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For the integral $$\int \sqrt{1-x^2} dx = \frac{1}{2} \left ( \arcsin(x) + x \sqrt{1-x^2} \right)$$ Now it was explained to me that geometrically I could take part of the integral as an area sector and the other half a triangle. I am having a hard time seeing how I can even get a triangle as I am summing the rectangles.

How could you even see that the angle must be $\arcsin(x)$?

part of circle

Compare to:

sector

This is not a good approximation of the integral. What happened to the red area?

3 Answers 3