Indefinite integral of $\sin(x)$ without using the derivative of $\cos(x)$ I can prove that 
$$\int\sin(x)dx=-\cos(x)+C$$
by using $\cos'(x)=-\sin(x)$ and $\sin'(x)=\cos(x)$. Are there other proofs not involving this (at least, not explicitly) ?
 A: It's important to notice that the definition of indefinite integral. Let $D \subset R$ and $f:D \mapsto R$ be a function. Then the indefinite integral of $f$ is defined as a function $F: D\mapsto R$ such that $F$ is differentiable on $D$ and $F'=f$ 
So there's no way to prove the indefinite integral without using its definition. But of course, it's possible to write the function $cos(x)$ in different ways i.e. cos$x$=$\frac{(e^{ix}+e^{-ix})}{2}$ or other ways and show the expressions are equal to $cos(x)$
A: Just use the taylor series and integrate term by term, you recognise the new Taylor series as $-\cos x$ 
A: Absolutely. You can integrate the exponential form
$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i} ,$$
and then return that result back into your desired integral.
Note that
$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}.$$ 
A: As pointed out in the comment, using  Weierstrass substitution,
$\displaystyle \tan\frac x2=t\implies \sin x=\frac{2t}{1+t^2}$ and $\displaystyle x=2\arctan t\implies dx=\frac2{1+t^2}dt$
$$\int \sin xdx=\int \frac{2t}{(1+t^2)^2}dt=\int \frac{du}{(1+u)^2}\text{ (putting }t^2=u)$$
$$=-\frac1{1+u}=-\frac1{1+t^2}=-2\cos^2\frac x2+C=C-1-\cos x$$
A: Let $x=\sec^{-1}u$ then we have
$$\sin(\sec^{-1}u)=\frac{\sqrt{u^2-1}}{|u|},dx=\frac{du}{|u|\sqrt{u^2-1}}$$
therefore
$$\int\sin x\,dx=\int\frac{du}{u^2}=\frac{-1}{u}=-\cos x.$$
Note that
$$\sec'x=\lim_{h\to0}\frac{\sec(x+h)-\sec x}{h}=\lim_{h\to0}\frac{\sin\frac{h}{2}\sin\frac{2x+h}{2}}{\frac{h}{2}\cos x\cos(x+h)}=\tan x\sec x.$$
