I will now explain how to apply the immediate integrals to solve integrals, at the same time as we use the properties of the integrals.
We will therefore begin by indicating each of the immediate integrals.
Índice de Contenidos
- 1 What are immediate integrals?
- 2 Immediate integrals of simple and composite functions
- 2.1 Immediate integral of a potential function
- 2.2 Immediate integral of an exponential function
- 2.3 Immediate integral of a logarithmic function
- 2.4 Immediate integral of the sine function
- 2.5 Immediate integral of the cosine function
- 2.6 Immediate integral of the tangent function
- 2.7 Immediate integral of a cotangent function
- 2.8 Immediate integral of a sine arc function
- 2.9 Immediate integral of an arc tangent function
- 3 Table of immediate integrals
- 4 Immediate integral exercises solved
- 5 Proposed Exercises
What are immediate integrals?
They are integral that are solved in a direct way, applying their corresponding formula. In order to apply the immediate integrals method, the function to be integrated must be transformed, by means of the integrals properties, so that it is the same as in each of the formulas.
The functions to be integrated can be of two types:
- Simple functions: Where only one x is involved
- Compound functions: Where instead of the x of the simple functions, a function and the multiplication by its derivative appears.
Immediate integrals of simple and composite functions
Next I leave you with all the immediate integrals of simple functions, that is to say, in those where only the x appears. The application of this type of integrals is done directly if it has the same shape as in each formula.
In order to have the same shape as in the formula, the properties of the integrals must be applied (we will see how below when we solve some integrals).
Compound functions are those that have the same shape as simple functions, but instead of an x, a function with more than one term appears and multiplication by its derivative.
To apply these immediate integrals, it is very important to identify that the derivative of the function exists or that it can be reached by transforming the function using the properties of the powers.
I leave you here the immediate integrals of simple and compound functions:
Immediate integral of a potential function
Where “a” must be other than -1.
Immediate integral of an exponential function
Immediate integral of a logarithmic function
Immediate integral of the sine function
Immediate integral of the cosine function
Immediate integral of the tangent function
Immediate integral of a cotangent function
Immediate integral of a sine arc function
Immediate integral of an arc tangent function
Table of immediate integrals
Here’s a table of immediate integrals so you can have them all more at hand:
Immediate integral exercises solved
Which integrals can be solved with immediate integrals?
For all those that can be transformed by the properties of the integrals, so that they remain the same as the formulas of the immediate integrals
The best way to learn to use both the immediate integrals of simple functions and the immediate integrals of compound functions is to practice.
That’s why we’re going to solve a few integrals where we’ll find all the possible cases and I’ll show you how to solve all the problems you may have, at the same time as we apply the properties of the integrals.
Exercise resolved 1
First we separate each term of the function into a new integral, applying one of the properties of the integrals:
Now at each integral, we pull out the constants:
Each of the remaining integrals, although it does not seem so, are solved with the immediate integral of the simple potential function, since the functions are of the form of x elevated to a number:
Nothing should be done to the first integral, but to the second and third integral, we must transform them so that they are in the same way. In the second integral, we raise the power to the numerator and change the sign and in the third, we put the root in its exponential form:
Now we can apply the formula of the immediate integral in each of the integrals:
We’ve already integrated. Now we’re going to operate. We make the sums that we have left in the exponents and denominators:
And finally we multiply the numbers that we have left, we leave all the exponents as positive or we pass them back to the root and add the constant:
Exercise resolved 2
This integral is solved with the immediate integral of the potential compound function:
We know that this is used because we see a function elevated to an exponent.
The function is that which is elevated to the exponent, which in this case is:
And its derivative therefore is:
Both functions are in the integral, in the same way that they are in the formula, so we can apply it without having to transform anything.
We apply it, operate and add the constant:
Exercise solved 3
We have an integral of a compound exponential function, which is solved with this immediate integral:
The function is what is raising the number e:
And its derivative is:
This time, the derivative does not appear in the integral
So what do we do? We can’t go on anymore?
