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Question Number 5490 by 123456 last updated on 16/May/16

$$\int \sqrt{r^2-x^2}dx=?$$$$\underset{-r}{\stackrel{x}{\int }}\sqrt{r^2-x^2}dx=??$$$$\underset{x}{\stackrel{y}{\int }}\sqrt{r^2-x^2}dx=???$$

Answered by FilupSmith last updated on 16/May/16

$$\mathrm{let}x=r\cdot \sin \left(\theta \right)\Rightarrow dx=\cos \left(\theta \right)d\theta$$$$\therefore \theta =\arcsin \left(\frac{x}{r}\right)$$$$\therefore \int \sqrt{r^2-x^2}dx=\int \left(\sqrt{r^2-r^2{\sin }^2\theta }\cos \theta \right)d\theta$$$$=\int \cos \theta \sqrt{r^2(1-{\sin }^2\theta )}d\theta$$$$=\int r\cos \theta \sqrt{{\cos }^2\theta }d\theta$$$$=r\int {\cos }^2\theta d\theta$$$${\cos }^2\left(x\right)=\frac{\cos \left(2x\right)+1}{2}$$$$=r\int \frac{1}{2}(\cos \left(2\theta \right)+1)d\theta$$$$=\frac{r}{2}(\frac{1}{2}\sin \left(2\theta \right)+\theta )+c$$$$\therefore \int \sqrt{r^2-x^2}dx=\frac{1}{4}r\cdot \sin (2\cdot \arcsin \left(\frac{x}{r}\right))+\frac{1}{2}r\cdot \arcsin \left(\frac{x}{r}\right)+c$$$$$$$$--------------$$$$\underset{-r}{\stackrel{x}{\int }}\sqrt{r^2-x^2}dx=\frac{r}{2}{[\frac{1}{2}\sin (2\cdot \arcsin \left(\frac{x}{r}\right))+\arcsin \left(\frac{x}{r}\right)]}_{-r}^x$$$$=\frac{r}{2}[\frac{1}{2}\sin (2\cdot \arcsin \left(\frac{x}{r}\right))+\arcsin \left(\frac{x}{r}\right)-\frac{1}{2}\sin (2\cdot \arcsin \left(\frac{-r}{r}\right))+r\cdot \arcsin \left(\frac{-r}{r}\right)]$$$$=\frac{r}{2}[\frac{1}{2}\sin (2\cdot \arcsin \left(\frac{x}{r}\right))+r\cdot \arcsin \left(\frac{x}{r}\right)-\frac{1}{2}\sin (-\pi )+\frac{\pi }{2}]$$$$\therefore \underset{-r}{\stackrel{x}{\int }}\sqrt{r^2-x^2}dx=\frac{r}{2}[\frac{1}{2}\sin (2\cdot \arcsin \left(\frac{x}{r}\right))+r\cdot \arcsin \left(\frac{x}{r}\right)+\frac{\pi }{2}]$$

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