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

Question

I want to find $$\int1+\sin^2x+\sin^4x+\sin^6x +...\text{dx}$$ My method was to interpret the integrand as a geometric series with first term $1$ and common ratio $\sin^2x$. Assuming $\sin x\ne1$, I reasoned the integrand should converge.

So the integrand should be equal to $$\frac{1}{1-\sin^2x}=\frac{1}{\cos^2x}=\sec^2{x}$$ $$\implies \int1+\sin^2x+\sin^4x+\sin^6x +...\text{dx}=\int\sec^2{x}~\text{dx}=\tan x$$ I would just like to know, is this right? If it is, it seems to me a rather beautiful result.

Answer 1

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.

Answer 2

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