Lecture. Indefinite integral: methods of integration.
Indefinite integral: methods of integration.
The lecture is devoted to methods for calculating the indefinite integral. We will consider three main methods: the first is direct integration, the second is the method of substitution or replacement of a variable, and the third is the method of integration in parts. Let’s look at each of these methods in more detail.
Identity transformations are the first part, and we start integrating with it. Then we use the properties of the indefinite integral and the tables of integrals.
Let’s consider the following example. The task is to calculate an indefinite integral. What do we pay attention to? There is no such an integral among the tabular ones, so we will try to use the direct integration method. To do this, we transform the integrand by dividing each term of the numerator by the denominator and get the integrand – this is the sum of three functions, and then we apply the properties of the indefinite integral – the linearity property.
We get a linear combination of integrals from table functions. All we need is to apply these formulas. Remember to write +C at the end since the answer is the family of all primitives.
We consider the second example related to a trigonometric function – the integral of tg^2(x). there is no such a tabular integral. To calculate the integral, we perform identity transformations – first, we replace tg^2(x) with the quotient sin^2(x)/cos^2(x). We transform the numerator sin^2(x) using the basic trigonometric identity, and then we divide the numerator by the denominator again. After that, we substitute the resulting expression instead of tg^2(x), apply the properties of the indefinite integral, and look at the table of integrals. Don’t forget to write +C. The problem is solved.
Let’s move on to the second method – the substitution method or the variable replacement method. This property of the indefinite integral is based on the invariance properties. This word comes from the Latin invariantis, which means immutability, independence from anything.
So, this property is formulated as follows: if you know one of the forms (from formulas) of an indefinite integral, you can use it, replacing the integrand, and the answer, and the independent variable in all places, what is the most important, with any other expression containing a single variable.
So, the formula of an indefinite integral remains unchanged, the dependent variable or independent variable is in this integral.
In this theorem, we write the condition: the integral of the function f(x) is given. It is equal to F(x)+c. Remember that this is equivalent to the fact that the derivative of the function F(x) equals f(x). We need to prove that the formula is valid if we substitute a new expression u instead of the variable x. By the properties of an indefinite integral, this is equivalent to the fact that the differential F(u) is equal to the integrand on the left side. Let’s prove it. We calculate the differential F(u). According to the differential formula, this is the derivative of this function multiplied by the differential of the independent variable. Remember that u is a function of the variable from x.
Next, we use the formula for the derivative of a complex function and get the last expression written here. Remember that F’(u) is f(u) and by the formula of the differential u’dx is the differential of the function u. As a result, we get f(u)du – integrand. The formula is proved.
This method is implemented in two ways: by introducing or without introducing a new variable.
Let’s consider the first method without replacing the variable. In this case, we use the formula for the differential of a function and any functions under the sign of the differential, in fact, we use this same formula in reverse order: first, we write the right side, then the left one. And we say that the function f(x) is introduced under the sign of the differential. In addition, we use the simplest properties of the function differential. We remember them from the course of differential calculus.
Let’s take the example of adding functions under the differential sign.
For example, dx. If we multiply by½, multiply by 2, and then add 2 under the sign of the differential, and then add a constant under the sign of the differential, it turns out that dx can be replaced with the differential by the linear expression 2x+3. Function 2x+3 was introduced under the sign of the differential.
Another example is cos(x)dx.
Cos(x) is a derivative of sin(x) (cos(x)=sin’(x)), so the function sin(x) can be entered under the sign of the differential. Let’s see how this is used when solving tasks.
We calculate the indefinite integral of cos(2x+3). We have just seen how the expression 2x+3 is introduced under the sign of the differential. We use this knowledge. Then the integral takes on a new form. 1/2 can be taken out of the sign of the integral – a constant multiplier, and then we use the invariance properties of the indefinite integral. As a result, it is a tabular integral, only under the integral sign (under the sine sign) is the expression 2x+3, plus c – arbitrary constants.
The second method is to replace a variable. In this case, the new expression is replaced with some variable, usually u, t, z, and so on. We see that the integral would be tabular if instead of 2x+3 it were x. We denote this expression with the new letter t. So the new variable is t=2x+3. To use the variable substitution method, we express the variable x from this expression, x becomes a function of the variable t, and we calculate the differential of this function using the differential formula – the derivative multiplied by dt. We see that dx is 1/2dt.
Let’s replace it with a variable, note that there is only one variable t under the integral. Then we take ½ out of the indefinite integral, and we get the table integral. We calculate it. And the next important step, please note that the answer must contain the same original variable x that was given in the task, so we replace t with 2x+3 and get the answer.
The second example.
Indefinite integral of xcos(x^2).
Method 1 is the introduction. We use the introduction of the function under the sign of the differential. We note that x is a derivative of the function x^2 with a coefficient of 1/2, so the function x^2 can be entered under the sign of the differential. So, xdx is 1/2 of the differential from x^2. We make the insertion under the sign of the differential, take out 1/2 out of the sign of the integral and get the table integral. We write the answer.
Let’s consider solving this task using the second way by replacing a variable.
We replace x^2 with the new variable t. In order to perform substitutions of x and dx, we must express x and calculate dx. So, x is the root of t, dx is the differential of the function root of t.
We calculate and get this dt divided by 2 roots of t. We make substitutions, perform the transformation, take 1/2 out of the sign of the integral, and get the table integral. Don’t forget to go back to the original variable x. And we get the final answer.
We consider the third method – the method of integration in parts.
This important theorem is stated as follows: if there exists an indefinite integral of the product of U and dV, there is an indefinite integral of the product of U and dV, which is equal to the difference between UV and integral VdU. This formula is called the formula for integration by parts.
Let’s consider the proof, it is not complicated. First, we use the formula for the differential of the product of two functions, then we integrate both parts of this equality – the right part is transformed to the sum of integrals. And then we replace the left part with UV using the absorption law of the indefinite integral. And then all we need is to express the integral UdV. We get the proved formula.
The piecemeal integration method is used if the integrand function is a product of some functions that are not related to each other by differentiation, such as x and e^x, x and cos(x), x and ln(x), etc. This method is also used for integrating elementary functions that are not included in the table of integrals. These are usually inverse logarithm functions, such as arccos(x), arcsin(x), arctg(x), arcctg(x), etc.
Let’s consider the following example: you need to calculate the indefinite integral of the function xcos(x). To use the formula, we write it out.
We see that the product UdV is to under the sign of the integral. We select any of the functions as U. There are some nuances here that we will not consider now. So, let’s choose the function x as the function U. The remaining multiplier of the integrand containing dx is expressed by dV.
To use the formula, we need to know the function V and the differential of the function U. We calculate dU by the differential formula, and restore V by integrating the function cos(x). So, dU is dx, and V is sin(x). Lets apply the formula. We multiply U by V – this is xsin(x), then we write the minus sign and under the sign of the integral sin(x) – this is V and dU – this is dx. We see that in the right part of this equality, the integral sin(x) dx is tabular. So the next step is to write down the answer. The integral is calculated.