Antiderivatives and Basic Integration Technique

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The Derivative: Unlocking Instantaneous Change

The derivative of a function at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. It is the instantaneous rate of change —the exact ratio of the instantaneous change in the dependent variable ($y$) to that of the independent variable ($x$).

Definition. A function $f(x)$ is differentiable at a point $a$ of its domain if its domain contains an open interval containing $a$, and the following limit exists: $\boxed{f'(a) = L = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}}$

Formal Definition: More rigorously, for every positive real number $\epsilon > 0$, there exists a positive real number $\delta > 0$, such that for every $h$ satisfying $0 < |h| < \delta$, the difference between the secant slopes at points sufficiently near a are guaranteed to be ε-close to the derivative: $\left| L - \frac{f(a+h) - f(a)}{h} \right| < \epsilon$

Differentiation Rules Reference

To avoid calculating limits from scratch every time, we use established differentiation rules. Let $f(x)$ and $g(x)$ be differentiable functions, and $n$ be a real number:

1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
2. Sum/Difference Rule: $\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)$
3. Product Rule: $\frac{d}{dx}(f(x) \cdot g(x)) = f'(x)g(x) + f(x)g'(x)$
4. Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$
5. Chain Rule: $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$

Common Transcendental Derivatives:

• Exponential & Logarithmic: $\frac{d}{dx}(e^x) = e^x \quad \Big| \quad \frac{d}{dx}(\ln x) = \frac{1}{x}$
• Trigonometric: $\frac{d}{dx}(\sin x) = \cos x \quad \Big| \quad \frac{d}{dx}(\cos x) = -\sin x \quad \Big| \quad \frac{d}{dx}(\tan x) = \sec^2 x$
• Inverse Trigonometric: $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}} \quad \Big| \quad \frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1 - x^2}} \quad \Big| \quad \frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}$

Antiderivatives

Antiderivatives are a fundamental concept in calculus, essentially acting as the inverse operation to taking a derivative.

Given a function $f(x)$, an antiderivative (also known as an indefinite integral), $F$, is a function that can be differentiated to obtain the original function. Symbolically, this means $F'(x) = f(x)$.

For example, $F(x) = x^3 - x + 7$ is an antiderivative of $f(x) = 3x^2 - 1$ because $\frac{d}{dx} (x^3 - x + 7) = 3x^2 - 1$

The process of finding antiderivatives is called integration. We express this symbolically using the integral sign: $\boxed{F(x) = \int f(x) \, dx}$

The Constant of Integration ($C$)

Because the derivative of any constant is zero, there are infinitely many antiderivatives for a given function, all differing only by a constant. For example, $x^3 - x + 7$ and $x^3 - x - 42$ are both antiderivatives of $3x^2 - 1$. Therefore, we always include $+ C$ (the constant of integration) to represent the entire family of possible antiderivatives.

Fundamental Integration Rules

These are the foundational formulas that map directly back to standard derivative rules.

• Power Rule: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$)

• Logarithmic Rule: $\int \frac{1}{x} \, dx = \ln\vert{}x\vert{} + C$
  Proof for $x < 0$: $\frac{d}{dx} \ln(-x) = \frac{1}{-x}(-1) = \frac{1}{x}$.

• Exponential Rules:
  $\int e^x \, dx = e^x + C$
  $\int e^{ax} \, dx = \frac{1}{a}e^{ax} + C$

• Trigonometric Rules:
  $\int \sin x \, dx = -\cos x + C, \int \cos x \, dx = \sin x + C$, \int \sec^2 x , dx = \tan x + C$

• Inverse Trigonometric Rules:
  $\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}x + C, \int \frac{1}{1+x^2} \, dx = \tan^{-1}x + C$

Algebraic Simplification First

Often, an integral looks complex but can be rewritten into basic power rules using algebra before integrating.

• Example: Expanding polynomials

$\int (x^2 - 3x) \, dx = \frac{x^3}{3} - \frac{3x^2}{2} + C$

• Example: Splitting fractions

$\int \left(\frac{x - 7}{x}\right) \, dx = \int \left(1 - \frac{7}{x}\right) \, dx = x - 7\ln\vert{}x\vert{} + C$

• Example: Fractional exponents

$\int \frac{x^2 + 4\sqrt[3]{x}}{x} \, dx = \int x \, dx + \int \frac{4x^{1/3}}{x} \, dx = \frac{x^2}{2} + \int 4x^{-2/3} \, dx$

$= \frac{x^2}{2} + 4 \left( \frac{x^{1/3}}{1/3} \right) + C = \frac{x^2}{2} + 12\sqrt[3]{x} + C$

Integration by Substitution (U-Substitution)

U-substitution undoes the Chain Rule. You look for a function $u$ and its derivative $du$ nested inside the integral.

General Steps

1. Choose a suitable substitution $u$.
2. Find the differential $du = u' \, dx$.
3. Rewrite the entire integral in terms of $u$ and $du$.
4. Evaluate the new integral.
5. Convert the $u$ back to the original variable $x$.

