There is no easy way for definition of $\pi$ .
One of the most simplest definition of $\pi$ is using the distance.
You must know $\pi$ is equivalent to the length of a semicircular circumference.
That is, $\pi$ is double the length of the circumference of the quadrant.
The graph $C$ is drawn by the function $y=f(x)$ on the interval $x\in[a,b]$ .
$f(x)$ is differentiable and $f'(x)$ is continuous on $[a,b]$ .
Then, the length $l$ of $C$ is
\[ l=\int_a^b \sqrt{1+f'(x)^2}dx . \]
The function of a circular whose center is origin and radius is 1 is
\[ x^2+y^2=1 . \]
Thus, the length of quadrant circumference is
\[ l=\int_0^1\frac{1}{\sqrt{1-x^2}}dx \]
because
\[ y=\sqrt{1-x^2} , \]
\[ y'=-\frac{x}{\sqrt{1-x^2}} , \]
\[ (y')^2=\frac{x^2}{1-x^2} , \]
therefore,
\[ l=\int_0^1\sqrt{1+\frac{x^2}{1-x^2}} dx . \]
We will define
\[ 2l= 2\int_0^1\frac{dx}{\sqrt{1-x^2}}=\pi . \]
