area under $\arccos x$ function What is the exact value of $k$ such that the area under the curve $\arccos(x)$ between $0$ and $k$ is exactly $0.5$? No calculator please. 
I am trying by symmetry the answer around $0.35$ but are there better methods out there?
 A: Hint
In order to compute the area under the curve, you first need to compyte the antiderivative of the function; so, you need $$I=\int \cos ^{-1}(x)dx$$ This integral is easily solved using integration by parts and $$I=x \cos ^{-1}(x)-\sqrt{1-x^2}$$ So, now, we have all elements to compute $$J=\int_0^k \cos ^{-1}(x)dx=-\sqrt{1-k^2}+k \cos ^{-1}(k)+1$$ and you want to find $k$ such that $J=\frac 12$ which means that you look for the solution of the equation $$f(k)=-\sqrt{1-k^2}+k \cos ^{-1}(k)+\frac{1}{2}=0$$ which cannot be solved analytically and so, numerical methods should be used and a calculator will be required.
A simpler root-finding method is Newton which, providing a reasonable guess $k_0$, will update it according to $$k_{n+1}=k_n-\frac{f(k_n)}{f'(k_n)}$$ In the case of your problem, Newton formula simplifies to $$k_{n+1}=\frac{2 \sqrt{1-k_n^2}-1}{2 \cos ^{-1}(k_n)}$$ A quite look at the graph of $f(k)$ shows that the root is around $k=0.5$; so, let us start Newton with $k_0=0.5$. The successive iterates will then be $0.349529$, $0.359986$, $0.360035$ which is the solution for six significant figures.
Added later
However, in your post you mention something around $k=0.35$ which, to me, means that you computed somehow the value of  $\cos ^{-1}(0.35)$. So, in the same spirit as Ross Millikan, you have all the elements required for approximating function $f(k)$ by a second order Taylor expansion around this value. This should give $$f(k)\approx -0.0121209+1.21323 (k-0.35)-0.533761 (k-0.35)^2+O\left((k-0.35)^3\right)$$ and solving the quadratic in $(k-0.35)$ will give $k=0.360035$.
Based on the same approach, a more attractive formula would have been $$k\approx a+\sqrt{1-a^2} \cos ^{-1}(a)-\sqrt{a^2+\sqrt{1-a^2}+\left(\sqrt{1-a^2} \cos
   ^{-1}(a)+a\right)^2-2}$$ in which you could replace $a=0.35$.
What is amazing is that, if you plug $a=0$ in the above formula, you get $$k \approx \frac{1}{2} \left(\pi -\sqrt{\pi ^2-4}\right) = 0.359433$$ For this value, which is a surprizingly good estimate that you could effectively obtain without using any calculator, $J \approx 0.4993$.
If you plug $a=\frac 12$, you get $$k \approx \frac{1}{6} \left(3+\sqrt{3} \pi -\sqrt{3 \left(-18+6 \sqrt{3}+2 \sqrt{3} \pi +\pi
   ^2\right)}\right)=0.360290.$$So close to the solution !
A: Analytically you are solving $k \arccos k -\sqrt {1-k^2}+1=\frac 12$, which Alpha gives as about $0.36$  Without a calculator, I would use the Taylor series $\arccos x \approx \frac \pi 2-x$ to get $k(\frac \pi 2-k)-\sqrt{1-k^2}+1=\frac 12, 1-k^2=(k(\frac \pi 2-k)+\frac 12)^2=1-k^2$  I would expand the square and isolate the linear term to write $k=stuff$ and use fixed point iteration.  It will be work without a calculator.
A: $$\int \cos^{-1}(x)=x\times \cos^{-1}(x)-\sqrt{1-x^2}$$
$$\int_0^k =k\times\cos^{-1}k-\sqrt{1-k^2}+1=0.5$$
I am pretty sure that there are only numerical solutions to the above problem.