Yes, you can go on. It can be arranged. In this case we are missing the 3 that corresponds to the derivative of the function. Being a constant we can add it (multiplying everything else), as long as we also divide the whole integral by the same constant.
That is, we multiply and divide by 3 so as not to alter the result of the integral:
Now we have the function and its derivative, so we can apply the formula of the immediate integral and it remains:
Exercise solved 4
We solve this integral with the formula of the immediate integral of the compound exponential function, since the 2 is elevated to a function that is:
The derivative of this function is:
That does not appear as such in the integral, since it lacks a 2. As it is a constant, we multiply and divide by 2, so as not to alter the result of the integral:
And now we can apply the formula of the immediate integral:
Exercise completed 5
This integral is solved with the formula of the immediate logarithmic compound integral, since we have a ratio of functions. According to this formula, the function is in the denominator, which is:
The derivative of the function is:
It does not appear as such in the integral but we can transform it so that it appears. First we take out the 5:
And now, we add the 12 we need, dividing by 12 also the whole integral, obviously:
Finally we apply the formula of the immediate integral:
Exercise solved 6
To solve this integral we will use the formula of the immediate integral of the compound sine function.
The function within the cosine is:
And its derivative:
Once again, we don’t have all the factors of the derivative. If you notice, I’ve broken down the derivative by taking common factor at 2, so that the factor (x-1) would show up more clearly. Therefore, we are missing a 2, so we add it and divide the integral by 2:
Now we apply the immediate integral formula to the breast:
Exercise resolved 7
We will solve this integral with the formula of the immediate integral of the compound cosine, since within the sine there is a function that is multiplying to another function that could be its derivative. Let’s go see him.
This is the function that is within the breast:
And its derivative is:
That doesn’t show up in the integral. In order for it to appear, the 3 that divides x³ is left over and the 4 is missing. Therefore, we first took the 3 out of the integral:
And now we multiply and divide by 4:
Now we can apply the formula of the immediate integral and add the constant to it:
Exercise resolved 8
This integral at first glance has no immediate integral form, but it is very similar to the formula of the immediate integral of the compound tangent:
The function that is within the tangent is:
And its derivative:
I have developed its derivative to leave the x in the form of a root. Thus, it is clearer than in the integral, we can obtain this derivative by adding a 2 in the denominator. Therefore, as we add it to the denominator, that is, it is dividing, we multiply the integral by 2:
And finally we apply the formula of the immediate integral of the tangent:
Exercise resolved 9
This integral is clearly seen to be solved with the other formula of the immediate integral of the tangent, where the function is:
And its derivative:
Which should appear in the numerator but does not appear, so we add it and divide by 7:
For the last step, apply the formula of the immediate integral and add the constant:
Exercise solved 10
This integral, if you realize, a priori, has no form of any of the immediate integrals. However, it is clear that we have a function (sen x), which is multiplied by its derivative (cos x).
In other words, we have an elevated function squared away that is:
Whose derivative is:
It is now clearer that we can apply the formula of the immediate integral of the potential compound function, where we have a function elevated to a number, multiplied by its derivative:
Therefore, by applying this formula and operating, we arrive at the result of the integral:
Exercise solved 11
This integral is a similar case to the previous one, since it does not directly resemble any of the immediate integral formulae either.
However, we have a function that is ln x and we also have its derivative, which is 1/x:
That is to say, we have a multiplication of a function by its derivative, so again we have to apply the formula of the immediate integral of the potential compound function:
where this time, the function is raised to 1.
Therefore, applying the formula of this immediate integral and operating remains:
As you can see, to know which formula of immediate integral should be used in each case, you have to look at which of them the integral to solve is similar to and if within the integral you have a function and its derivative or in the worst case, if you do not have the derivative, if you can obtain it by adding or removing constants.
What happens if the derivative of the function does not exist and it is not possible to obtain it?
Then you cannot integrate by this method of integration and you would have to know which one depending on the form of the integral.
Solves the following integrals by the method of integration of immediate integrals: