Integrating $\sqrt{1-x^2}$ by interpreting it geometrically as an area within a circle For the integral $$\int \sqrt{1-x^2} dx = \frac{1}{2} \left ( \arcsin(x) + x \sqrt{1-x^2}  \right)$$
Now it was explained to me that geometrically I could take part of the integral as an area sector and the other half a triangle. I am having a hard time seeing how I can even get a triangle as I am summing the rectangles.
How could you even see that the angle must be $\arcsin(x)$? 

Compare to: 

This is not a good approximation of the integral. What happened to the red area? 
 A: Note that the integral in question $$\int_0^b{\sqrt{1-x^2} ~\mathrm dx}$$ is the shaded area in the integral below. Now compare a direct computation from the figure below with the result you have got through the integral.
Join the center of the circle to the point $(b,f(b))$ to see a sector and a triangle!
$\hskip{1.5in}$
A: Draw the line from the center of the circle to your point $(b,f(b))$.  That line splits your big blue region into two parts: a triangle below the line, and a sector above. 
The area of the triangle is clearly $\frac{1}{2}b\sqrt{1-b^2}$.
For the area of the sector, the angle of that sector is $\arcsin b$. This is because the complementary angle (below the line) has cosine equal to $b$. Or else you can see directly, by drawing a perpendicular from $(b,f(b))$ to the $y$-axis, that the angle of the sector has "opposite" side, and therefore sine, equal to $b$.  So the area of the sector is $\frac{1}{2}\arcsin b$.
By integration, the area of the blue region is $\int_0^b\sqrt{1-x^2}dx$. We conclude that
$$\int_0^b\sqrt{1-x^2}\,dx=\frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\arcsin b.\qquad\qquad(\ast)$$
Now in $(\ast)$, change the $b$ to $x$, and (to make me feel good) the dummy variable of integration to $t$.
We have
$$\int_0^x\sqrt{1-t^2}\,dt=\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}\arcsin x.$$
This says that the right-hand side is an antiderivative of $\sqrt{1-x^2}$. All antiderivatives are obtained by adding a constant of integration.
Note that the geometric derivation is not quite complete, since the picture does not deal directly with negative $b$. But that is not hard to do. The simplest way is to make the change of variable $z=-x$.
A: You probably were told to put it this way:
The coordinates $(\sin \theta, \cos \theta)$ can be defined as those who delimit an area of $\frac{\theta}{2}$ in the unit circle when joined with its origin. 
So the area in this image would be

$$A = \dfrac{\theta}{2}$$
What you're being told is that
$$\int\limits_0^x {\sqrt {1 - {t^2}} dt}  = \frac{\pi }{4} -\frac{\theta }{2} + \frac{{\sin \theta \cos \theta }}{2}$$
Note that $\dfrac{{\sin \theta \cos \theta }}{2}$ is the area of the triangle; $ \dfrac{\theta }{2}$ area of the sector and $\dfrac{\pi }{4} $ the area of the circle's 1st quadrant.
So expressing in terms of $y = \sin \theta$ you get that
$$\int\limits_0^x {\sqrt {1 - {t^2}} dt}  =  \frac{{{{\sin }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{2}$$
And in terms of $x = \cos \theta$ you get that
$$\int\limits_0^x {\sqrt {1 - {t^2}} dt}  = \frac{\pi }{4}- \frac{{{{\cos }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{2}$$
Note that your solution is incorrect. It should be
$$\int {\sqrt {1 - {x^2}} dx}  = \frac{{{{\sin }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{2} + C$$
