Why is the Maclaurin series for arccos centred at origin?

I imagine I have made an error somewhere, but am unable to see where I have gone wrong.

If I graph $\int_0^x\frac{-1}{\sqrt{1-t^2}}dt$ I end up with a graph centred at the origin, rather than at $(0,\frac{\pi}{2})$ as $\cos^{-1}\left(x\right)$ should be. This approach works perfectly fine for $\sin^{-1}\left(x\right)$ when I graph $\int_0^x\frac{1}{\sqrt{1-t^2}}dt$ so I am confused as to what is going wrong.

• Per definition Maclaurin series are always centered at zero. – gammatester Nov 23 '17 at 10:41
• Could you explain what you mean by pre definition Maclaurin series? – Benjamin Nov 23 '17 at 10:42
• A Maclaurin series are is a Taylor series at 0, see en.wikipedia.org/wiki/Taylor_series. Is your question actually why $\arccos(0)=\tfrac{\pi}{2}?$ – gammatester Nov 23 '17 at 10:45
• @gammatester: He means that the graph goes through $(0,0)$ instead of $(0,\pi/2)$. – Hans Lundmark Nov 23 '17 at 10:49
• $$\int_{0}^{\varepsilon}1\,dx = \varepsilon,$$ there is nothing strange. – Jack D'Aurizio Nov 23 '17 at 10:50

That's because $$\int_0^x \frac{-dt}{\sqrt{1-t^2}} = \bigl[\arccos t \bigr]_0^x = \arccos x - \arccos 0 = \arccos x - \frac{\pi}{2} .$$ So you're not drawing the graph of $\arccos x$ but of $\arccos x - \frac{\pi}{2}$.
• Well, you rather want just $\pi/2$ plus the integral from $0$ to $x$, if you're really going to use it to obtain the Maclaurin expansion (by expanding the binomial and integrating termwise). – Hans Lundmark Nov 23 '17 at 10:53