Integration of Root What will be the integral
$$\int_0^1\sqrt{1-y^2}dy$$
I initially substituted $y =\sin t$ to solve (but couldn't since I'm bad at trigonometry).
 A: If $y= \sin(t)$ then $dy= \cos(t)\,dt$ and $\sqrt{1- y^2}= \sqrt{1- \sin^2(t)}= \sqrt{\cos^2(t)}= \cos(t)$ so the integral becomes $\int \cos^2(t)dt$.
In order to integrate that you need the trig identity $\cos^2(t)= \frac{1}{2}(1+ \cos(2t))$ so the integral becomes $\frac{1}{2}\int (1+ \cos(2t)) dt.$
A: Gah.  If $x=\sqrt{1-y^2}$ then $x^2=1-y^2$ or $x^2+y^2 =1$.  So the curve is the quarter unit circle in the first quadrant.  It has area equal to $\frac{1}{4}\pi$ (because the radius is $1$.)
This is probably the whole point of the problem.
A: By substituting $x=\sin(t)$ for $t\in [0,\frac{\pi}{2}]$ and using $\frac{dx}{dt}\sin(t)=\cos(t)$ and $\sin^2(t)+\cos^2(t)=1$ you get:
$$\int_0 ^1\sqrt{1-x^2}dx=\int_0^{\frac{\pi}{2}}\cos^2(t)dt$$
because $\cos$ is non-negative in the given domain. Now integrating by parts and using $\sin^2(t)+\cos^2(t)=1$ again You get:
$$\int_0^{\frac{\pi}{2}}\cos^2(t)dt=\sin(t)\cos(t)\rbrack _0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\sin^2(t)dt=\int_0^{\frac{\pi}{2}}(1-\cos^2(t))dt$$ and so
$$2\int_0^{\frac{\pi}{2}}\cos^2(t)dt=\int_0^{\frac{\pi}{2}}1dt=\frac{\pi}{2}$$
from which it follows:
$$\int_0^1\sqrt{1-x^2 }dx=\frac{\pi}{4}$$
thus proving this is the area of the quarter of the unit circle.
A: If you are not able to work around trig substitutions, you can use other elementary functions to substitute as well.
$$\sqrt{1 - y^2} \to y\sqrt{\frac{1}{y^2} - 1}$$
Let $u = y^2, du = 2y\cdot dy$
$u(0) = 0^2 = 0$
$u(1) = 1^2 = 1$
Hence, we have
$$y\sqrt{\frac{1}{y^2} - 1} \cdot dy \overset{u = y^2}{\to} \frac{1}{2}\sqrt{\frac{1}{u} - 1} \cdot du \to \frac{1}{2}\sqrt{\frac{1 - u}{u}} \cdot du \to  \frac{1}{2}\frac{1 - u}{\sqrt{u}\sqrt{1 - u}} \cdot du \to \frac{1}{2}\frac{1 - u}{\sqrt{u - u^2}} \cdot du$$.
We substitute again, $w = u - u^2 = \frac{1}{4} - \big(u - \frac{1}{2}\big)^2, dw = 1- 2u \cdot du$.
$w(0) = 0$
$w(1) = 0$
$$\frac{1}{2}\frac{1 - u}{\sqrt{u - u^2}} \cdot du \to \frac{1}{2} \big[\frac{1}{2}\frac{1-2u + 1}{\sqrt{\frac{1}{4} - \big(u - \frac{1}{2}\big)^2}}\cdot du\big] \overset{w = \frac{1}{4} - \big(u - \frac{1}{2}\big)^2}{\to} \frac{1}{4}\big[\frac{dw}{\sqrt{w}} + \frac{du}{\sqrt{\frac{1}{4} - \big(u - \frac{1}{2}\big)^2}}\big]$$
Hence,
$$\frac{1}{4}\big[\int_{0}^{0}{\frac{dw}{\sqrt{w}}} + \int_{0}^{1}{\frac{du}{\sqrt{\frac{1}{4} - \big(u - \frac{1}{2}\big)^2}}}\big]$$
$$\to \frac{1}{4} \big{[} \sin^{-1}{(2u - 1)} \big{]}_{0}^{1}$$
$$= \frac{\pi}{4}$$
