# Proof that $\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$?

In my textbook appears that $\displaystyle\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$

But where does this equation come from?

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How do you define $\pi$? – Dylan Moreland May 10 '12 at 0:04
@DylanMoreland $\pi$ is the mathematical constant $3.141592\dots$ – Garmen1778 May 10 '12 at 0:07
I had no doubt that you were referring to this, but for many this integral would be the definition of $\pi/2$, and hence of $\pi$. I figure you want some long chain of equalities, starting on the left side of what you wrote and ending up on the right. What should the penultimate expression look like? – Dylan Moreland May 10 '12 at 0:10
@Garmen: that's not a definition. Those digits had to have come from somewhere. (As Dylan and the others say, since the integrand is a semicircle, and with the interpretation of the integral as an area, you have a basic definition of $\pi$.) – J. M. May 10 '12 at 0:26
I did a grammatical edit. In English one may say "This comes from X" or "Where does this come from", or "This came from X" or "Where did this come from", but never "Where does this comes from" or "Where did this came from". – Michael Hardy May 10 '12 at 12:53

The equation $y = +\sqrt{1 - x^2}$ for $-1\leq x\leq1$ describes the top half of a circle of radius $1$. The area between this curve and the $x$-axis is therefore $\pi/2$. On the other hand, you can compute the area under this curve by doing the integral $$\int_{-1}^1 \sqrt{1 - x^2}\,dx.$$

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Those "area" answers are probably the best ones. On the other hand, if $\pi$ is defined in some way other than area, we can pursue the standard trigonometric substitution: $x = \sin \theta$, $-\pi/2 \le \theta \le \pi/2$ to get: \begin{align} \int_{-1}^1\sqrt{1-x^2}\,dx &= \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta = \int_{-\pi/2}^{\pi/2}\frac{1+\cos(2\theta)}{2}\,d\theta \\ &=\frac{1}{2}\int_{-\pi/2}^{\pi/2}d\theta + \frac{1}{2}\int_{-\pi/2}^{\pi/2} \cos(2\theta)\,d\theta \\ &= \frac{\pi}{2} + 0 = \frac{\pi}{2} . \end{align}

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Yes, fixed..... – GEdgar May 10 '12 at 15:56

Not the "cleverest" method, like the above, - but works!

$$\int_{-1}^1 \sqrt{1-x^2} \ dx$$

To compute that integral, one may substitute $x=\sin{t}$, and get:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-\sin^2{t}} \ d(\sin{t})=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos{t}\cos{t} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{t} \ dt$$

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{t} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+\cos{2t}}{2} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dt}{2}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos{2t}}{2}=$$

$$= \frac{\pi}{2}+\frac{1}{2}\sin{t}\cos{t}{\huge{|}}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2}+0=\frac{\pi}{2}$$

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$f(x)=\sqrt{1-x^2}$ is the graph of the upper half of the circle $x^2+y^2=1$. To see this, just square both side.

Thus,

$$\int_{-1}^1 \sqrt{1-x^2}dx$$ is the integral which represents the area under half of circle of radius 1. That is the are of half disk, thus $\frac{1}{2} \pi 1^2$

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