Why does the integrated cosine series become the sine series?

The integrated cosine series becomes the sine series because integrating the even-power terms

x^{2n}

of the cosine series produces odd-power terms

x^{2n+1}

. Combined with the factorial simplification

(2n)!(2n+1) = (2n+1)!

, the resulting series exactly matches the standard Maclaurin series for

\sin x

Conditions

Write the cosine series: $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$ .
Integrate term-by-term to get $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \frac{x^{2n+1}}{2n+1}$ .
Simplify the denominator using $(2n)!(2n+1) = (2n+1)!$ to obtain $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ .
Compare this result to the known Maclaurin series for $\sin x$ , which is $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ .
Conclude that the integrated series is exactly the sine series.

The speaker says this power series is recognizable as the power series for sine of x.

Assuming that the derivative of the cosine series is the sine series, rather than the integral.
Forgetting the alternating sign $(-1)^n$ when comparing the integrated series to the sine series.