Does $\int1+\sin^2x+\sin^4x+\sin^6x+...\text{dx}=\tan{x}$?

HINT: For any nonnegative $a <1$ the sum $1+a+a^2+ \ldots = \frac{1}{1-a}$. So setting $a=\sin^2x$, we note: $1+\sin^2 x+\sin^4 x + \ldots = \frac{1}{1-\sin^2 x} = \sec^2 x = \frac{d(\tan x)}{dx}$.

Can you finish from here.

ETA: Nevermind just reread you already got this. YES you are correct.

The series converges except when $\sin^2x=1$, but in that case $\tan x$ also diverges. All you missed is the $+C$.