What's the Maclaurin series for $\arcsin(x)$?
I solved the problem by using a known series: $\frac{1}{\sqrt{1-x^2}}$, but the solution I got is wrong. Also, I'm not sure what to do with the constant of integration $C$. Where is my mistake?
$$ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} +... $$ $$ \int\frac{1}{\sqrt{1-x^2}}dx = \int1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} +... dx$$ $$ \arcsin(x) + C = x + \frac{2x^3}{3} + \frac{3x^5}{24} + \frac{5x^7}{112}+... \tag{what happens to $C$?}$$
The right solution is: $$ x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} +... $$
You did a great job:
Just small mistakes.
To find constant of integration, substitute known value of $\arcsin(x)$, $x=0$ is a good choice. You need to have a well defined interval while dealing with inverse trigonometric functions.
The formula to integrate is $$\int x^n \, dx=\frac{x^{n+1}}{n+1}$$
You're integrating the right hand side incorrectly. The integral of $\dfrac{x^2}{2}$ isn't $\dfrac{2x^3}{3}$, it's $\dfrac{x^3}{2\cdot 3} = \dfrac{x^3}{6}.$ :)))