Find Pi using integral

Question

I am just started learning calculus and wonder why:

$$\int_0^1 \frac{4}{1+x^2}$$ Allows to find $\pi$?

It would be great if someone could provide very detailed explanation.

Answer 1

Hint: $$\int_{0}^{1} \frac{1}{x^2+1}dx=[\text{arctan}(x)]^1_{0}$$

Answer 2

As Nicky Hekster shows, we have

$$\int_0^1\frac1{1+x^2}\ dx=4\arctan(x)\bigg]_0^1=\pi$$

Then, as you may recall, we have the geometric series:

$$\frac1{1-r}=1+r+r^2+r^3+\dots$$

By setting $r=-x^2$, we get

$$\frac1{1+x^2}=1-x^2+x^4-x^6+\dots$$

Integrating and you'll get

$$\int_0^1\frac1{1+x^2}\ dx=\int_0^11-x^2+x^4-x^6+\dots\ dx=x-\frac13x^3+\frac15x^5-\frac17x^7+\dots\bigg]_0^1$$

That is, we end up with the famous Leibniz formula for $\pi$:

$$\frac\pi4=1-\frac13+\frac15-\frac17+\dots$$

Which is just too beautiful! Is it not? From here, using different special angles, we can arrive at various different formulas for $\pi$ that converge much faster than the one above. Alternatively, you could also apply an Euler sum to the above series expansion as I have described in this answer:

How to prove this $\pi$ formula?