• Example A: $\int x^3(x^4 + 2)^5 \, dx$

Let $u = x^4 + 2$, then $du = 4x^3 \, dx \implies \frac{du}{4} = x^3 \, dx$.

Substitute: $\int u^5 \frac{du}{4} = \frac{1}{4} \int u^5 \, du = \frac{u^6}{24} + C$.

Revert: $\int x^3(x^4 + 2)^5 \, dx = \frac{(x^4 + 2)^6}{24} + C$.

• Example B: $\int \frac{1}{x \ln x} \, dx$

Let $u = \ln x$, then $du = \frac{1}{x} \, dx$.

Substitute: $\int \frac{1}{u} \, du = \ln\vert{}u\vert{} + C$.

Revert: $\int \frac{1}{x \ln x} \, dx = \ln\vert{}\ln x\vert{} + C$.

• Example C: $\int \frac{x}{1+x^2} \, dx$

Let $u = 1 + x^2$, then $du = 2x \, dx \implies \frac{du}{2} = x \, dx$.

Substitute: $\int \frac{1}{2u} \, du = \frac{1}{2}\ln\vert{}u\vert{} + C$.

Revert: $\int \frac{x}{1+x^2} \, dx = \frac{1}{2}\ln\vert{}1+x^2\vert{} + C$. Besides, absolute value bars can be dropped since $1+x^2$ is always positive.

• Example D: $\int x e^{x^2} \, dx$

Let $u = x^2$, then $du = 2x \, dx \implies \frac{du}{2} = x \, dx$.

Substitute: $\int \frac{1}{2} e^u \, du = \frac{1}{2} e^u + C$.

Revert: $\int x e^{x^2} \, dx = \frac{1}{2}e^{x^2} + C$.

• Example E: $\int \sin x \cos x \, dx$

Let $u = \sin x$, then $du = \cos x \, dx$.

Substitute: $\int u \, du = \frac{u^2}{2} + C$.

Revert: $\int \sin x \cos x \, dx = \frac{1}{2}\sin^2 x + C$.

Using trig identities like $\sin^2 x = 1 - \cos^2 x$, this can also be expressed as $-\frac{1}{2}\cos^2 x + C'$. Both are correct, they just differ by a constant.

Integration by Parts

Integration by parts undoes the Product Rule. It is used when two different types of functions are multiplied together.

The Formula: $\boxed{\int u \, dv = uv - \int v \, du}$

• Example: $\int x \cos x \, dx$

Let $u = x \implies du = dx$ and $dv = \cos x \, dx \implies v = \sin x$.

Apply formula $\boxed{\int u \, dv = uv - \int v \, du}$: $\int x \cos x \, dx = x \sin x - \int \sin x \, dx$.

Final answer: $\int x \cos x \, dx = x \sin x + \cos x + C$.

Because indefinite integrals always include a constant of integration ($C$), two completely correct answers to the same problem might look entirely different until you apply algebra or trigonometric identities.

$\int \sin x \cos x \, dx$

Using $u$-substitution with $u = \sin x$ and $du = \cos x \, dx$, we get: $\int u \, du = \frac{u^2}{2} + C = \frac{1}{2}\sin^2 x + C$

At first glance, this is the final answer.

However, using the Pythagorean identity ($\sin^2 x = 1 - \cos^2 x$), we can rewrite it as $= \frac{1}{2}(1 - \cos^2 x) + C= \frac{1}{2} - \frac{1}{2}\cos^2 x + C$

Because $C$ is just an arbitrary unknown constant, adding $\frac{1}{2}$ to it simply creates a new arbitrary constant. We can absorb the $\frac{1}{2}$ into a new constant, $C'$: $= -\frac{1}{2}\cos^2 x + C'$

Both $\frac{1}{2}\sin^2 x + C$ and $-\frac{1}{2}\cos^2 x + C'$ are mathematically valid antiderivatives of $\sin x \cos x$. They just represent different starting points in the same family of curves.

Uniqueness of Antiderivatives

The example above perfectly illustrates a core theorem of calculus: two antiderivatives of the same function are essentially the same curve shifted up or down on a graph. They are unique up to a constant.

Theorem: Let $F(x)$ and $G(x)$ be two different antiderivatives of the same function $f(x)$. If $F'(x) = G'(x) = f(x)$, then they are unique up to a constant, i.e., $F(x) = G(x) + C$.

Proof:

1. Consider the difference between the two functions: $H(x) = F(x) - G(x)$.
2. Take the derivative of both sides: $H'(x) = F'(x) - G'(x)$.
3. Because both are antiderivatives of $f(x)$, we substitute: $H'(x) = f(x) - f(x) = 0$.
4. The only type of function that has a derivative of zero everywhere is a horizontal line, that is, a constant.
   Therefore, $H(x) = C$.
5. Substituting $H(x)$ back in gives $F(x) - G(x) = C$, which means $F(x) = G(x) + C \ \blacksquare$

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