Calculating this series integral $\int (\frac{x^{2n-2}}{(2n-2)!})\cdot \frac{x^{2n-1}}{(2n-1)!}$ I've calculated this integral and would like the communities feedback and support towards my solution:
$$\int \left(1 \cdot \frac{x^2}{2!} \cdot \frac{x^4}{4!} \cdots\frac{x^{2n-2}}{(2n-2)!}\right)\cdot \left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\cdots\frac{x^{2n-1}}{(2n-1)!}\right)\,\mathrm{d}x$$
My attempt:
I performed substitution where $u = \left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\cdot\frac{x^{2n-1}}{(2n-1)!}\right)$.
Then taking the derivative of both sides $\mathrm{d}u = \left(1 \cdot \frac{x^2}{2!} \cdot \frac{x^4}{4!} \cdots\frac{x^{2n-2}}{(2n-2)!}\right)\mathrm{d}x$.
This yields $$\int u\space \mathrm{d}u = \frac{u^2}{2}+C.$$
Then substituting back in the variable in $u$:
$$\frac{1}{2}\left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\frac{x^{2n-1}}{(2n-1)!}\right)^2$$
Although I'm not entirely sure how to distrbute the $2$ properly. Do I multiply the single $x$ onto $\frac{x^3}{2!}$, then distribute the power, and multiply the $2$ on the numerator using this fraction rule like this
$$\left(\frac{2(x^6)}{3!} \cdot \frac{2(x^7)}{5!}\frac{2(x^{2n+1)}}{(2n-1)!}\right) ?$$

Please keep in mind: the derivative of a product is not the product of the derivatives!
In this case you should simply multiply both factors obtaining
$$\int\prod_{k=0}^{2n-1}\frac {x^k}{k!}dx=\frac1 {\prod_{k=0}^{2n-1} k!}\int x^{\sum_{k=0}^{2n-1}k}dx=\frac1 {\prod_{k=0}^{2n-1} k!}\int x^{n (2n-1)}dx=
\frac{x^{n (2n-1)+1}}{(n (2n-1)+1)\prod_{k=0}^{2n-1} k!}.
$$
