∫Calculus I

Antiderivative Formulas

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Why This Matters

Antiderivatives are the foundation of integral calculus. Every definite integral you evaluate, every area problem you solve, and every differential equation you tackle depends on your ability to reverse the differentiation process. You're being tested not just on memorizing formulas, but on recognizing which formula applies when you see a function and understanding why each formula works.

These formulas fall into distinct families: power functions, exponential functions, trigonometric functions, and inverse trigonometric functions. Each family follows predictable patterns rooted in differentiation. Every antiderivative formula is just a derivative rule read backwards. If you know what differentiation rule each one reverses, you can reconstruct any formula you forget on exam day.

Power Functions: The Foundation

The power rule for derivatives reverses cleanly into the power rule for antiderivatives. Since $\frac{d}{dx}[x^n] = nx^{n-1}$ , we "undo" this by increasing the exponent and dividing.

Power Rule for Integration

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ where $n \neq -1$ — this handles polynomials, roots (rewrite as fractional exponents), and negative powers
The restriction $n \neq -1$ exists because plugging in $n = -1$ creates division by zero; that case needs its own formula
Always add $+ C$ to represent the constant of integration — the family of all functions whose derivative equals $x^n$

Natural Logarithm Rule

$\int \frac{1}{x} \, dx = \ln|x| + C$ — this fills the gap left by the power rule when $n = -1$
The absolute value bars are essential because $\ln(x)$ is only defined for positive $x$ , but $\frac{1}{x}$ exists for all $x \neq 0$
Recognize disguised forms like $\int x^{-1} \, dx$ — this is the same integral rewritten

Compare: Power Rule vs. Natural Log Rule — both handle expressions of the form $x^n$ , but the log rule is the special case when $n = -1$ . If you see $\int x^{-1} \, dx$ , don't mistakenly apply the power rule. That's a common error.

Exponential Functions: Self-Replicating Integrals

Exponential functions have the remarkable property that differentiation and integration preserve their form. The base determines the scaling factor.

Base $e$ Exponential

$\int e^x \, dx = e^x + C$ — the function $e^x$ is its own antiderivative, making it unique among all functions
Watch for chain rule variants — $\int e^{kx} \, dx = \frac{1}{k}e^{kx} + C$ requires compensating for the inner derivative $k$

General Exponential Base

$\int a^x \, dx = \frac{a^x}{\ln(a)} + C$ where $a > 0$ and $a \neq 1$ — the natural log of the base appears as a scaling factor
This formula reverses the derivative rule $\frac{d}{dx}[a^x] = a^x \ln(a)$ , so you divide by $\ln(a)$ to compensate
When $a = e$ , note that $\ln(e) = 1$ , so this formula reduces to the simpler $e^x$ case

Compare: $\int e^x \, dx$ vs. $\int a^x \, dx$ — the $e^x$ formula is cleaner because $\ln(e) = 1$ eliminates the denominator. On exams, you can convert to base $e$ when helpful using $a^x = e^{x \ln(a)}$ .

Basic Trigonometric Functions: Sine and Cosine Cycle

Trigonometric antiderivatives follow from the cyclic nature of trig derivatives. Since sine and cosine are derivatives of each other (with sign changes), their antiderivatives swap roles.

Sine Function

$\int \sin(x) \, dx = -\cos(x) + C$ — the negative sign appears because $\frac{d}{dx}[\cos(x)] = -\sin(x)$ , not $+\sin(x)$
To remember: integrating sine gives cosine with a sign flip

Cosine Function

$\int \cos(x) \, dx = \sin(x) + C$ — no sign change here since $\frac{d}{dx}[\sin(x)] = \cos(x)$ directly
Verify by differentiating: $\frac{d}{dx}[\sin(x)] = \cos(x)$ ✓

Secant Squared Function

$\int \sec^2(x) \, dx = \tan(x) + C$ — this reverses the derivative $\frac{d}{dx}[\tan(x)] = \sec^2(x)$
Recognize equivalent forms like $\int \frac{1}{\cos^2(x)} \, dx$ — same integral, different notation

Cosecant Squared Function

$\int \csc^2(x) \, dx = -\cot(x) + C$ — this reverses $\frac{d}{dx}[\cot(x)] = -\csc^2(x)$
Note the negative sign, similar to the sine/cosine relationship

Secant-Tangent Product

$\int \sec(x)\tan(x) \, dx = \sec(x) + C$ — this reverses $\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)$

Cosecant-Cotangent Product

$\int \csc(x)\cot(x) \, dx = -\csc(x) + C$ — this reverses $\frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x)$

Compare: $\int \sin(x) \, dx$ vs. $\int \cos(x) \, dx$ — both produce the other trig function, but only sine's antiderivative picks up a negative sign. Track signs carefully; this is a top source of errors on exams. A good pattern to notice: the "co-" functions (cosine, cosecant, cotangent) tend to carry negative signs in their derivative and antiderivative formulas.

