﻿ Resolution of irrational integrals by trigonometric substitution.

# Resolution of irrational integrals by trigonometric substitution.

Now I will explain step by step how to solve the irrational integrals by trigonometric substitution. I’ll explain it to you step by step while we work out an example.

I recommend that you take paper and pencil and solve it at the same time, as there are many steps and it is very easy to get lost.

Let’s go over there!

## Trigonometric substitutions in irrational integrals. Variable changes

I will begin by indicating what are the changes of variable to be made depending on the radical and the terms we have inside.

These variable changes are called trigonometric substitutions because trigonometric ratios are used in expressions.

We have three cases:

### First case of trigonometric substitution

When in the function to be integrated the square root of a number squared minus a variable x squared appears, i.e. in this way: The trigonometric substitution to be done in this case is to equal the variable x to the number multiplied by the sine of t: Then we derive in both members to obtain the equivalence of dx as a function of “t”: ### Second case of trigonometric substitution

When in the function to be integrated the square root of a number squared plus a variable x squared appears, that is, in this way: In this case, the trigonometric substitution to be made is to equal the variable x to the number multiplied by the tangent of t: To find the equivalence of dx as a function of “t”, we derive from both members: ### Third case of trigonometric substitution

The third case of trigonometric substitution is when the square root of variable x squared minus a number squared appears in the function to be integrated: The trigonometric substitution to be performed is to equal x to the number divided by the sine of t: We derive in both limbs to obtain dx as a function of “t”: There are other ways to solve irrational integrals.

## Resolved exercise of an irrational integral by means of trigonometric substitution

We are going to solve an example of an irrational integral in which we have to make some of the variable changes proposed in the previous section.

We have the following integral: This is the integral of an irrational function, since we have a radical, but we do not have to know what the method of resolution or the change of variable to be carried out is. Other possibilities have to be ruled out first.

If the integral had been this way: It would have been solved in a simpler way, with immediate integrals, but we do not have the 2x and we cannot add it either, so we cannot solve the integral by this method.

Nor can we solve the integral in parts.

So, there is no choice but to use the integration method with changes of trigonometric substitution variables.

In this case we have the square root of a number less x squared. If we look at the cases above, the one that most closely resembles it is the first case, but for that, we need to have a number squared:

We don’t have any high square numbers, so the first key to begin to solve the integral is to retouch it, so that the first term appears squared so that we can make the proposed variable change.

Since we want the number to appear squared, we can put 2 as the root of 2 and square it. Thus, it is still 2 (we have not changed the number) but it already appears squared, which is what we were looking for: Now we can change the variable, where “a” is the root of 2: We replace the new value of x and dx in our integral: Now inside the root we operate and solve the squares: We’ve extracted a factor common to 2, inside the root: And we take the 2 out of the root, which is then multiplied to the root as the root of 2: ### Trigonometric changes to simplify the calculations

At this point, we have to make several trigonometric changes to simplify the calculations.

From the fundamental theorem of trigonometry, we clear cosine from t: And if you realize it, it’s just like one of the radicals we have. Therefore, we replace that radical with cosine of t. In this step I have also multiplied the roots of 2, whose result is 2 and taken the constant out of the integral. That’s how I got it: We multiply the two cosines that we have left within the integral: And we make another trigonometric change, which will make our life easier when it comes to integration: We replace it in our integral: Now we separate the fraction into a sum of two fractions (one for each summand of the numerator and keeping the same denominator) and we put the integral of that sum of fractions as a sum of integrals: We pull the constants out of each integral: And finally, we integrated the two immediate integrals that we have left (well, the 2t cosine integral is almost immediate): We operate to eliminate the parentheses and we’re left with it: We now come to another trigonometric change, to simplify the result, which is as follows: And we replace it in our result: And we operated on the constants: ### Undo variable change

We can no longer simplify the result by making trigonometric changes, nor by operating with the constants.

Therefore, we are going to put the solution in function of x, which is how the result must be given, since we have changed the variable in order to integrate it, but the result must always be given in function of x.

To do this, we must search for an expression in function of x, for each of the terms in function of t, that is, we must put t, sine of t and cosine of t, in function of x and we will obtain this from the change of variable we made at the beginning.

We carry out this variable change: From here we can clear sinus t, which we have in our solution: We already have a sine expression of t as a function of x.

From this last expression, we clear t and obtain its experiment as a function of x It remains for us to find the expression for cosine of t. that we are going to obtain from the fundamental theorem: We clear cosine of t and it’s clear: As we have previously obtained the expression in function of x for the sine of t, we substitute it in the cosine expression and it remains: So we already have the x-dependent expression of the cosine of t.

Now we replace t, sen t and cos t in the solution and finally we come to the solution of the irrational integral in function of x: So far, the integral has been resolved, but we can further simplify the result.

I’ll go on and show you how it’s done. We can solve the square inside the root: Within the root we perform the sum of fractions: We take the denominator out of the root: Now we multiply the remaining fractions in the second term and add + C, because it cannot be simplified any further: We’re finally done! 🙂

As you can see, to solve irrational integrals of this type, several trigonometric changes must be taken into account. The good news is that it’s always the same changes (that’s why I’ve turned them blue) and the procedure is always the same.