Solve the upper bound of the integral for arccosine
Sophia Terry 
The function arccosine can defined as
$$\arccos(x) = \int_{1}^{x} -\frac{1}{\sqrt{1 - t^2}} \, \mathrm{d}t$$
Let $\theta$ be a known angle such that
$$\theta = \int_{1}^{x} -\frac{1}{\sqrt{1 - t^2}} \, \mathrm{d}t$$
Is it possible to solve for the upper bound $x$ in the second integral?
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$\begingroup$Okay, so from the comment section what you actually are trying to do is approximate the value of ${\cos(\theta)}$ by using the integral ${\theta=-\int_{1}^{x}\frac{1}{\sqrt{1-t^2}}dt}$. As you know, rearranging for ${x}$ gives
$${x=\cos(\theta)}$$
In all honesty, the best way of approximating ${\cos(\theta)}$ is probably to just use a Taylor Series (there are probably more advanced methods out there, but eh - the Taylor Series is a simple elementary approach, and should give you a fairly good approximation quite quickly).
If you want to use this integral though, a simple algorithmic approach would be to vary ${x}$ up and down, and approximate the integral using area approximation methods. You want to essentially find the ${x}$ that gives you the area closest to ${\theta}$. For example, if ${\theta}$ was ${0.5}$, then
$${-\int_{1}^{0.8}\frac{1}{\sqrt{1-t^2}}dt\approx 0.64...}$$
So we increase ${x}$ a little bit (since this is above ${0.5}$):
$${-\int_{1}^{0.9}\frac{1}{\sqrt{1-t^2}}dt\approx 0.451...}$$
So we decrease ${x}$ a little....
$${-\int_{1}^{0.85}\frac{1}{\sqrt{1-t^2}}dt\approx 0.554...}$$
Then increase again, then decrease, etc etc etc. You see the pattern. Provided your method of approximating the area is good enough, eventually you should get a value that is closer and closer to the true value of ${x}$, and thus get a value for ${x=\cos(\theta)}$. Note that ${\cos(0.5)\approx 0.877...}$ so you can see this was getting us closer and closer.
It is really very, very dependant though on how good your area approximations are. I'm sure there are probably ways to speed this method up and make it fancier, but this is just a base method using the integral. Like I said - for any serious ${\cos}$ function calculating though, you want to use the Taylor Series.
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