Finding the area of shaded region Consider a circle with the equation $x^2 + y^2 = 1$. Now find the area of the shaded region which we denote by $u$.
image of circle:

I'm trying to approach this question and would like support on my integral calculation. The difficulty I'm having is with substitution.
For example, given that:
$$u = (OM)(MP)+2\int^{1}_{x} \sqrt{1-x^2} \space dx$$
I should be getting:
$u = xy + [x\sqrt{1-x^2}-\cos^{-1}x]^{1}_{x}$
Although, I currently lack the technical skills to get this answer.
My approach:
I thought that I could substitute back in $x$ for $\sin(x)$ to get $\sqrt{1-\sin^2(x)}$, then when $u = \sin(x) \implies du = \cos(x)$, so that I get:
$$2\int^{1}_{x} \sqrt{1-\sin^2(x)}\cdot \cos(x) \space dx$$
However, this does not lead me close to the answer.
I have also tried inputting this equation into symbolab, and get:
$\arcsin \left(x\right)+\frac{1}{2}\sin \left(2\arcsin \left(x\right)\right)+2C$
Which is also wrong. What might be the approach towards this?
In Progress:
I've currently rearranged my substitution and it looks promising:
$u = 1-\sin^2 (x) \implies du = -cos(x)dx \implies-arcos(x)du = dx$
I cannot manage with this approach neither, as I'll have to integrate the square root.
Looking at the equation, I'll need something like:
$secant(x) -arcos(x)$ to convert the secant into $x\sqrt{x^2-1}$
 A: You may integrate by parts directly
\begin{align}
\int_{x}^{1}\sqrt{1-t^2}dt
& =\int_{x}^{1}\frac{\sqrt{1-t^2}}{2t}d(t^2)
= -\frac12 x\sqrt{1-x^2}+\frac12 \cos^{-1}x 
\end{align}
which, with $y=\sqrt{1-x^2}$, leads to the area
$$u =xy + 2 \int_{x}^{1}\sqrt{1-t^2}dt =\cos^{-1}x$$
A: Alternative method:
The area of the shaded region is $\frac{1}{2}r^2\theta,\ $ where $r$ is the radius of the circle and $\theta\ $ is the angle in radians that the region makes at the centre of the circle $O$.
$r = 1,\ $ therefore the area is $\frac{1}{2}\times1^2\times \theta = \frac{1}{2}\theta.$
Now, $\angle POM = \frac{\theta}{2},\ $ so we can express $\frac{\theta}{2}\ $ in terms of the $x-$coordinate of $M = (M_x,M_y)$, $M_x$ say. Looking at the triangle $OPM,\ $ we know that $OP=1,\ $ therefore $\frac{\theta}{2} = \cos^{-1}\left(\frac{M_x}{1}\right) = \cos^{-1}(M_x) \implies \theta = 2\cos^{-1}(M_x).$
So area of shaded region $= \frac{1}{2} \times 2\cos^{-1}(M_x) = \cos^{-1}(M_x).$
A: (I assume that $(x,y)$ is in the first open quadrant).
The area of a circular sector with angle $\alpha$ is
$${\frak A}=\alpha/2$$
(in particular, if $\alpha=2\pi$, we get an area equal to $\pi$ in conformity with formula $\pi R^2$ for $R=1$).
As, in triangle $OMP$,
$$\alpha/2=\operatorname{atan}\dfrac{MP}{MO}=\operatorname{atan}\dfrac{y}{x}$$
As a consequence:
$${\frak A}\text{=}\operatorname{atan}\dfrac{y}{x}$$
