Here is a possible beginning. Note first that
$$ \int_{-1}^1 f(t)^a \cos(at) dt= 2I_a:= 2\int_{0}^1 f(t)^a \cos(at) dt. $$
Hence, it suffices to investigate $I_a(t)$. I_a$. Let me first assume that $f'(t) <0$ on $(0,1)$. (Note that if $f'(t_0)=0$ for some $t_0\in (0,1)$ then $f'(t)=0$ on $[0,t_0]$.) This means that the map $t\mapsto f$ is one-to-one. We regard $t$ as a function of $f$. Then the change in variables formula implies.
$$I_a= \int_1^{f(0)} f^a \cos(a t)\frac{dt}{df} df $$
I can make this formula friendlier to the 21st century mathematician by changing notations,
$$t \longleftrightarrow \phi,\;\;\; f \longleftrightarrow x$$
and we can rewrite the above as
$$ I_a= \int_1^{x_0} x^a\frac{d\phi}{dx} \cos( a \phi(x) ) dx = \frac{1}{a}\int_1^{x_0} x^a \frac{d}{dx}\Bigl( \sin\bigl(\; a\phi(x)\;\bigr) \Bigr)dx$$
$$=\frac{1}{a}\Bigl( x^a\sin\bigl( a\phi(x)\bigr)\;\Bigr)\Bigr|^{x_0}_1- \int_1^{x_0}x^{a-1}\sin\bigl(\; a\phi(x) \;\bigr)dx. $$
Now observe that $\phi(x_0) =0$, $\phi(1)=1$, so the first term above goes to zero as $a\to\infty$.
At this point it may be useful to look in some books on asymptotics of integrals. A good place to start is
Bleistein & Handelsman: Asymptotic expansions of Integrals, Dover
Also you need to keep in mind that
$$\frac{d\phi}{dx}< 0,\;\;\forall x\in (1,x_0)$$
$$\lim_{x\nearrow x_0} \frac{d\phi}{dx}=-\infty. $$
