Maclaurin Series : $f(x) = 3\sin^2(x)$ Use a Maclaurin series in this table to obtain the Maclaurin series for the given function.
$$f(x) = 3\sin^2(x)$$
[Hint: Use $\sin^2(x) = \frac{1}{2}\cdot(1-\cos(2x))$]
Series Sum from $0$ to $\infty$ = ?

Alternatively, consider that:
$f(x) = 3\sin^2(x) \implies f'(x) = 6\sin(x)\cos(x) = 3\sin(2x)$
The series for $f'(x)$, we can easily find:
$\displaystyle f'(x) = 3\cdot\sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k + 1}x^{2k + 1}}{(2k + 1)!}$
Thus $\displaystyle f(x) = \int f'(x) dx = \int 3\cdot\sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k + 1}x^{2k + 1}}{(2k + 1)!}$

$f(x) \displaystyle = 3\cdot\sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k + 1}x^{2k + 2}}{(2k 
+ 2)!}$

