x + \tan x)}{\sec x + \tan x} = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x}\]

Factor out $\sec x$ from the numerator:

\[\frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \sec x\]

Therefore:

\[\int \sec x\,dx = \ln|\sec x + \tan x| + C\]

2. Proof of the Half-Angle Equivalence:

We show that $\sec x + \tan x = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$.

Express $\sec x$ and $\tan x$ in terms of sine and cosine:

\[\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x}\]

Write $1$, $\sin x$, and $\cos x$ using half-angle formulas:

$1 + \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2$
$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)$

Substitute these into the fraction:

\[\sec x + \tan x = \frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)} = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\]

Divide both numerator and denominator by $\cos\frac{x}{2}$:

\[\sec x + \tan x = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\]

Since $\tan\frac{\pi}{4} = 1$, apply the compound angle formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:

\[\frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\]

Taking logarithms yields the alternative standard formula:

\[\int \sec x\,dx = \ln\left|\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right| + C\]

The Cosecant Integral and its Half-Angle Form

1. Verification by Differentiation:

Let us differentiate $\ln|\csc x - \cot x|$:

\[\frac{d}{dx}[\ln|\csc x - \cot x|] = \frac{\frac{d}{dx}(\csc x - \cot x)}{\csc x - \cot x} = \frac{-\csc x \cot x - (-\csc^2 x)}{\csc x - \cot x}\]

\[\frac{d}{dx}[\ln|\csc x - \cot x|] = \frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x} = \frac{\csc x(\csc x - \cot x)}{\csc x - \cot x} = \csc x\]

Therefore:

\[\int \csc x\,dx = \ln|\csc x - \cot x| + C\]

2. Proof of the Half-Angle Equivalence:

We show that $\csc x - \cot x = \tan\frac{x}{2}$.

Express in terms of sine and cosine:

\[\csc x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1 - \cos x}{\sin x}\]

Apply the half-angle identities $1 - \cos x = 2\sin^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$:

\[\csc x - \cot x = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2}\]

Taking logarithms yields the second standard form:

\[\int \csc x\,dx = \ln\left|\tan\frac{x}{2}\right| + C\]

(Note: Since $\csc x - \cot x = \frac{1}{\csc x + \cot x}$, it is also written as $-\ln|\csc x + \cot x| + C$.)

Three Fundamental Theorems of Indefinite Integration

To compute integrals of combinations and transformations of functions, we rely on three structural theorems. Each theorem is proved from first principles using the corresponding rules of differentiation.

Theorem 1: Factoring Out a Scalar Constant

For any non-zero real constant $k \neq 0$:

\[\int k f(x)\,dx = k \int f(x)\,dx\]

First-Principles Proof

Let $F(x)$ be an antiderivative of $f(x)$, so that:

\[\frac{d}{dx}[F(x)] = f(x)\]

Consider the function $k \cdot F(x)$. By the constant multiple rule of differentiation:

\[\frac{d}{dx}[k \cdot F(x)] = k \cdot \frac{d}{dx}[F(x)] = k \cdot f(x)\]

By the fundamental definition of the indefinite integral:

\[\int k f(x)\,dx = k \cdot F(x) + C_1\]

Now examine the right-hand side:

\[k \int f(x)\,dx = k(F(x) + C_2) = k \cdot F(x) + k C_2\]

Because $C_2$ is an arbitrary constant and $k \neq 0$, the product $k C_2$ is simply an arbitrary constant. Both expressions describe the identical family of functions. Hence:

\[\int k f(x)\,dx = k \int f(x)\,dx\]

Theorem 2: Addition Distribution (Sum and Difference Rule)

The integral of a sum or difference of two functions equals the sum or difference of their individual integrals:

\[\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx\]

First-Principles Proof

Let $F(x)$ and $G(x)$ be antiderivatives of $f(x)$ and $g(x)$ respectively, so that:

\[\frac{d}{dx}[F(x)] = f(x) \quad \text{and} \quad \frac{d}{dx}[G(x)] = g(x)\]

Consider the combined function $F(x) \pm G(x)$. By the sum/difference rule of differentiation:

\[\frac{d}{dx}[F(x) \pm G(x)] = \frac{d}{dx}[F(x)] \pm \frac{d}{dx}[G(x)] = f(x) \pm g(x)\]

By the definition of the indefinite integral:

\[\int [f(x) \pm g(x)]\,dx = [F(x) \pm G(x)] + C\]

Evaluating the right-hand side:

\[\int f(x)\,dx \pm \int g(x)\,dx = (F(x) + C_1) \pm (G(x) + C_2) = [F(x) \pm G(x)] + (C_1 \pm C_2)\]

Because $C_1$ and $C_2$ are independent arbitrary constants, $C_1 \pm C_2$ represents an arbitrary constant. Both expressions represent the identical family of functions.

