How does integrating trigonometric functions work? I tried searching Google but all I get is what integrating each trigonometric function equals to.
Nothing tells me why.
I can sort of see why:
${d\over dx} \sin(x) = \cos(x)$
And...
${d\over dx} \cos(x) = -\sin(x)$
Purely by looking at the two graphs. But I want to know the maths behind it. How does integrating both functions work?
Why the results?
I've also seen from Wolfram Alpha that trigonometric functions can be written as series, but is there any article I can read that explains how these series are derived?
I can't seem to find any.
UPDATE: Wow. Thank you everyone for sharing your knowledge. You have no idea how much you've helped :).
 A: $$
\begin{split}
\frac{d [\sin x]}{dx}
 &= \lim_{h \to 0^+} \frac{\sin(x+h) - \sin x}{h} \\
 &= \lim_{h \to 0^+} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \\
 &= \lim_{h \to 0^+} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h} \\
 &= \sin x \left(\lim_{h \to 0^+} \frac{\cos h - 1}{h} \right)
  + \cos x \left( \lim_{h \to 0^+} \frac{\sin h}{h} \right)\\
 &= \sin x \cdot 0 + \cos x \cdot 1 \\
 &= \cos x.
\end{split}
$$
UPDATE
To finish this off you need a proof that
$$\lim_{h \to 0^+} \frac{\cos h - 1}{h} = 0 \quad \text{and} \quad
\lim_{h \to 0^+} \frac{\sin h}{h} = 1.$$
Here is a creative way to show the first limit assuming the second limit. Note that $$\lim_{h \to 0^+} \frac{\cos h -1}{h} \times \frac{\cos h +1}{h} = \lim_{h \to 0^+} \frac{\cos^2 h -1}{h} = \lim_{h \to 0^+} \frac{-\sin^2 h}{h^2} = -1.$$ Since $$\lim_{h \to 0^+} \frac{\cos h +1}{h} = +\infty$$ we must conclude that $$\lim_{h \to 0^+} \frac{\cos h -1}{h} = 0.$$
Now the only thing left is to show the second limit, which is a geometric proof to establish the inequality $$\cos h \le \frac{\sin h}{h} \le 1,$$ and then using the Squeezing Theorem. You can see the details, for example, here.
A: I'll assume you're aware that $\frac d{dx} e^{bx} = be^{bx}$ where $b \in \Bbb R$.  It turns out that this also holds when $b$ is a complex number.  Using this and the definition of the trig functions in terms of the exponential function it's easy to see that
$$\begin{align}\frac d{dx}\cos(x) &= \frac d{dx}\left(\frac{e^{ix}+e^{-ix}}{2}\right) \\ &= \frac 12\left(\frac d{dx}e^{ix}+\frac d{dx}e^{-ix}\right) \\ &= \frac 12\left(ie^{ix}-ie^{-ix}\right) \\ &= \frac 12\left(\frac iiie^{ix}-\frac iiie^{-ix}\right) \\ &= -\frac{e^{ix}-e^{-ix}}{2i}\ \\ &= -\sin(x)\end{align}$$
A: The most important
limit formula for trig functions is
$\lim_{h \to 0} \frac{\sin h}{h}
= 1$.
You can convince yourself
of this by looking
at a triangle with angle $h$
in a circle of radius $1$.
The best way to show this
analytically is probably
to use the power series,
but the diagram of
the small angle in the circle
is where it all comes from.
Once you have this,
the formulas for
the derivatives of
the trig functions
readily follow.
Another result is that
$\lim_{h \to 0} \frac{\cos h - 1}{h}
= 0
$.
But this follows from 
the $\sin$ limit combined with
$\cos h - 1
=-2\sin^2(h/2)
$
so
$\lim_{h \to 0} \frac{\cos h - 1}{h}
=\lim_{h \to 0} \frac{-2\sin^2(h/2)}{h}
=\lim_{h \to 0} \sin(h/2)\frac{-\sin(h/2)}{h/2}
= 0
$
since the right term
goes to $-1$
and the left term
goes to zero.
