Using geometry to show that $\int_{0}^{b} \sqrt{1-x^2} dx = \frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\sin^{-1}b$

Question

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.

Answer 1

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.