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This page should give you an idea on finding area for circles and solving complicated integrals that require trigonometric substitution.
(Be warned, immense and complicated math talk ahead.)
![Calculus circle derivation area](https://static.wixstatic.com/media/ccbe32_8014433d44cd47f2b0c9e0c3c3b18be8~mv2_d_2040_2640_s_2.jpg/v1/fill/w_600,h_776,al_c,q_85,usm_0.66_1.00_0.01,enc_avif,quality_auto/circle%201%20%20%20%20001.jpg)
![Calculus circle area derivation](https://static.wixstatic.com/media/ccbe32_72a21084378341448817817a6c85a64e~mv2_d_2040_2640_s_2.jpg/v1/fill/w_600,h_776,al_c,q_85,usm_0.66_1.00_0.01,enc_avif,quality_auto/circle%202%20%20%20%20%20002.jpg)
![Calculus circle area derivation](https://static.wixstatic.com/media/ccbe32_a0704598516a4621918d4cebdfd3ec59~mv2_d_2040_2640_s_2.jpg/v1/fill/w_600,h_776,al_c,q_85,usm_0.66_1.00_0.01,enc_avif,quality_auto/circle%203%20%20%20%20%20%20%20%20003.jpg)
My friend actually did this without calculus. He used trigonometry to do it, which is quite impressive.
Here I simply moved the circle on the graph one radius length to the right. So the half circle is in the upper right quadrant in the graph. After that I did the same process as before.
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