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:

integrales irracionales

The trigonometric substitution to be done in this case is to equal the variable x to the number multiplied by the sine of t:

integration of irrational functions

Then we derive in both members to obtain the equivalence of dx as a function of “t”:

how to solve trigonometric integrals step by step

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:

integral by substitution

In this case, the trigonometric substitution to be made is to equal the variable x to the number multiplied by the tangent of t:

substitution trigonometrica

To find the equivalence of dx as a function of “t”, we derive from both members:

integrals por substitucion trigonometrica

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:

integrals de funciones irracionales

The trigonometric substitution to be performed is to equal x to the number divided by the sine of t:

integral replacement

We derive in both limbs to obtain dx as a function of “t”:

integral trigonometric replacement

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:

substitucion trigonétrica integrales

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:

resolution of irrational integrals

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:

which is an irrational function

Now we can change the variable, where “a” is the root of 2:

integral irrational

We replace the new value of x and dx in our integral:

integrate by variable change

Now inside the root we operate and solve the squares:

integral root of x

We’ve extracted a factor common to 2, inside the root:

substitucion trigonometrica

And we take the 2 out of the root, which is then multiplied to the root as the root of 2:

integration by trigonometric substitution

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:

resolver integrales por sustitucion trigonometrica

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:

integrales cambio de variable

We multiply the two cosines that we have left within the integral:

integrals by substitution

And we make another trigonometric change, which will make our life easier when it comes to integration:

integrals by trigonometric substitution examples

We replace it in our integral:

integrals by trigonometric substitution solved exercises

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:

integral by substitution exercises solved step by step

We pull the constants out of each integral:

integrals by trigonometric substitution solved step by step

And finally, we integrated the two immediate integrals that we have left (well, the 2t cosine integral is almost immediate):

integral irrational exercises solved

We operate to eliminate the parentheses and we’re left with it:

integrals irrational variable change

We now come to another trigonometric change, to simplify the result, which is as follows:

integrals de funciones irracionales casos

And we replace it in our result:

irrational integral changes

And we operated on the constants:

integrals irrational cases

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:

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From here we can clear sinus t, which we have in our solution:

integrales de irracionales

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

integrals indefinite of irrational functions

It remains for us to find the expression for cosine of t. that we are going to obtain from the fundamental theorem:

integrales indefinidas irracionales

We clear cosine of t and it’s clear:

exercises of irrational indefinite integrals

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:

integrals of some irrational functions

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:

integral irrational functions

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:

integrals irrational types

Within the root we perform the sum of fractions:

integrals de funciones irracionales ejercicios resueltos

We take the denominator out of the root:

integrales irracionales binomias

Now we multiply the remaining fractions in the second term and add + C, because it cannot be simplified any further:

integrales irracionales

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.