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The question is to prove that area of a circle with radius $r$ is $\pi r^2$ using integral. I tried to write $$A=\int\limits_{-r}^{r}2\sqrt{r^2-x^2}\ dx$$ but I don't know what to do next.

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Depending on the curriculum where you are, integrals of this form may be treated in second semester calculus, when techniques of integration beyond basic "u-substitution" are discussed. If you haven't had "trig substitutions" yet, this will seem mysterious. (In first semester, the issue is dodged by just asking students to recognize that the radical represents the expression for a semi-circle with radius $ \ r \ $ centered on the origin.) – RecklessReckoner Jul 15 '14 at 17:28
If you are genuinely interested in the topic, I might suggest this question as further reading. – Chris Jul 16 '14 at 4:29
up vote 12 down vote accepted

$$2\sqrt{r^2-x^2}=2r\sqrt{1-\left(\frac xr\right)^2}$$


$$x=r\sin t\;,\;\;dx=r\cos t\,dt\implies 2r\sqrt{1-\left(\frac xr\right)^2}dx=2r^2\cos^2t\,dt\implies$$

$$\int\limits_{-r}^r2\sqrt{r^2-x^2}dx=2r^2\int\limits_{-\pi/2}^{\pi/2}\cos^2t\,dt=\left.r^2\left(t+\sin t\cos t\right)\right|_{-\pi/2}^{\pi/2}=\pi r^2$$

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You're missing dt in the "dx = r cos t dt" – Michael Jul 15 '14 at 21:03
Thanks Michael, edited it (awaiting approval). – CompuChip Jul 15 '14 at 21:13
Thank you very much! – Mathematician171 Jul 15 '14 at 21:24

Hint: try a trigonometric substitution. In particular, try setting $x = r \sin \theta$.

Also, note the identity: $$ \cos^2 \theta = \frac 12 (1 + \cos(2\theta)) $$

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