Weierstrass Substitution I was reading up about the Weierstrass Substitution and don't understand what 'No generality is lost' means in this context. 
In integral calculus, the tangent half-angle substitution is a substitution used for finding antiderivatives, and hence definite integrals, of rational functions of trigonometric functions. No generality is lost by taking these to be rational functions of the sine and cosine
Also, while I'm here, I might as-well ask an additional question! When is it the correct circumstances to implement this technique? 
 A: The Weierstrauss substitution is useful when you have a function you want to integrate that's a rational function in terms of trig functions.  The without loss of generality means that if your function involves things like $\tan x,\csc x$, etc, then you can just write $\tan x = \frac{\sin x}{\cos x}$ and $\csc x = \frac{1}{\sin x}$, etc, to get it all in terms of $\sin$ and $\cos$.
Now, the Weierstrauss substitution doesn't give us the answer directly.  Imagine we have some function $f$ (you can choose one from here if you can't think of any) such that:
$$f(x) = \frac{N(\sin x,\cos x)}{D(\sin x,\cos x)}$$
By this, I mean that both the numerator ($N$) and denominator $(D)$ are functions of $\sin x,\cos x$ - specifically polynomial in terms of them.
So, if we have an $f$ like that, we can apply the Weierstrauss substution, and it will rewrite $f$ as:
$$f(x) = \frac{N_1(x)}{D_1(x)}$$
Now, these are new polynomials, but they're still polynomials, and instead of being polynomial of trig functions (like $\sin^8x\cos 5 x+3\cos x-6\sin^2 x$ or some ugly stuff), it's just a polynomial in terms of $x$.  Combined, $f$ is a rational function in terms of $x$, which is great for integration, because of the partial fraction decomposition.  This says that if:
$$f(x) = \frac{N(x)}{D(x)}$$ is a rational function, then we can factor the denominator into just its roots (so $(x-r)$ for a root $r$), and irreducible quadratics (so $(x^2+ax+b)$ where it has no zeros in $\mathbb R$).  There's some technical details involving repeated roots, but essentially we can write:
$$f(x) = \frac{A_1}{x-r_1}+\dots +\frac{A_n}{x-r_n}+\frac{B_1 x+C_1}{(x^2+d_1x+e_1)}+\dots +\frac{B_kx+C_k}{(x^2+d_kx+e_k)}$$
This is just saying we can "break up the denominator" completely.  This is great, because we can now integrate termwise, using either completing the square and $\arctan$ (for the irreducible quadratics), or just $\ln$ (for the roots).
So, the Weierstrauss substution converts integrals of rational functions of $\sin,\cos$ into just integrals of rational functions, and we can do integrals of rational functions REALLY well.
