First substitute $u= x^{\alpha}$ into the integral to see your problem reduces to asking the same question about the integral $\displaystyle \int^{\infty}_0 \dfrac{\sin x}{x^a} dx$ for $a\in (0,1).$
To show that is Riemann integrable, we consider the integral over $(0,1)$ and $[1,\infty)$ separately. To show the integral over $[1,\infty)$ exists, integrate by parts. You'll have
$$ \int^R_1 \frac{\sin x}{x^a} dx = -x^{-a} \cos x \mid^R_1 - a \int^R_1 \frac{\cos x}{x^{a+1}} dx$$
and it should be easy to finish from there.
To see $\displaystyle \int^1_0 \frac{\sin x}{x^a} dx$ exists note that $\sin x \sim x$ and $\displaystyle \int^1_0 \frac{x}{x^a} dx$ is finite.
To show it is not Lebesgue integrable, it suffices to show $$\int^{\infty}_0 \frac{|\sin x|}{x^a} dx = \sum_{k=0}^{\infty} (-1)^k \int^{(k+1)\pi}_{k\pi} \frac{\sin x}{x^a} dx = \sum_{k=0} \int^{\pi}_0 \frac{\sin x}{(x+k\pi)^a} dx$$ diverges, which is not hard with a basic estimate on the last integral. Note that the integral over $[0,\pi]$ is certainly greater than the integral over $[\pi/4,3\pi/4]$ and on that interval we have $\sin x \geq \frac{1}{\sqrt{2}}$ so
$$\int^{\pi}_0 \frac{\sin x}{(x+k\pi)^a} dx > \frac{1}{\sqrt{2}} \int^{3\pi/4}_{\pi/4} \frac{1}{( x+k\pi)^a} dx > \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2} \frac{1}{( 3\pi/4 + k\pi)^a}= \frac{1}{2\sqrt{2}\pi^{a-1}}\frac{1}{(k+3/4)^a} .$$
Your proof is valid for a bounded function defined on the closed, bounded interval $[a,b]$, despite the apparent simplicity. It also relies on the fact that the Riemann and Lebesgue integrals are the same for step functions, as mentioned by Tony Piccolo.
The other proof you mention is also valid but takes you on a more roundabout path because it strings together a number of results, each of which is not altogether trivial to prove.
A bounded, measurable function defined on a set of finite
measure is Lebesgue integrable.
If a sequence of measurable functions converges almost everywhere
to $f$, then the limit function $f$ is measurable.
If $f$ is Riemann integrable on $[a,b]$, then there exists a
sequence of simple (measurable) functions converging almost everywhere
to $f$.
Adding even more complexity, the proof of the third statement that I know also uses the fact that the set of discontinuities of a Riemann integrable function must be of measure zero. It begins by constructing a partition of $[a,b]$ with dyadic intervals:
$$I_{n,k} = \begin{cases}[a + (b-a)\frac{k-1}{2^n}, a + (b-a)\frac{k}{2^n})\,\,\, k = 1 , \ldots, 2^n -1 \\ [a + (b-a)\frac{2^n-1}{2^n}, b] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, k = 2^n\end{cases} $$
With $m_k = \inf_{I_{n,k}}f(x)$, we can construct the sequence of simple functions
$$\phi_n(x) = \sum_{k=1}^{2^n} m_k \, \chi_{I_{n,k}}(x)$$
The sequence is increasing and, since $f$ is bounded, by monotonicity is convergent to some function $\phi$ such that $\phi_n(x) \uparrow \phi(x) \leqslant f(x).$ If $x \in I_{n,k}$ then the oscillation of $f$ over that interval satisfies
$$\sup_{u,v \in I_{n,k}}|f(u) - f(v)| \geqslant f(x) - \phi_n(x) \geqslant f(x) - \phi(x).$$
Thus, $f(x) \neq \phi(x)$ only at points where $f$ is not continuous, which must belong to a measure zero set if $f$ is Riemann integrable. Therefore, $f$ is almost everywhere the limit of a sequence of simple functions and is measurable.
I would agree that the second approach is "pretty complicated". However, I think it is not uncommon to find theorems with a variety of proofs ranging in complexity.
Best Answer
To compute the Lebesgue integral of the function $f:[0,5]\to\mathbb R$ directly, one can note that $f_n\leqslant f\leqslant g_n$ for every $n\geqslant1$, where the simple functions $f_n:[0,5]\to\mathbb R$ and $g_n:[0,5]\to\mathbb R$ are defined by $$ f_n(x)=n^{-1}\lfloor nx\rfloor,\qquad g_n(x)=n^{-1}\lfloor nx\rfloor+n^{-1}. $$ By definition of the Lebesgue integral of simple functions, $$ \int_{[0,5]} f_n\,\mathrm d\lambda=n^{-1}\sum_{k=0}^{5n-1}k\cdot\lambda([kn^{-1},(k+1)n^{-1})), $$ that is, $$ \int_{[0,5]} f_n\,\mathrm d\lambda=n^{-2}\sum_{k=0}^{5n-1}k=\frac{5(5n-1)}{2n}. $$ Likewise, $$ \int_{[0,5]} g_n\,\mathrm d\lambda=n^{-1}\sum_{k=0}^{5n-1}(k+1)\cdot\lambda([kn^{-1},(k+1)n^{-1}))=\frac{5(5n+1)}{2n}. $$ Since the integrals of $f_n$ and $g_n$ converge to the same limit, $f$ is Lebesgue integrable and $$ \int_{[0,5]} f\,\mathrm d\lambda=\lim_{n\to\infty}\frac{5(5n\pm1)}{2n}=\frac{25}2. $$