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Now to

This can be done using various substitutions. We will look at a hyperbolic substitution in the end of this page. It can also be done using the substitution

But that is begging the question, because we basically have to know the answer to come up with that idea. We could also do the substitution

But that will require us to know, or figure out, that

which is quite easy to find. You can get it from

in a couple of steps. Worse is that after the substitution we end up with the necessity to find the integral of sec *θ*, and that is not all too straight forward.

The above can be found in various sources, so we will not go through that here, but instead we will find the answer to our integral using (simple) complex analysis.

**Using complex analysis**

We will use complex analysis, or rather just complex numbers and a bit of calculus in a rather straight forward way. Let us first pull out a factor of -1 from the square root, and then the rule from the previous page, to get

Technically we are done, but it is in a not all too useful form. To fix this we need to look at the arcsine part. We want it to be able to “absorb” the *i*.

To start with, if

then

(We put the constant of integration back later on.) From a previous section we have that

So

or

where

this gives us

or

This we can solve using the quadratic formula to get

Substituting back *e ^{I}* and adding back the constant of integration this will give us

Here we let the constant absorb the ln(*i*). Next comes the question of which sign to use. We may to start with noting that

and thus that

If *z* is larger than one the logarithm is not negative, since

and we can also see that this function is increasing, but less and less fast. The function

should be the derivative of the above logarithmic expression. We can see that is s positive, and decreasing as x>0, and the only option is thus that the integral is the positive variant of the logarithm. So we get

Of symmetry reasons we also have

as shown in the graph below.

**The inverse of the answer**

Let us look at the inverse function

We get

and thus

This is one of the hyperbolic functions. You can find more on those here. So this means that

so

**Using hyperbolic functions directly**

If we happen to know our hyperbolic functions we can do the substitution

We also have that

This gives us

**The above generalized a bit**

Now, we get

or

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