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Integrals of Inverse Trigonometric Functions

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With the general formulas for the derivatives of inverse trigonometric functions, we can use integration by parts with substitutions to derive the corresponding formulas for their integrals.
Integrals of Inverse Trig Functions
Integrals of inverse trigonometric functions can be challenging to solve for, as methods for their integration are not as straightforward as many other types of integrals. However, knowing the identities of the derivatives of these inverse trig functions will help us to derive their corresponding integrals. We will, therefore, need to couple what we know in terms of the identities of derivatives of inverse trig functions with the method of integrating by parts to develop general formulas for corresponding integrals for these same inverse trig functions.

Derivatives of Inverse Trig Functions
At some point, you may have seen the following table that depicts derivatives of inverse trigonometric functions:

Derivatives of inverse trig functions table

Integrating Inverse Trig Functions
We can use these inverse trig derivative identities coupled with the method of integrating by parts to derive formulas for integrals for these inverse trig functions.

The Integral of Inverse Tangent
Let’s first look at the integral of an inverse tangent.

Let’s use the inverse tangent tan-1 x as an example. Given the formula for the derivative of this inverse trig function (shown in the table of derivatives), let’s use the method for integrating by parts, where ∫ udv = uv – ∫ vdu, to derive a corresponding formula for the integral of inverse tan-1 x or ∫ tan-1 xdx. Look again at the derivative of the inverse tangent:

dertan

We must find corresponding values for u, du and for v, dv to insert into ∫ udv = uv – ∫ vdu.

As we wish to integrate tan-1 xdx, we set u = tan-1 x, and given the formula for its derivative, we set:

du

We can set dv = dx and, therefore ,say that v = ∫ dx = x.

Now we have all the components we need for our integration by parts. Substituting our corresponding u, du, v and dv into ∫ udv = uv – ∫ vdu, we’ll have:

inttan

The only thing left to do will be to integrate the far-right side:

far

In this case, we’ll have to make some easy substitutions, where w = 1 + x2 and dw = 2x dx. By setting up the integral as follows:

int2

and then integrating this and then making the reverse substitution, where w = 1 + x2, we have:

int3

 
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