By mathematical induction, this distribution property extends across any finite linear combination:

\[\int [k_1 f_1(x) + k_2 f_2(x) + \dots + k_n f_n(x)]\,dx = k_1 \int f_1(x)\,dx + k_2 \int f_2(x)\,dx + \dots + k_n \int f_n(x)\,dx\]

Theorem 3: Linear Transformation of the Argument

If $\int f(x)\,dx = F(x) + C$, then for any non-zero real constant $a$ ($a \neq 0$) and any real constant $b$:

\[\int f(ax + b)\,dx = \frac{1}{a} F(ax + b) + C\]

First-Principles Proof

To find $\int f(ax + b)\,dx$, we seek a function whose derivative with respect to $x$ is precisely $f(ax + b)$.

Consider the candidate function:

\[\Phi(x) = \frac{1}{a} F(ax + b)\]

Let us differentiate $\Phi(x)$ with respect to $x$.

Let $u(x) = ax + b$. By the Chain Rule of differentiation:

\[\frac{d}{dx}[\Phi(x)] = \frac{d}{dx}\left[ \frac{1}{a} F(u) \right] = \frac{1}{a} \cdot \frac{d}{du}[F(u)] \cdot \frac{du}{dx}\]

Evaluate each term:

Since $F$ is an antiderivative of $f$, we have $\frac{d}{du}[F(u)] = f(u) = f(ax + b)$.
Differentiating the linear argument: $\frac{du}{dx} = \frac{d}{dx}[ax + b] = a$.

Substitute these into the derivative:

\[\frac{d}{dx}[\Phi(x)] = \frac{1}{a} \cdot f(ax + b) \cdot a\]

Since $a \neq 0$, the factor $a$ in the numerator and denominator cancels completely:

\[\frac{d}{dx}[\Phi(x)] = f(ax + b)\]

Because the derivative of $\frac{1}{a} F(ax + b)$ with respect to $x$ is $f(ax + b)$, by the fundamental definition of the indefinite integral:

\[\int f(ax + b)\,dx = \frac{1}{a} F(ax + b) + C\]

This completes the proof.

Critical Warning: Failure for Non-Linear Arguments

Theorem 3 is valid only and strictly when the inner argument is linear ($ax + b$). If the inner argument contains $x^2, \sqrt{x},$ or any non-linear expression, this rule fails completely.

For instance, students commonly make the following mistake:

\[\int \cos(x^2)\,dx \neq \frac{\sin(x^2)}{2x} + C \quad \text{(FALSE!)}\]

To see why this is wrong, differentiate the proposed answer using the quotient rule:

\[\frac{d}{dx}\left[ \frac{\sin(x^2)}{2x} \right] = \frac{(2x \cdot \cos(x^2) \cdot 2x) - (\sin(x^2) \cdot 2)}{4x^2} = \cos(x^2) - \frac{\sin(x^2)}{2x^2} \neq \cos(x^2)\]

The extra variable $x$ in the denominator produces an unwanted cross-term. The constant $\frac{1}{a}$ can only factor outside the derivative safely because $a$ is a constant.

Direct Corollaries of the Linear Scaling Theorem

Applying Theorem 3 to our basic table gives immediate integration formulas for linear arguments:

Linear Power and Reciprocal:

\[\int (ax + b)^n\,dx = \frac{(ax + b)^{n+1}}{a(n + 1)} + C \quad (n \neq -1)\]

\[\int \frac{1}{ax + b}\,dx = \frac{1}{a}\ln|ax + b| + C\]

Linear Exponential:

\[\int e^{ax + b}\,dx = \frac{1}{a}e^{ax + b} + C, \quad \int k^{ax + b}\,dx = \frac{k^{ax + b}}{a \ln k} + C\]

