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Now let us look at
The integrand is simply describing a half circle with radius one. So we already know
To find the indefinite integral, let us do the following substitution
We get
Here we used a couple of common trig rules. But
and
so
This gives us
The graph of the integrand vs. the integral is shown below.
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
The area of the sector OAB, if the angle AOB is α, is
But the angle COA=θ is
So
This means that
So the gray area ABC is
This means that our primitive function could be written as
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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