How Integration Runs Differentiation Backwards

Differentiation turns a distance formula into a speed formula. But often we know the speed and want the distance, or know how fast a population grows and want its size. Working back from a derivative to the original function is integration.

This part covers integration as antidifferentiation, standard formulae and properties, substitution, and integrals simplified with trigonometric identities.

**An integral of $f(x)$ is any function $F(x)$ with $F^{\prime }(x)=f(x)$, and because constants vanish on differentiation, all such functions differ by a constant, so the indefinite integral is written $\int f(x)dx=F(x)+C$.

Notation:**

- $\int$ — the integral sign
- $f(x)$ — the integrand
- $dx$ — the variable of integration
- $C$ — the constant of integration, any real number

Worked example. Since $\frac{d}{dx}(x^3)=\frac{d}{dx}(x^3+7)=\frac{d}{dx}(x^3-2)=3x^2$,

$$\int 3x^{2}dx=x^3+C$$

Geometric meaning. The curves $y=x^3+C$ are parallel copies shifted up or down; at each $x$ they all have the same slope $3x^2$.

Worked example — fixing the constant. A curve has slope $2x$ and passes through $(1,5)$. Then $y=x^2+C$, and $5=1+C$ gives $C=4$, so $y=x^2+4$.

An everyday example. Knowing a car's speed at every moment tells you how far it has travelled, but not where it started — the starting point plays the role of $C$.

The substance. **Leaving out $+C$ is a genuine error**, because it names only one of infinitely many antiderivatives.

Every standard derivative gives a standard integral, and integrals can be split over sums and differences, with constant factors taken outside.

Standard formulae:

- $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$, for $n\ne -1$
- $\int \frac{1}{x}dx=\log |x|+C$
- $\int e^{x}dx=e^x+C$, and $\int a^{x}dx=\frac{a^x}{\log a}+C$
- $\int \sin xdx=-\cos x+C$, and $\int \cos xdx=\sin x+C$
- $\int {\sec }^{2}xdx=\tan x+C$, and $\int {\text{cosec}}^{2}xdx=-\cot x+C$
- $\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x+C$, and $\int \frac{dx}{1+x^2}={\tan }^{-1}x+C$

Properties:

- $\int [f(x)\pm g(x)]dx=\int f(x)dx\pm \int g(x)dx$
- $\int kf(x)dx=k\int f(x)dx$ for a constant $k$

Worked example.

$$\int \left(4x^3+\frac{2}{x}-3\cos x\right)dx=x^4+2\log |x|-3\sin x+C$$

Differentiating the answer gives back $4x^3+\frac{2}{x}-3\cos x$.

An everyday example. A multiplication table read backwards gives division facts — likewise, a table of derivatives read backwards gives integral formulae.

The substance. **The power rule excludes $n=-1$**, which is why $\int \frac{1}{x}dx$ gives $\log |x|$ instead of a power.

**When an integrand contains a function together with its derivative, replace that function by $t$, replace $dx$ using $dt=g^{\prime }(x)dx$, integrate in $t$, and substitute back.

The idea.** If $t=g(x)$, then

$$\int f(g(x))g^{\prime }(x)dx=\int f(t)dt$$

Worked example 1. $\int 2x\cos (x^2)dx$. Put $t=x^2$, so $dt=2xdx$:

$$\int \cos tdt=\sin t+C=\sin (x^2)+C$$

Worked example 2. $\int \frac{2x+3}{x^2+3x+5}dx$. Put $t=x^2+3x+5$, so $dt=(2x+3)dx$:

$$\int \frac{dt}{t}=\log |x^2+3x+5|+C$$

Worked example 3 — a standard result. For $\int \tan xdx=\int \frac{\sin x}{\cos x}dx$, put $t=\cos x$, so $dt=-\sin xdx$:

$$-\int \frac{dt}{t}=-\log |\cos x|+C=\log |\sec x|+C$$

An everyday example. Converting a recipe from cups to grams, scaling it in grams, and converting back is the same move as changing the variable, integrating, and substituting back.

The substance. **The substitution must replace every $x$, including $dx$** — a leftover $x$ means a different substitution is needed.

Rewrite powers and products of trigonometric functions with identities, such as double-angle or product-to-sum formulae, until every term matches a standard integral.

Useful identities:

- ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ and ${\cos }^{2}x=\frac{1+\cos 2x}{2}$
- $2\sin A\cos B=\sin (A+B)+\sin (A-B)$
- ${\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}$

Worked example 1.

$$\int {\cos }^{2}xdx=\int \frac{1+\cos 2x}{2}dx=\frac{x}{2}+\frac{\sin 2x}{4}+C$$

Worked example 2.

$$\int \sin 3x\cos 2xdx=\frac{1}{2}\int (\sin 5x+\sin x)dx=-\frac{\cos 5x}{10}-\frac{\cos x}{2}+C$$

Worked example 3.

$$\int {\sin }^{3}xdx=\int \frac{3\sin x-\sin 3x}{4}dx=-\frac{3}{4}\cos x+\frac{1}{12}\cos 3x+C$$

An everyday example. Untangling a knotted charger cable before plugging it in saves effort later — simplifying with identities first makes the integral straightforward.

The substance. Two correct answers can look different — trigonometric antiderivatives may differ only by a constant absorbed into $C$.