Linear Trigonometric Forms:

\[\int \sin(ax + b)\,dx = -\frac{1}{a}\cos(ax + b) + C\]

\[\int \cos(ax + b)\,dx = \frac{1}{a}\sin(ax + b) + C\]

\[\int \sec^2(ax + b)\,dx = \frac{1}{a}\tan(ax + b) + C\]

\[\int \sec(ax + b)\,dx = \frac{1}{a}\ln|\sec(ax + b) + \tan(ax + b)| + C\]

Linear Inverse Trigonometric Forms:

\[\int \frac{dx}{\sqrt{1 - (ax + b)^2}} = \frac{1}{a}\arcsin(ax + b) + C\]

\[\int \frac{dx}{1 + (ax + b)^2} = \frac{1}{a}\arctan(ax + b) + C\]

Illustrative Worked Examples

Example 1

Evaluate:

\[I = \int \left( (3x - 5)^4 - \frac{4}{2x + 1} + 6e^{4x - 1} \right)\,dx\]

Solution:

Distribute the integral using Theorem 1 and Theorem 2:

\[I = \int (3x - 5)^4\,dx - 4\int \frac{1}{2x + 1}\,dx + 6\int e^{4x - 1}\,dx\]

Apply Theorem 3 to each individual term:

Term 1: For $(3x - 5)^4$, the argument is linear with $a = 3$:

\[\int (3x - 5)^4\,dx = \frac{(3x - 5)^5}{3 \cdot 5} = \frac{(3x - 5)^5}{15}\]

Term 2: For $\frac{1}{2x + 1}$, the denominator is linear with $a = 2$:

\[\int \frac{1}{2x + 1}\,dx = \frac{1}{2}\ln|2x + 1|\]

Term 3: For $e^{4x - 1}$, the exponent is linear with $a = 4$:

\[\int e^{4x - 1}\,dx = \frac{1}{4}e^{4x - 1}\]

Combining the evaluated terms:

\[I = \frac{(3x - 5)^5}{15} - 4\left(\frac{1}{2}\ln|2x + 1|\right) + 6\left(\frac{1}{4}e^{4x - 1}\right) + C\]

\[I = \frac{(3x - 5)^5}{15} - 2\ln|2x + 1| + \frac{3}{2}e^{4x - 1} + C\]

Example 2

Evaluate:

\[I = \int \left( \sin(5x - 2) + \sec^2(3x + 4) - \sec(2x)\tan(2x) \right)\,dx\]

Solution:

Distribute the integral across the terms:

\[I = \int \sin(5x - 2)\,dx + \int \sec^2(3x + 4)\,dx - \int \sec(2x)\tan(2x)\,dx\]

Apply the standard formulas with linear scaling ($u = ax + b$):

Term 1: For $\sin(5x - 2)$, the coefficient of $x$ is $a = 5$:

\[\int \sin(5x - 2)\,dx = -\frac{1}{5}\cos(5x - 2)\]

Term 2: For $\sec^2(3x + 4)$, the coefficient of $x$ is $a = 3$:

\[\int \sec^2(3x + 4)\,dx = \frac{1}{3}\tan(3x + 4)\]

Term 3: For $\sec(2x)\tan(2x)$, the coefficient of $x$ is $a = 2$:

\[\int \sec(2x)\tan(2x)\,dx = \frac{1}{2}\sec(2x)\]

Combining the terms:

\[I = -\frac{1}{5}\cos(5x - 2) + \frac{1}{3}\tan(3x + 4) - \frac{1}{2}\sec(2x) + C\]

Example 3

Evaluate:

\[I = \int \cos^2(3x)\,dx\]

Solution:

We cannot integrate $\cos^2(3x)$ directly as a power because the argument inside the squaring operation is a function of $x$. Instead, we use the standard trigonometric power-reduction identity:

\[\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\]

Setting $\theta = 3x$ gives $2\theta = 6x$:

\[\cos^2(3x) = \frac{1 + \cos(6x)}{2}\]

Substitute this identity into the integral:

\[I = \int \frac{1 + \cos(6x)}{2}\,dx = \frac{1}{2}\int 1\,dx + \frac{1}{2}\int \cos(6x)\,dx\]

Now evaluate each term:

$\int 1\,dx = x$
By Theorem 3 with $a = 6$, $\int \cos(6x)\,dx = \frac{1}{6}\sin(6x)$

Therefore:

\[I = \frac{1}{2}x + \frac{1}{2}\left(\frac{1}{6}\sin(6x)\right) + C = \frac{1}{2}x + \frac{1}{12}\sin(6x) + C\]

