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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 $\arcsin(x)$?

part of circle

Compare to:


This is not a good approximation of the integral. What happened to the red area?

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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!

$\hskip{1.5in}$enter image description here

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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 $z=-x$.

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Beat me to it. (by 55 seconds!) +1 – user21436 Feb 14 '12 at 21:19
Why? That's not how the Riemann Sum in Cartesian plane works. I actually don't see why the triangle would cover the rest of the circle. – Unh Feb 14 '12 at 21:21
@Unh: It’s not supposed to cover the rest of the circle. $\int_0^b\sqrt{1-x^2}dx$ gives you the blue area in your first picture. Don’t think about Riemann sums here; just think about the area that you’re calculating. – Brian M. Scott Feb 14 '12 at 21:28
There is no direct connection with Riemann sum. The integral (definite integral, which can be taken as having a variable upper limit) gives you the area. Decomposition as described above, plus formulas from geometry, compute the same area. So they are equal. – André Nicolas Feb 14 '12 at 21:30
Oh I see what you mean. The sector + triangle IS the Riemann Sum I am referring to. Does the triangle "disappear" once we reach to the point of "infintessimally small"? How does that get us arcsin(x) though? I mean "x" is the angle measure from the x-axis, not from the y-axis. – Unh Feb 14 '12 at 21:34

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

enter image description here

$$A = \dfrac{\theta}{2}$$

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

$$\int {\sqrt {1 - {x^2}} dx} = \frac{{{{\sin }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{2} + C$$

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How did you substitute $\arcsin(x) = \theta$? – Unh Feb 14 '12 at 21:38
@Peter Please use the display style with restraint. I'll edit it for now. Your questions also had this discrepancy. I am pointing it out in a friendly fashion. No offence. – user21436 Feb 14 '12 at 21:40
@KannappanSampath I corrected the error but don't understand the display style remark. – Pedro Tamaroff Feb 14 '12 at 21:42
Oh I understand now. You traced out an imaginary triangle and that line perp to the radial line also happens to length b. I see npow. – Unh Feb 14 '12 at 21:44
@Unh I added the more natural apporach. Start with an area of $\pi/4$, substract the sector $\theta / 2$ and add the triangle $\cos \theta \sin \theta /2$ – Pedro Tamaroff Feb 14 '12 at 21:46

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