How to integrate $1/(1 + x^2)$ on a desert island? Classically, one would integrate $1/(1 + x^2)$ using trigonometric substitution rules, thus obtaining:
$$ \int \frac{1}{1 + x^2} \; \textrm{d}x = \arctan(x) + C $$
Sadly, you're stuck on a desert island, and must determine an algebraic expression, or a series of algebraic expressions of the integral of $1/(1 + x^2)$ to survive. However, you never took the time to remember the more complicated trigonometric identities. All you know about the trigonometric functions is what you can see by drawing out a circle, and comparing the graphs of $\sin(x)$ and $\cos(x)$:
$$ \frac{\sin(x)}{\cos(x)} = \tan(x)$$
$$ \sin(x)^2 + \cos(x)^2 = 1 $$
$$ \sin(2n\pi) = 0, \;\cos(2n\pi) = 1, \;\sin(n\pi/2) = 1, \;\cos(n\pi/2) = 0 $$
$$ \frac{\textrm{d}}{\textrm{d}x}\sin(x) = \cos(x) $$
$$ \frac{\textrm{d}}{\textrm{d}x}\cos(x) = -\sin(x) $$
Essentially, you must derive the integral of of $1/(1 + x^2)$ from first principles. What should you do? 
(To answer the question, you can make assumptions about stuff you might know which is not excluded by the above. e.g. you could claim that you know about Taylor series approximations of functions, and thus go about finding a solution in that way. The fewer assumptions you can make, the better your answer.) 
 A: The triangle with vertices $(0,0)$, $(1,x)$, and  $(1, x + \Delta x)$ has area $\tfrac12 \Delta x$. Shrinking it by a factor $1/\sqrt{1+x^2}$ from the origin yields a triangle with one vertex on the unit circle and area $\tfrac12 (1+x^2)^{-1} \Delta x$. Therefore the area of the unit disc cut out by the triangle with vertices $(0,0), (1,0), (1,x)$ is $$\frac12 \int_0^x\frac{\mathrm{d}t}{1+t^2}$$ and this equals half the angle at the origin since the unit disc has total area $\pi$.
A: By defining $\tan(x)=\frac{\sin(x)}{\cos(x)}$ you may notice that this function fulfills the differential equation $f'(x)=1+f^2(x)$. 
It follows that the computation of $\int_{a}^{b}\frac{dx}{1+x^2}$ is horribly simplified by the substitution $x=\tan(t)$, and that's it.
A: We only assume the Pythagorean Theorem, namely $$\sin^2 x+\cos^2x=1\tag{1},$$
as well as integration via substitution and differential calculus. Via $(1)$ we may prove that $$\tan^2x+1=\sec^2x\tag{2}$$
so it becomes evident that $$\frac{\frac{d}{dx}\tan x}{1+\tan^2x}=1.\tag{3}$$
Thus, using $t\mapsto\tan x$ and $(3)$,
$$\int\frac{dt}{1+t^2}=\int \frac{\frac{d}{dx}\tan x}{1+\tan^2x}dx=\int dx=x.$$
Since $t=\tan x$, $x=\arctan t$, and we are done.
A: On this desert island, is it conceivable that, even though you have admittedly forgotten all about trigonometric identities, you still vaguely remember that there are such things as complex numbers, and contour integrals?
Then (in a wild moment, half-crazed by your isolation), if faced with the unfamiliar integral $\int_0^t\frac{du}{1+u^2}$, where $t \geqslant 0$, you might think of writing it as:
$$
\int_0^t\frac{du}{1+u^2} = \frac{1}{2}\int_0^t\left(\frac{1}{1+iu} + \frac{1}{1-iu}\right)du = \frac{1}{2i}\int_\gamma\frac{dz}{z},
$$
where $\gamma$ is the contour given by $\gamma(u) = 1/(1 + iu)$ for $-t \leqslant u \leqslant t$, which traverses the straight line segment between $1-it$ and $1+it$.
Then you might remember something about complex logarithms, a little of your lost knowledge of trigonometry might come flooding back - and all of a sudden, you might see that the value of the integral is $\arctan(t)$.
(The camera zooms back, and we see a ragged and unshaven figure capering over the sand, shouting "Eureka!")
A: Here is what you can do if your arsenal of functions is limited. First of all show that no rational function can have its derivative equal to $1/(1+x^2)$. It is slightly harder to show that no algebraic function can have its derivative equal to $1/(1+x^2)$.
Thus the anti-derivative must be a new kind of transcendental function and that's a great thing because we have found something new and exciting on the desert island. Define a new function $$f(x) =\int_{0}^{x}\frac{dt}{1+t^2}\tag{1}$$ and one can immediately see that $f$ is odd, continuous, differentiable and strictly increasing on $\mathbb {R} $. Next we can note that $$\int_{1}^{x}\frac{dt}{1+t^2}=\int_{1/x}^{1}\frac{dz}{1+z^2}$$ via substitution $t=1/z$ for $x>1$. Letting $x\to\infty $ we can see that $$\int_{1}^{\infty}\frac{dt}{1+t^2}$$ is convergent and equals $\int_{0}^{1}(1+t^2)^{-1}\,dt$.
In other words the integral $$\int_{0}^{\infty} \frac{dt} {1+t^2}$$ exists so that $f$ is bounded with $$\lim_{x\to\pm\infty} f(x) =\pm 2\int_{0}^{1}\frac{dt}{1+t^2}$$ Let $$\alpha=2\int_{0}^{1}\frac{dt}{1+t^2}=\int_{0}^{\infty}\frac{dt}{1+t^2}\tag{2}$$ and then we can show that $\alpha$ is the least positive value for which $\cos \alpha=0$.
To do so we put $t=x/\sqrt{1-x^2}$ in the integral defining $\alpha$ and get $$\alpha=\int_{0}^{1} \frac{dx} {\sqrt{1-x^2}}\tag{3}$$ and using integration by parts one can show that the above integral equals $$2\int_{0}^{1}\sqrt{1-x^2}\,dx\tag{4}$$ Using integral $(3)$ and $(4)$ it can be easily seen that circumference of a circle of unit radius is $4\alpha$ and area of a circle of unit radius is $2\alpha$.
And thus we can identify $\pi$ with $2\alpha $ depending upon your chosen definition of circular functions and one immediately gets $\sin\alpha=1,\cos\alpha=0$.
The next step is obviously to invert $f$ and study the inverse function on $(-\pi/2,\pi/2) $ and it will be seen that $f^{-1}(x)=\tan x$. It is also instructive exercise to prove the functional equation $$f(x) +f(y) =f\left(\frac{x+y} {1-xy}\right)\tag{5}$$ for $|x|<1,|y|<1$.
Using these ingredients you can now develop a full theory of circular functions of a real variable based on definition $(1)$.
