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How to find $\int \sqrt{a^2-x^2} dx\;,$ where $a$ is a constant?

It appears to be $\frac{\pi a}{2}\;,$ but how do I get there?

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    Yes, you’ll need to find the antiderivative, but it’s still better to state the problem correctly. (Actually, if the limits are $-a$ and $a$ you *don’t* need to find the antiderivative: $y=\sqrt{a^2-x^2}$ is the upper half of$a$circle of radius $a$ centred at the origin, and $\int_{-a}^a\sqrt{a^2-x^2}dx$ is the area between that half-circle and the $x$-axis, or half the area of$a$circle of radius $a$.)2012-04-30

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One way of finding the definite integral is to consider the shape of the graph. If $y = \sqrt{a^2-x^2}$ then $y^2=a^2-x^2$, so $x^2+y^2=a^2$. Apply the Pythagorean theorem: that's the equation of a circle of radius $a$. $y = \pm\sqrt{a^2-x^2}$ is the whole circle; $y = \sqrt{a^2-x^2}$ is the top half of the circle. If you know that the area of the whole circle is $\pi a^2$, then the area of the top half is $ \int_{-a}^a \sqrt{a^2-x^2} \, dx = \frac{\pi a^2}{2}. $

Generally, if you have $ a^2-x^2 $ in an integral, you can use $x = a\sin\theta$ and $dx = a\cos\theta\,d\theta$, and then $a^2-x^2$ becomes $a^2\cos^2\theta$.

If you have $ a^2 + x^2 $ then you can use $x = a\tan\theta$, $dx = a\sec^2\theta\,d\theta$, $a^2+x^2=a^2\sec^2\theta$.

If you have $ x^2 - a^2 $ then you can use $x=a\sec\theta$, $dx=a\sec\theta\tan\theta\,d\theta$, $x^2 - a^2 = a^2\tan^2\theta$.

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    3:21:45 versus 3:21:35. =)2012-04-30
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It might be enlightening for you to think the following.

The function to integrate is $\sqrt{a^2-x^2}=y$.

This gives that

$y^2+x^2=a^2$

I really hope you know this is the equation for a circle of radius $a$ centered at $(0,0)$. Since you're integrating from $-a$ to $a$, you're calculating the area of half a circle of radius $a$. This means that the integral

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

gives the area of half a circle of radius $a$, which is $\dfrac 1 2\pi a^2$.