A function $F(x)$ is called an antiderivative of a function $f(x)$ if it is a solution of the differential equation $\frac{d}{dx}F(x)=f(x)$. For instance, $F(x)=\frac{1}{2}x^2$ is an antiderivative of $f(x)=x$. There can be more than one (actually infinitely many) antiderivative of a function $f(x)$ but they all differ by a constant. In other words, if $F(x)$ and $G(x)$ are antiderivatives of $f(x)$, then $G(x)=F(x)+C$, where $C$ is a constant. In fact, one can easily check that for any constant $C$, $\frac{1}{2}x^2+C$ is an antiderivative of $x$ since the derivative of any constant is zero. On the other hand, if $F(x)$ and $G(x)$ are both antiderivatives of $f(x)$, they differ only by a constant. This can be easily seen: Since $F(x)$ and $G(x)$ are both antiderivatives of $f(x)$,
$$\frac{d}{dx}F(x)=\frac{d}{dx}G(x)=f(x),$$
ans so
$$\frac{d}{dx}(F(x)-G(x))=0.$$
Hence, $F(x)=G(x)+C$, where $C$ is a constant.

If $F(x)$ is an antiderivative of a function $f(x)$,
$$F(x)+C,$$
where $C$ is an arbitrary constant is called the indefinite integral of $f(x)$ and is denoted by $\int f(x)dx$, i.e.
$$\int f(x)dx=F(x)+C.$$
Example. Find the indefinite integral of each of the following functions.

1. $f(x)=\sin x$.

Solution. Let $F(x)$ be an antiderivative of $f(x)$. Then $F'(x)=f(x)$. If we let $y=F(x)$, then
$$\frac{dy}{dx}=\sin x.$$
We know one such function $y$ which satisfies the equaion $\frac{dy}{dx}=\sin x$. It is $y=-\cos x$. So, the indefinite integral of $\sin x$ is
$$\int \sin xdx=-\cos x+C,$$
where $C$ is an arbitrary constant.

2. $f(x)=\frac{1}{x}$.

Solution. An antidetivative of $\frac{1}{x}$ is $\ln x$. Note that $f(x)=\frac{1}{x}$ is defined for $x<0$ while $\ln x$ is not, so in fact the antiderivative should be written as $\ln|x|$ instead of $\ln x$. Hence, the indefinte integral is
$$\int\frac{1}{x}dx=\ln |x|+C,$$
where $C$ is an arbitrary constant.

3. $f(x)=x^n$, where $n\ne -1$.

Solution. $F(x)=\frac{1}{n+1}x^{n+1}$ is an antiderivative of $f(x)=x^n$. Hence, the indefinite integral of $x^n$ is
$$\int x^n dx=\frac{1}{n+1}x^{n+1}+C,$$
where $C$ is an arbitrary constant.

The most general solution of the differential equation
$$\frac{dy}{dx}=f(x)$$
is the indefinite integral
$$y=\int f(x)dx=F(x)+C,$$
where $C$ is an arbitrary constant. With an additional condition, the arbitrary constant $C$ may be determined. Such a condition is called an initial condition.

Example. Solve the differential equation $\frac{dy}{dx}=\sin x$ with the condition $y(0)=1$.

Solution. In the previous example, we find that the general solution is given by
$$y(x)=-\cos x+C,$$
where $C$ is an arbitrary constant. The condition $y(0)=1$ implies that $-\cos 0+C=1$ i.e. $-1+C=1$. So, we obtain $C=2$. Therefore, the solution we seek is
$$y(x)=-\cos x+2.$$

Indefinite integrals satisfy the following properties:
\begin{align*}
\int (f(x)+g(x))dx&=\int f(x)dx+\int g(x)dx,\\
\int c f(x)dx&=c\int f(x)dx,
\end{align*}
where $c$ is a constant. For this we say indefinite integrals are linear. If you have studied linear algebra, you know that a linear map is a map from a vector space to another vector space which preserves vector space operations (vector addition and scalar multiplication). It turns out that functions may be treated as vectors and the indefinite integral $\int f(x)dx$ may be considered as a linear map. This sort of abstract treatment is important in advanced mathematics, physics and engineering. The linearity of indefinite integrals can be used to find the indefinite integral of a complicated function. For example,

Example. Find the indefinite integral of
$$f(x)=1-x^3+12x^5.$$

Solution.
\begin{align*}
\int (1-x^3+12x^5)dx&=\int dx-\int x^3dx+12\int x^5dx\\
&=x-\frac{x^4}{4}+\frac{12}{6}x^6+C\\
&=x-\frac{x^4}{4}+2x^6+C,
\end{align*}
where $C$ is an arbitrary constant.

Some Important Formulas

\begin{align}
\int x^ndx&=\frac{x^n}{n+1}+C\ (n\ne -1\ \mbox{is a rational number})\\
\int \sin kxdx&=-\frac{\cos kx}{k}+C\ (k\ne 0\ \mbox{is a constant})\\
\int \cos kx dx&=\frac{\sin kx}{k}+C\ (k\ne 0\ \mbox{is a constant})\\
\int \sec^2 xdx&=\tan x+C\\
\int \csc^2 xdx&=-\cot x+C\\
\int \sec x\tan xdx&=\sec x+C\\
\int \csc x\cot xdx&=-\csc x+C
\end{align}

Example. [Initial Value Problem] A balloon is ascending at the constant speed of 12 ft/sec is at a height 80 ft above the ground when a package is dropped. How long does it take for the package to reach the ground?

Solution. In order to answer the question we need to have $h(t)$, the motion i.e. the position (height) function of the falling package. We don’t have it yet. If we know the velocity $v(t)$ of the falling package, we would be abel to find $h(t)$ by solving the differential equation
\begin{equation}
\label{eq:height}
\frac{dh(t)}{dt}=v(t).
\end{equation}
However, we don’t have it either. Instead, what is known is the acceleration $a(t)$ of the freely falling package which is constant
$$a(t)=-9.8\mathrm{m/sec^2}=-32\mathrm{ft/sec^2}.$$
The velocity $v(t)$ can be found by solving the differential equation
\begin{equation}
\label{eq:velocity}
\frac{dv(t)}{dt}=a(t)
\end{equation}
with $a(t)=-32\mathrm{ft/sec^2}$. The solution of \eqref{eq:velocity} is the indefinite integral
\begin{align*}
v(t)&=\int -32dt\\
&=-32t+C_1,
\end{align*}
where $C_1$ is a constant. At the time the package was dropped from the balloon, the balloon was ascending at the rate $12\mathrm{ft/sec}$. So the package’s initial velocity is $v(0)=12\mathrm{ft/sec}$. Using this we find $C_1=12$ and so
$$v(t)=-32t+12.$$
Now, we are ready to find $h(t)$. The solution of the differential equation \eqref{eq:height} with $v(t)=-32t+12$ is the indefinite integral
\begin{align*}
h(t)&=\int (-32t+12)dt\\
&=-16t^2+12t+C_2,
\end{align*}
where $C_2$ is a constant. At the time the package was dropped from the balloon the height was 80ft, i.e. $h(0)=80\mathrm{ft}$. Using this we find $C_2=80$. Hence, we have
$$h(t)=-16t^2+12t+80.$$
Setting $h(t)=0$, we obtain the quadratic equation
$$-16t^2+12t+80=0$$
whose solutions are $t_1\approx -1.89$ or $t_2\approx 2.64$. Therefore, it takes 2.64 seconds for the package to reach the ground.