Maclaurin Series Expansion of $\ln(1+\sin x)$ Hello can anyone help with this question 
Show that the Maclaurin series of the function 
$$\ln(1+\sin x)$$
up to the term in $x^4$ is 
$$x-x^2/2 + x^3/6 - x^4/12 + \ldots$$
So I know the expansion for $\ln(1+x)= x - x^2 + x^3/3 +\dots$ and that of $\sin x= x - x^3/3!+x^5/5!-\dots$ hence I tried by substituting the first two terms of $\sin x$ into the expansion of $\ln(1+x)$ to get $\ln(1+x-x^3/6)$ up to the $x^4$ term of the expansion of $\ln(1+x)$. But I got stucked with the algebra so I will value the help anyone can provide.
 A: What you tried is, however, an interesting attempt to solve the given task but not quite right. It is in fact (and please don't ask me why) correct up a few terms if you finish what you tried. Anyway, this is not the standard way of finding a MacLaurin Series of a given function. 
Recall, a MacLaurin Series Expansion is a Taylor Series Expansion centered at $0$. By Taylor's Theorem we know that the series expansion is then given by

$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\tag1$$

Since you are only asked to find the expansion up to the $x^4$-term we only need to compute the first four derivatives and evaluate them at $0$. Thus, we obtain
\begin{align*}
&f(x)=\ln(1+\sin x),&&f(0)=\ln(1+0)=0\\
&f^{(1)}(x)=\frac{\cos x}{1+\sin x},&&f^{(1)}(0)=\frac1{1+0}=1\\
&f^{(2)}(x)=-\frac1{1+\sin x},&&f^{(2)}(0)=-\frac1{1+0}=-1\\
&f^{(3)}(x)=\frac{\cos x}{(1+\sin x)^2},&&f^{(3)}(0)=\frac1{(1+0)^2}=1\\
&f^{(4)}(x)=-\frac{1+\sin x+\cos^2x}{(1+\sin x)^3},&&f^{(4)}(0)=-\frac{1+0+1}{(1+0)^3}=-2
\end{align*}
Plugging these values in $(1)$ we obtain
\begin{align*}
\ln(1+\sin x)&=f(0)+f^{(1)}(0)x+\frac{f^{(2)}(0)}{2}x^2+\frac{f^{(3)}(0)}{6}x^3+\frac{f^{(4)}(0)}{24}x^4+\cdots\\
&=0+1\cdot x-\frac12x^2+\frac16x^3-\frac2{24}x^4+\cdots\\
&=x-\frac{x^2}2+\frac{x^3}6-\frac{x^4}{12}+\cdots
\end{align*}

$$\therefore~\ln(1+\sin x)~=~x-\frac{x^2}2+\frac{x^3}6-\frac{x^4}{12}+\cdots$$

In a similiar way you can obtain the MacLaurin Series Expansions for $sin x$ or $\ln(1+x)$. Just substituting one into another isn't the afterall the exspected way to do this but rather computing the derivatives at $0$.