Summary of Core Concepts

Concept	Mathematical Form	Meaning & Significance
Verification Principle	$\frac{d}{dx}[F(x) + C] = f(x)$	Differentiating the right-hand side validates any integration formula immediately.
Standard Power & Log	$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$, $\int \frac{1}{x}\,dx = \ln\\|x\\| + C$	Fundamental algebraic primitives; absolute value preserves domain for negative $x$.
Secant Integral	$\int \sec x\,dx = \ln\\|\sec x + \tan x\\| = \ln\\|\tan(\frac{\pi}{4} + \frac{x}{2})\\|$	Direct verification via differentiation and half-angle trigonometric identity.
Cosecant Integral	$\int \csc x\,dx = \ln\\|\csc x - \cot x\\| = \ln\\|\tan\frac{x}{2}\\|$	Direct verification via differentiation and half-angle trigonometric identity.
Theorem 1 (Scalar Multiple)	$\int k f(x)\,dx = k \int f(x)\,dx$	Constants factor outside the integral operator (proved via constant multiple rule).
Theorem 2 (Sum Rule)	$\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$	Integrals distribute across sums and differences (proved via derivative sum rule).
Theorem 3 (Linear Scaling)	$\int f(ax + b)\,dx = \frac{1}{a} F(ax + b) + C$	Linear argument scaling divides by the coefficient $a$ (strictly fails for non-linear arguments).

Integrand \(f(x)\)	Indefinite Integral \(\int f(x)\,dx\)	Domain / Restriction
\(x^n\)	\(\frac{x^{n+1}}{n+1} + C\)	\(n \neq -1\)
\(1\)	\(x + C\)	Constant function (\(x^0\))
\(\frac{1}{x}\)	$\ln	x
\(e^x\)	\(e^x + C\)	Natural exponential
\(a^x\)	\(\frac{a^x}{\ln a} + C\)	\(a > 0, a \neq 1\)
\(\cos x\)	\(\sin x + C\)	Standard trigonometric
\(\sin x\)	\(-\cos x + C\)	Standard trigonometric
\(\sec^2 x\)	\(\tan x + C\)	Standard trigonometric
\(\csc^2 x\)	\(-\cot x + C\)	Standard trigonometric
\(\sec x \tan x\)	\(\sec x + C\)	Standard trigonometric
\(\csc x \cot x\)	\(-\csc x + C\)	Standard trigonometric
\(\tan x\)	$\ln	\sec x
\(\cot x\)	$\ln	\sin x
\(\sec x\)	$\ln	\sec x + \tan x
\(\csc x\)	$\ln	\csc x - \cot x
\(\frac{1}{\sqrt{1 - x^2}}\)	\(\arcsin x + C = -\arccos x + C'\)	$
\(\frac{1}{1 + x^2}\)	\(\arctan x + C = -\operatorname{arccot} x + C'\)	\(x \in \mathbb{R}\)
**$\frac{1}{	x	\sqrt{x^2 - 1}}$**

Concept	Mathematical Form	Meaning & Significance
Verification Principle	\(\frac{d}{dx}[F(x) + C] = f(x)\)	Differentiating the right-hand side validates any integration formula immediately.
Standard Power & Log	\(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\), \(\int \frac{1}{x}\,dx = \ln\\|x\\| + C\)	Fundamental algebraic primitives; absolute value preserves domain for negative \(x\).
Secant Integral	\(\int \sec x\,dx = \ln\\|\sec x + \tan x\\| = \ln\\|\tan(\frac{\pi}{4} + \frac{x}{2})\\|\)	Direct verification via differentiation and half-angle trigonometric identity.
Cosecant Integral	\(\int \csc x\,dx = \ln\\|\csc x - \cot x\\| = \ln\\|\tan\frac{x}{2}\\|\)	Direct verification via differentiation and half-angle trigonometric identity.
Theorem 1 (Scalar Multiple)	\(\int k f(x)\,dx = k \int f(x)\,dx\)	Constants factor outside the integral operator (proved via constant multiple rule).
Theorem 2 (Sum Rule)	\(\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx\)	Integrals distribute across sums and differences (proved via derivative sum rule).
Theorem 3 (Linear Scaling)	\(\int f(ax + b)\,dx = \frac{1}{a} F(ax + b) + C\)	Linear argument scaling divides by the coefficient \(a\) (strictly fails for non-linear arguments).