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).

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,
$$\frac{1}{4}\big[\int_{0}^{0}{\frac{dw}{\sqrt{w}}} + \int_{0}^{1}{\frac{du}{\sqrt{\frac{1}{4} - \big(u - \frac{1}{2}\big)^2}}}\big]$$
$$\to \frac{1}{4} \big{[} \sin^{-1}{(2u - 1)} \big{]}_{0}^{1}$$
$$= \frac{\pi}{4}$$
