I integrated $\arccos x$ properly, but I can't seem to figure out the bounds of integration? Okay my problem is: 
$$ \int_0^{1/2} \arccos x \, dx $$
According to wolfram, I performed the indefinite integral correctly. Here is my math:
$$ u = \arccos x$$
$$ du = \frac{-1}{\sqrt {1-x^2}} $$
$$ v = x $$
$$dv = dx $$
---> $$ x\arccos x + \int \frac x {\sqrt{1-x^2}} \, dx $$
So at this point I move on from using the part integration technique, and I use U-sub for the remaining integral.
$$ u = -x^2 + 1 $$
$$ \frac {-1} 2 \, du = dx $$
and I end up with the final indefinite integral result of:
$$x\arccos x - \sqrt{-x^2 +1} $$
Now my problem is.. according to wolfram I am write on the money with the integral. But when I plug in my values, my answer is different from wolfram's, and the books.
From $0$ to $1/2$:
$$ \frac 1 2 \arccos \frac 1 2 - \sqrt{-(1/2)^2 + 1} = -0.604$$
$$0\arccos 0 - \sqrt{-(0)^2 + 1 } = -1 $$
$ -0.604 - (-1) = 0.395 $ but the answer is $0.657$ does anyone know why I get the bounds wrong? I'm simply plugging it in but I don't understand how my math is wrong :( 

$$\int_{0}^{\frac{1}{2}}\arccos(x)dx$$
$u=\arccos(x)\implies du=\dfrac{-1}{\sqrt{1-x^2}}dx$ and $dv=dx\implies v=x$:
$$uv-\int v\,du=x\arccos(x)\big\vert_0^{\frac{1}{2}}+\int_0^{\frac 1 2}\dfrac{x}{\sqrt{1-x^2}} \, dx$$
Now, $u$-sub $u=1-x^2\implies \frac{du}{-2}=x\,dx$ so that $$x\arccos(x)\big\vert_0^{\frac{1}{2}}+\int_0^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}}\,dx =x\arccos(x)\big\vert_0^{\frac{1}{2}}-\frac{1}{2}\int u^{-\frac{1}{2}}du$$ $$=x\arccos(x)\big\vert_0^{\frac{1}{2}}-\sqrt{1-x^2} \big\vert_0^{\frac{1}{2}}$$
Up to here, you've done it correctly. However, $$x\arccos(x)\big\vert_{0}^{\frac{1}{2}}=\frac{1}{2}\arccos(\frac{1}{2})-0\arccos(0)=\frac{1}{2}\cdot\frac{\pi}{3}=\frac{\pi}{6}$$
and $$\sqrt{1-x^2}\big\vert_{0}^{\frac{1}{2}}=\sqrt{\frac{3}{4}}-1$$
So, the final answer is $$\boxed{\frac{\pi}{6}-\frac{\sqrt{3}}{2}+1}\approx 0.657573$$
Overall, your problem was simply stating that $(\frac{1}{2})\arccos(\frac{1}{2}) - \sqrt{-(\frac{1}{2})^2 + 1} = -.604$ when in fact $(\frac{1}{2}) \arccos(\frac{1}{2}) - \sqrt{-(\frac{1}{2})^2 + 1}\approx -0.342$
