Математика. Материалы курса

     

Our class is devoted to calculating the indefinite integral.

There is a method, a general method: the method of substitution or replacement of a variable. This method includes a technique that we call adding a function under the sign of the differential.

What do we mean? Imagine that there is some expression under the integral sign, we write R(f(x)).

What does it mean? This is an expression where the argument is the function f(x). Also, the derivative of this function, f‘(x), is a separate factor in the entire integrand.

It is really important to be able to notice this.

What should we do? We introduce a new variable f(x), and then f‘(x) dx is the differential of this new variable. The form of the integral becomes much simpler, and we use some new methods for it.

The purpose of this transformation is to simplify, to reduce this integral either to a fractional-rational function, or to some type of integrals that we have already learned to solve.

Let's take some examples. I can reveal you a little secret: how can we learn to calculate the integral, what do we need to know?

You need to know the tables – the table of derivatives and the table of integrals. The calculus of integrals is not related to information search, it is related to the use of these tables.

We should see that there is a function and its derivative. There is no task to find the derivative. You should just see this pair: the function and its derivative. This is a very important point.

Let's take the first example: here is the problem. How can we see it? There is a function - the natural logarithm of x, and there is a multiplier - a unit divided by x.

It is not highlighted as a multiplier here, and it is the derivative of the natural logarithm. That's what you should take into consideration.

We note that the first multiplier contains the logarithm function x, and the second multiplier is the derivative of the logarithm x.

We take the logarithm of x as the variable t, and then the integral is transformed.

It becomes much simpler.

This is almost a tabular integral.

We calculate the integral from 1 - this is t, from t^2 - this is 1/3 t ^3 plus c (don't forget!) and we return to the variable x, t is the logarithm of x. Here is the final answer.

Let's take another example.

Any transformations are almost impossible here.

What type of integral is this?

We should notice (what do we see?) the function of the arc sine.

There is also a derivative of this function.

It is sufficient to transform the integrand as follows.

The second multiplier here is just the derivative of the arc sine, so we take the arc sine x as t.

That's it! The integral is converted to the simplest form: the root of t, we write t to the power of½. The table integral of the power function. We apply the formula, return to the arc sine x, and we do not forget about t. It is easy to write down it as a root. That’s it! This is the answer.

A specific case: you have a function of the variable e under the integral to the power of x and then there is nothing, and then dx.

It is easy to reduce it to the previous case: we can divide by e to the power of x and multiply by e to the power of x, and then we take e to the power of x as t. The integral is transformed to a new form, where there is no exponential function, and the problem may become much simpler.

Let's take an example.

Here is dx divided by e to the power of x minus 1. It seems to be a simple integral, but it is not tabular. How can we solve it?

It is quite easy to transform.

It's pretty simple. We see that the integrand is a function that depends on e to the power of x.

What should we do?

We divide by e to the power of x and multiply by e to the power of x.

If there were a multiplier e to the power of x, we wouldn't even do this, we would immediately write that t is e to the power of x, then dt is e to the power of x multiplied by dt.

We consider the new variable t and get the integral of a fractional-rational function. 

There are different ways to calculate this integral. Let's perform the calculation using identical transformations.

What should we do? We write in the numerator one minus t plus t instead of 1 and then divide the numerator by the denominator. We divide one minus t by the denominator, we get minus one by t and divide the second term by reducing it by t, and we get a fraction: one divided by t minus 1

Then these are practically table integrals. The first one is just tabular, and we put t minus 1 instead of t under the differential sign for the second integral.

That’s it! We use the table integral, which is the natural logarithm of the module t, then we return to the variable x and perform simple transformations.

We get the answer.