Circle equation with sine without parametric equation I had to integrate an area delimited by a quarter of a circle, something like this:
http://www.wolframalpha.com/input/?i=integrate+10+-+sqrt%2864+-+x%5E2%29+dx+from+0+to+5
Which comes from the equation:
$$x^2 + (y-10)^2 = 8^2$$
I wondered if it was possible to express the same curve but using sin instead?
I read somewhere (cannot find it again) you could use something like:
$$y = 10 - 8∗\cos(xπ/(2∗8))$$
But it seems just wrong? Can you confirm that it's not possible to do a perfectly circle-shaped form using sin/cos using "classic" equations (not parametric equations).
[EDIT]
I'm not interested in polar coordinates either... I want to know if it's possible to have a final equation in the form of:
$$y = A + B * \sin(C * x)$$ for some values of A/B/C. You are free to add some cos() in there :)
[EDIT2]
Apparently the question I should have asked is: "Is it possible to express parts of a circle as a cartesian equation involving sin/cos" 
 A: Recall polar coordinate equivalents. 
Here, radius is $r=8,\;$ 
$x = r\cos \theta = 8\cos \theta$, and 
$y = r\sin\theta = 8\sin\theta$.
When you expand $x^2 + (y-10)^2 = 8^2$, we can use the identity $$\sin^2\theta + \cos^2 \theta = 1$$

This gives us 
$$\begin{align}x^2 + (y-10)^2 = 8^2 & \iff x^2 + y^2 - 20 y + 100  = 64 \\ \\& \iff 64(\cos^2\theta + \sin^2\theta) - 20(8\sin\theta) +100 = 64 \\ \\ &\iff 20 (8\sin \theta) = 100 \\ \\ & \iff 8\sin\theta = 5 \\ \\&\iff \cdots \end{align}$$
EDIT:
No, you can't express the equation of a circle with $\sin, \cos$ in cartesian notation, and neither of your posted expressions represent the posted equation: expressing, e.g., $y$ as a function of $x$ where $x$ is in the argument of $\sin$ or $\cos$.
A: Your problem is to find the area of a quarter of a circle enclosed in a circumference which is given by:
$$x^2 + (y-8)^2 = 8^2.$$
First of all, it would be useful to make a translation by defining $z = y-8$ so your circumference become:
$$x^2 + z^2 = 8^2,$$
with the same area, of course. Then, the area you request is given by:
$$A = \int \int_R 1 \, dxdy,$$
where $R$ is the region in the $(x,z)$ plane where your circle lies and it's given by:
$$R \equiv \{ (x,z) \in \mathbb{R}^2, \ 0<x<R, \ x^2 + z^2 = 8^2\}, $$
which can be rewritten more conveniently by using polar coordinates as follows:
$$ x = \rho \cos \theta, \quad y = \rho \sin \theta, \quad \rho \in [0,8], \quad \theta \in[0, \pi/2], $$
then:
$$A = \int \int_R |J(\rho,\theta)| d\rho d\theta = \int^{\pi/2}_0 \int^8_0 \rho \, d\rho d\theta = \frac{\pi 8^2}{4},$$
as you would expect. 
Cheers!
I hope this is useful to you.
A: If you really wish to get rid of trigonometric functions, you can always expand the integrand as an infinite Taylor series built at x=0, compute the antiderivative and compute the integral. But you must understand that it is supposed to be an infinite series and that the result will then be dependent on the number of terms you use.  
I took the problem as you submitted it to Wolfram Alpha and I followed the steps I described above. The antiderivative is
2 + x^2 / 16 + x^4 / 4096 + x^6 / 524288 + 5 x^8 / 268435456 + ....  
Concerning the integral, depending on the number of terms, its value is successively 10.0000, 12.6042, 12.7568, 12.7780, 12.7821, 12.7830, 12.7832, 12.7833. The last number coincides with the approximate value given by Wolfram Alpha .
