Oscillating integral converges to zero? An integral like, say, 
$$\int_0^1 \cos[ nf(x)]~dx$$
with some function $f$ which is well behaved, and maybe almost everywhere non-zero, should be very small for large $n$ since the positive and negative contributions should about cancel each other whenever the oscillation frequency is high enough. 
Is there a way to formalise this?
For $f(x)=x$, one could compute $$\int_0^1 \cos nx ~dx=\int_0^1 \frac{d}{dx}\left(\frac{\sin nx}{n}\right) dx=\frac{\sin n}{n}$$ but how could one argue when there is no expression for the antiderivative available? For certain cases a substitution $y=f(x)$ and some integration by parts might help, but it seems that the restrictions on $f$ imposed by that are stronger than necessary.
 A: If we assume that $f:(0,1)\to\mathbb{R}$ has a continuous and positive derivative, its inverse function $g(x)$ is an increasing $C^1$ function and:
$$ c_n = \int_{0}^{1}\cos(nf(x))\,dx = \int_{f(0)}^{f(1)} g'(y) \cos(ny)\,dx $$
converges to zero as $n\to +\infty$ by the Riemann-Lebesgue lemma (series version), since $g'(y)$, as a continuous (or just positive) derivative, is obviously an integrable function over $(f(0),f(1))$, and its integral equals one. If we know something more about the regularity of $g(x)$ ($g\in C^2,C^3$, $C^\infty$ or $C^\omega$) and the size of $f(1)-f(0)$ we may also estimate how fast $c_n$ converges to zero by using integration by parts. 
I fear we cannot drop the assumptions on the monotonicity of $f(x)$, since the following nightmare holds: there are some differentiable functions on $(a,b)$ whose derivative is not Riemann-integrable. 
However, the above argument should be enough for concrete applications.
For instance, $\int_{0}^{1}\cos(ne^x)\,dx$ behaves like $\frac{\pi}{2n}$ (despite some odd-looking oscillations) while $\int_{0}^{1}\cos(n\log x)\,dx $ is very regular, it is exactly equal to $\frac{1}{1+n^2}$. On the other hand, $\int_{0}^{1}\cos(n\sin(\pi x))\,dx = J_0(n)$ decays just like $\frac{1}{\sqrt{n}}$.
