# The integral of sqrt(1-x^2) dx

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Now let us look at

$\int {\sqrt {1 - {x^2}} \,dx}$

The integrand simply describes a half circle with a radius of one. So we already know

$\int\limits_{ - 1}^1 {\sqrt {1 - {x^2}} } dx = \frac{1}{2}\pi \cdot {1^2} = \frac{\pi }{2}$

To find the indefinite integral, let us do the following substitution

$\left| \begin{gathered} x = \sin \alpha \hfill \\ dx = \cos \alpha \;d\alpha \hfill \\ \end{gathered} \right.$

We get

$\begin{gathered} \int {\sqrt {1 - {x^2}} \,dx} \hfill \\ \quad = \int {\sqrt {1 - {{\sin }^2}\alpha } \cos \alpha \;d\alpha \,} \hfill \\ \quad = \int {{{\cos }^2}\alpha \;dy\,} \hfill \\ \quad = \int {\frac{{1 + \cos 2\alpha }}{2}} \hfill \\ \quad = \frac{1}{2}\int {(1 + \cos 2\alpha )dy} \hfill \\ \quad = \frac{1}{2}\left( {\alpha + \frac{{\sin 2\alpha }}{2}} \right) + C \hfill \\ \quad = \frac{1}{2}\alpha + \frac{1}{4}\sin 2\alpha + C \hfill \\ \end{gathered}$

Here we used a couple of common trig rules. But

$\sin 2\alpha = 2\sin \alpha \cos \alpha$

and

$x = \sin \alpha \Rightarrow \alpha = {\sin ^{ - 1}}x$

so

$\begin{gathered} \sin 2\alpha = 2\sin \alpha \cos \alpha \hfill \\ \quad = 2x\cos {\sin ^{ - 1}}x \hfill \\ \quad = 2x\sqrt {1 - {x^2}} \hfill \\ \end{gathered}$

This gives us

$\begin{gathered} \int {\sqrt {1 - {x^2}} \,dx} \hfill \\ \quad = \frac{1}{2}\alpha + \frac{1}{4}\sin 2\alpha + C \hfill \\ \quad = \frac{1}{2}{\sin ^{ - 1}}x + \frac{1}{2}x\sqrt {1 - {x^2}} + C \hfill \\ \quad = \frac{1}{2}({\sin ^{ - 1}}x + x\sqrt {1 - {x^2}} ) \hfill \\ \end{gathered}$

The graph of the integrand vs. the integral is shown below.

Using geometry

We could also find this integral by looking at the half-circular disk.

If the length OC=x and AC=y we have that the triangle OAC has the area

$\frac{1}{2}xy = \frac{1}{2}x\sqrt {1 - {x^2}}$

The area of the sector OAB, if the angle AOB is α, is

${\text{Sector OAB = }}\frac{\alpha }{{2\pi }} \cdot \pi = \frac{\alpha }{2}$

But the angle COA=θ is

$\theta = {\cos ^{ - 1}}x$

So

$\alpha = \pi - \theta$

This means that

$\begin{gathered} {\text{Sector OAB = }}\frac{\alpha }{2} = \frac{\pi }{2} - \frac{1}{2}\theta \hfill \\ \quad = \frac{\pi }{2} - \frac{1}{2}{\cos ^{ - 1}}x \hfill \\ \end{gathered}$

So the grey area ABC is

$\int\limits_{ - 1}^x {\sqrt {1 - {x^2}} dx = } \frac{1}{2}x\sqrt {1 - {x^2}} + \frac{\pi }{2} - \frac{1}{2}{\cos ^{ - 1}}x$

This means that our primitive function could be written as

$\int {\sqrt {1 - {x^2}} \,dx}= \frac{1}{2}x\sqrt {1 - {x^2}} - \frac{1}{2}{\cos ^{ - 1}}x + C$

But that is not what we previously got. But sine and cos are just phase-shifted against each other so the inverse should work (as it does). Below is the integrand and the two solutions, with C=π/2 in the second case.

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