How to change speed of audio samples without changing pitch?
This question (about "time-scaling" audio) is closely related to pitch shifting, which is time-scaling combined with resampling.  But changing the speed without changing pitch is only time-scaling, so there is no resampling involved (contrary to what thomas has suggested).

There are frequency-domain methods (phase-vocoder and sinusoidal modeling) that can change speed for an orchestral mix (or some other broadbanded non-periodic sound) without glitches, but they can become a little "phasey" in their stretched sound.

If it's monophonic, like a solo or a single note or string or voice, a less expensive time-domain approach is sufficient and can sound very good, under this monophonic condition.  If it's a single note or tone, the signal is quasiperiodic.  That means, in the vicinity of some time, $t_0$, that

$$ x(t) \approx x(t - \tau(t_0)) \qquad \text{for } t \approx t_0 $$

So $\tau(t_0) &gt; 0$ is the estimated period of the quasiperiodic waveform in the neighborhood of $t_0$.  This estimated period $\tau(t_0)$ is estimated by finding the value of $\tau$ that minimizes something like

$$ Q_x(\tau, t_0) = \int \Big|x(t) - x(t-\tau)\Big|^2 w(t-t_0) \, dt $$

where the window $w(t-t_0) \ge 0$ is centered around the time $t_0$.  If there are multiple candidates of $\tau$ that minimize $Q_x(\tau, t_0)$, usually it is best to pick the smallest $\tau$ in which $Q_x(\tau, t_0)$ is small.

Now imagine copying your note $x(t)$ and offsetting (delaying) the copy in time by the amount $\tau(t_0)$.  That delayed copy would be $x(t - \tau(t_0))$.  Then you have two identical waveforms, except for the offset of $\tau(t_0)$ but around the time $t_0$ the two waveforms will look almost the same because the offset is exactly one period, based on the estimate of the period around time $t_0$.

Then you can cross-fade (using a Smoothstep function $S(t)$) from the original to the offset copy, the cross-fade will not have any nasty cancellation or "destructive interference" because the waveforms are lined up.  And the result will be the same note, but one period longer.  (Longer by $\tau(t_0)$ seconds.)

$$ y(t) = \Big(1-S\big(\tfrac{t-t_0}{\tau(t_o)}\big) \Big) \, x(t) \ + \ S\big(\tfrac{t-t_0}{\tau(t_o)}\big) \, x(t - \tau(t_0)) $$

(That won't stretch it long enough to notice a difference, but if you repeat this operation many times per second, you can stretch a patch of sound by even a factor of two.)