The statement trying to be proved:
Let $f(x)$ be a continous function, show that for any $a > 0$, we must have
$$\int_{0}^a x^3f(x^2)dx-\frac{1}{2}\int_{0}^{a^2} xf(x)dx=0$$
So my approach is this:
Let,
$$u=x^2$$ $$du=2x\cdot dx$$ $$dx=\frac{du}{2x}$$
Substituting $u$ to $\int_{0}^a x^3f(x)dx$,
$$\int_{0}^a x^3f(x)dx$$
$$=\int_{0^2}^{a^2} x^3f(u)\frac{du}{2x}$$ $$=\frac{1}{2}\int_{0^2}^{a^2} x^2f(u)du$$ $$=\frac{1}{2}\int_{0^2}^{a^2} uf(u)du$$
Since the choice of variable is arbitrary, $$=\frac{1}{2}\int_{0^2}^{a^2} xf(x)dx$$
Hence, $$\int_{0}^a x^3f(x)dx=\frac{1}{2}\int_{0^2}^{a^2} xf(x)dx$$ $$\int_{0}^a x^3f(x)dx-\frac{1}{2}\int_{0^2}^{a^2} xf(x)dx=0$$
Are the steps that I used valid? I am particularly concerned whether my statement "Since the choice of variable is arbitrary" is correct or no.
I also think that the steps I use are not rigorous enough. Is there perhaps any other rigorous approach to this problem?
No. You cannot substitute only some instances of $x$ by $u$, i.e. this $\int_{0^2}^{a^2} x^3f(u)\frac{du}{2x}$ makes no sense. And even if you think of $x$ as $u^{1/2}$ in that term then $$ \frac{1}{2}\int_{0^2}^{a^2} x^2f(u)du =\frac{1}{2}\int_{0^2}^{a^2} u^2f(u)du $$ becomes wrong.
A correct proof via the substitution rule is simple and straightforward: the function $g: [0, a] \to [0, a^2]$, $g(x) = x^2$ has a continuous derivative, therefore is $$ \int_0^a x^3 f(x^2) \, dx = \frac 12 \int_0^a g(x) f(g(x)) g'(x) \, dx = \frac 12 \int_0^{a^2} u f(u) \, du. $$ Or if you prefer the $u = x^2$, $du = 2x dx$ notation: $$ \int_0^a x^3 f(x^2) \, dx = \frac 12 \int_0^a x^2 f(x^2) \, 2x dx = \frac 12 \int_0^{a^2} u f(u) \, du. $$ (Note that there is no need to divide by $x$, which would be problematic at $x=0$.)
That is correct: $$ \frac 12 \int_0^{a^2} u f(u) \, du = \frac 12 \int_0^{a^2} x f(x) \, dx $$ because the naming of the integration variable does not matter.