Find which differentiable function is determined by this power series 
Given the series $$\sum\limits_{n=1}^\infty \dfrac{1}{n(n+1)} (\sin(x))^n$$ for all $x \in \mathbb{R}$, find an interval on which it determines a differentiable function of $x$, together with an expression for its derivative in terms of standard function.  

My Attempt
Using comparison test , we can see that the above series converges for all $|\sin(x)| \leq 1 \iff \forall x \in \mathbb{R}$.
now if we differentiate the above series we get; 
$$\sum\limits_{n=1}^\infty \dfrac{\cos(x)}{(n+1)} (\sin(x))^{n-1}$$ which if we rewrite it is;
$$\cos(x) \sum\limits_{n=0}^\infty \dfrac{1}{(n+2)} (\sin(x))^{n}$$  
But I can't make out what function this power series relates to? 
Any help would be much appreciated.
 A: Hint 
Rewrite, using $y=\sin(x)$ and partial fraction decomposition,  $$\sum\limits_{n=1}^\infty \dfrac {\sin(x)^n}{n(n+1)}=\sum\limits_{n=1}^\infty \dfrac {y^n}{n(n+1)}=\sum\limits_{n=1}^\infty \dfrac {y^n}{n}-\dfrac {1}{y}\sum\limits_{n=1}^\infty \dfrac {y^{n+1}}{n+1}$$ Now recognize that each of the sum is the antiderivative of quite well known sums which are simple geometric progressions. 
When you finish the work with these two sums, replace $y$ by $\sin(x)$. You will end with a rather simple expression.
I am sure that you can take from here
A: It turns into a differential equation. Can you solve the ODE you get by your method?
EDIT:
Consider $f'(x)+\cos(x)f(x)$:
$$\frac{\cos(x)}{2}+\cos(x)\sum_{n=1}^{\infty}\frac{1}{n+2}(\sin(x))^n+\cos(x)\sum_{n=1}^{\infty}\frac{1}{n(n+1)}(\sin(x))^n$$
$$=\frac{\cos(x)}{2}+\cos(x)\sum_{n=1}^{\infty}(\frac{1}{n+2}-\frac{1}{n(n+1)})(\sin(x))^n$$
Notice that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$
So, you can get rid of all powers of $n$ that are greater than $1$ in the denominator.
Then use the identity:
$$\frac{1}{1-t}=1+t+t^2+\cdots \text{ for } |t|<1$$
to derive the formulas you need for your power series. Notice that if we set $t=\sin(x)$ things become clearer.
Can you take it from here?
Second EDIT:
Actually I just realized that turning it into an ODE doesn't simplify things. Even though it works, but it's just a step that is unnecessary.
Just use partial fractions the same way as explained.
A: Consider:
\begin{align}
\sum_{n \ge 0} u^n
  &= \frac{1}{1- u} \\
\int_0^u \sum_{n \ge 0} t^n \, \mathrm{d} t
  &= \sum_{n \ge 1} \frac{t^n}{n} \\
  &= -\ln (1 - u)
\end{align}
Another round of integration gets the right denominator, divide by $u$ for the numerator, and all this is valid for $\lvert u \rvert < 1$ (with a removable singularity at 0). Substitute $\sin x$ when you are done.
