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

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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Let $x=a\sin\,u$... – J. M. Apr 30 '12 at 3:08
@J.M. how do I know to use that substitution? – Jiew Meng Apr 30 '12 at 3:10
Your integral is an indefinite integral, so its value cannot be a specific number. Presumably it’s supposed to be a definite integral; what are the limits of integration? I suspect that they’re $-a$ and $a$, in which case the correct answer is $\pi a^2/2$. – Brian M. Scott Apr 30 '12 at 3:11
Square roots are usually nasty when taking antiderivatives. You try to make a substitution that will allow you to get rid of the radical. Here, using a slight modification of $\sin^2\theta+\cos^2\theta=1$, or $\cos^2\theta=1-\sin^2\theta$, will allow you to do that. – David Mitra Apr 30 '12 at 3:14
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 $a$.) – Brian M. Scott Apr 30 '12 at 3:18

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. =) – Pedro Tamaroff Apr 30 '12 at 3:24

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 $\dfrac 1 2\pi a^2$.

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