Advanced Trigonometric Functions: Logarithmic Results

Some trigonometric integrals produce logarithmic results rather than other trig functions. These arise from rewriting the integrand and applying substitution.

Tangent Function

$\int \tan(x) \, dx = -\ln|\cos(x)| + C$ — equivalently written as $\ln|\sec(x)| + C$
Derived by rewriting $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and using $u$ -substitution with $u = \cos(x)$ , so $du = -\sin(x)\,dx$
The negative sign reflects that $\frac{d}{dx}[\cos(x)] = -\sin(x)$

Compare: $\int \tan(x) \, dx$ vs. $\int \sec^2(x) \, dx$ — both involve tangent and secant, but one yields a logarithm while the other yields tangent directly. The squared secant is the cleaner case.

Inverse Trigonometric Functions: Recognizing the Patterns

These formulas produce inverse trig functions and arise from specific algebraic forms. The key is pattern recognition — spot the characteristic denominators.

Arctangent Pattern

$\int \frac{1}{1+x^2} \, dx = \arctan(x) + C$ — this reverses $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$
The $1 + x^2$ denominator (sum with no square root) is your signal to use this formula
Generalizes to $\int \frac{1}{a^2+x^2} \, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$

Arcsine Pattern

$\int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin(x) + C$ — this reverses $\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}$
The $\sqrt{1-x^2}$ denominator (difference under a square root) signals this formula; note the domain restriction $|x| < 1$
Generalizes to $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$

Compare: Arctangent vs. Arcsine patterns — both have " $1$ " and " $x^2$ " in the denominator, but arctangent has addition ( $1 + x^2$ ) while arcsine has subtraction under a square root ( $\sqrt{1-x^2}$ ). Memorize these signatures; they appear frequently on multiple choice.

Quick Reference Table

Family	Formula	Result
Power ( $n \neq -1$ )	$\int x^n \, dx$	$\frac{x^{n+1}}{n+1} + C$
Power ( $n = -1$ )	$\int \frac{1}{x} \, dx$	$\ln\\|x\\| + C$
Exponential (base $e$ )	$\int e^x \, dx$	$e^x + C$
Exponential (general)	$\int a^x \, dx$	$\frac{a^x}{\ln(a)} + C$
Trig	$\int \sin(x) \, dx$	$-\cos(x) + C$
Trig	$\int \cos(x) \, dx$	$\sin(x) + C$
Trig	$\int \sec^2(x) \, dx$	$\tan(x) + C$
Trig	$\int \csc^2(x) \, dx$	$-\cot(x) + C$
Trig	$\int \sec(x)\tan(x) \, dx$	$\sec(x) + C$
Trig	$\int \csc(x)\cot(x) \, dx$	$-\csc(x) + C$
Trig (log result)	$\int \tan(x) \, dx$	$-\ln\\|\cos(x)\\| + C$
Inverse trig	$\int \frac{1}{1+x^2} \, dx$	$\arctan(x) + C$
Inverse trig	$\int \frac{1}{\sqrt{1-x^2}} \, dx$	$\arcsin(x) + C$

Self-Check Questions

Why does the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ fail when $n = -1$ , and which formula handles that case instead?
Compare $\int e^x \, dx$ and $\int 2^x \, dx$ : what's the difference in their antiderivatives, and why does one have a simpler form?
If you see $\int \frac{1}{4 + x^2} \, dx$ on an exam, which antiderivative formula applies, and how would you adjust for the "4" instead of "1"?
Both $\int \sin(x) \, dx$ and $\int \cos(x) \, dx$ produce the other function. Which one picks up a negative sign, and how can you verify your answer?
Evaluate $\int \frac{1}{\sqrt{9-x^2}} \, dx$ . Identify the pattern, state the formula, and explain what adjustment the "9" requires.