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$$ \pi = \int_{-\sqrt{2}}^{\sqrt{2}} \sqrt{2-x^2} dx $$

So, thinking I was going to discover something amazing I reasoned that this is equal to pi, and that all I had to do to get the exact value of pi was to find the indefinite integral.

$$ \int \sqrt{2-x^2} dx = \space ? $$

Of course, I couldn't, not even with partial integrals.

So now I'm wondering whether this is possible, but I don't think so.

:)

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    $\begingroup$ Integrands of the form $\sqrt{a^2 - x^2}$ can be integrated using a standard trigonometric substitution of the form $x = a \sin \theta$, $dx = a \cos \theta \, d\theta$. $\endgroup$
    – heropup
    Jan 21, 2014 at 21:55
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$
    – kukac67
    Jan 21, 2014 at 21:59
  • $\begingroup$ Your integral is the area of a semi-circle of radius $ \sqrt2$. $\endgroup$ Jan 21, 2014 at 22:01

2 Answers 2

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Use trigonometric substitution, with $x = \sqrt 2 \sin\theta$.

Then $\,dx = \sqrt 2 \cos\theta\,d\theta$.

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As an alternative to trigonometric substitution, you can also argue that this integral is equal to $\pi$ geometrically. As you probably are aware, the definite integral of a non-negative function $f$ on $[a,b]$ is the area of the region bounded by the curves $y = f(x)$, $y = 0$, $x = a$, and $x = b$. For $f(x) = \sqrt{2 - x^2}$, the curve $y = f(x)$ with $-\sqrt 2 \leq x \leq \sqrt 2$ is the semicircle $\{(x,y) \mid x^2 + y^2 = 2 \,\&\, y \geq 0 \}$. So your integral equals half the area within the whole circle $x^2 + y^2 = 2$, which is simply $\frac{1}{2}(2\pi) = \pi$.

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