<s>[INST] Using geometry to show that $\int_{0}^{b} \sqrt{1-x^2} dx = \frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\sin^{-1}b$ 
Show that for $0\leq b\leq1$,
$$\int_{0}^{b} \sqrt{1-x^2} dx = \frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\sin^{-1}b$$
by using geometry.

This is what I have thus far:
$\int_{0}^{b} \sqrt{1-x^2} dx$
$$=y=\sqrt{1-x^2}$$
$$=y^2=1-x^2$$
$$=x^2+y^2=1$$

From the picture above, to find the area we add the area of the sector and the area of the triangle.
Area of sector: $\frac{θ}{2}r^2$
Area of triangle: $\frac{1}{2}bh$
Side note: $(h=a)$
I have no clue what to do next. Please help.
 [/INST] The area of the triangle with sides $b$ and $a=\sqrt{1-b^2}$ forming the right angle is
$$
\frac 12b\sqrt{1-b^2}\ .
$$
The angle, marked in black, can be extracted from the triangle with sides $a,b,r=1$, it is $\arcsin (b/1)$. So the sector has area
$$\frac 12\arcsin b\cdot 1^2\ .
$$
Adding the two areas we get the answer.
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