I will appreciate any comment or review to the proof below. Thank you.
Let $f\in C^1([a,\infty),\Bbb R)$ for some $a\in\Bbb R$ with $f'$ strictly increasing and with $\lim_{x\to\infty}f'(x)=\infty$. Prove that $\int_a^\infty\sin(f(x))\mathrm dx$ converges.(Hint: use the substitution $y=f(x)$ and the inverse function theorem).
What I did
With the substitution $y=f(x)$ we have
$$\int_a^\infty \sin(f(x))\mathrm dx=\int_{f(a)}^{f(\infty)}\frac{\sin y}{f'(x)}\mathrm dy$$
If $f'$ is strictly increasing then at most it have a zero at some point $x=c$, what implies that $f$ is strictly convex and, at most, have a minimum in $x=c$.
Without lose of generality suppose that $f'$ have a zero in $x=c$.
Then $f$ is invertible in the regions $[a,c]$ and $[c,\infty)$ and by the inverse function theorem
$$(f^{-1})'(y)=\frac1{f'(x)}\tag{1}$$
Then
$$\int_a^\infty \sin(f(x))\mathrm dx=\int_{a}^{c}\sin (f(x))\mathrm dx+\int_{f(c)}^{f(\infty)}(f^{-1})'(y)\sin y\,\mathrm dy$$
Where clearly
$$\left|\int_{a}^{c}\sin (f(x))\mathrm dx\right|<\infty$$
because $\sin (f)$ is continuous in $[a,c]$.
Now observe that for $x>c$ $f$ is strictly increasing and positive by it convexity and the positivity of $f'$ in this region. Hence $f(\infty)=\infty$, $f^{-1}$ is strictly concave, strictly increasing and positive, what imply that $(f^{-1})'$ is strictly decreasing and positive.
And by (1) we can see that $(f^{-1})'(y)\to 0$ as $y\to\infty$. Then, putting all together, we can write
$$\int_{f(c)}^\infty (f^{-1})'(y)\sin y\,\mathrm dy=\int_{f(c)}^{M\pi/2}(f^{-1})'(y)\sin y\,\mathrm dy+\sum_{k=0}^\infty a_k$$
for some $M\in\Bbb N$ such that $M\pi/2>f(c)$, where
$$a_k:=\int\limits_{(M+k)\pi/2}^{(M+k+1)\pi/2}(f^{-1})'(y)\sin y\,\mathrm dy$$
Then $(a_k)$ is an alternating sequence where $(|a_k|)\to 0$ so the series converges and clearly
$$\left|\int_{f(c)}^{M\pi/2}(f^{-1})'(y)\sin y\,\mathrm dy\right|<\infty$$
because $(f^{-1})'$ is bounded. Hence the integral $\int_a^\infty \sin (f(x))\mathrm dx$ converges.$\Box$