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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