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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.

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

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  • $\begingroup$ Oh thank you. This makes sense. So surely then the graph I want is the integral from 1 to x $\endgroup$ – Benjamin Nov 23 '17 at 10:52
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    $\begingroup$ 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). $\endgroup$ – Hans Lundmark Nov 23 '17 at 10:53

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