Previous page : The integral of one over sqrt(x^2+1) dx

Next page : The integral of sqrt(x^2-1) dx

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 COB=*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.

Previous page : The integral of one over sqrt(x^2+1) dx

Next page : The integral of sqrt(x^2-1) dxLast modified: