Practical lesson 1. Integration of irrational expressions (part 1)
Integration of irrational expressions
At today’s lesson, we speak about another very important class of functions and how to integrate them. Integration of irrational expressions, which contain roots of some degrees. What should we do? Suppose there is a rational function under the integral of expressions containing roots from a fractional-rational function, from a fractional-linear function. Yes, it is clear that c and d do not turn to 0 at the same time, just like a and b. So, well, a special case is when the denominator of this fraction is 1. What do we do? We introduce a new variable, so that the root expression is to such a power that all the roots are extracted.
What do we do for this? N is considered the least common multiple of the numbers p, q, and so on. These are exponents of the roots. Well, exactly the same in this particular case. Well, of course, the easiest way is to see how it works in practice. Here, look, under the sign of the root is a linear expression 2x + 1 and here is the root to the power of 3, here is the root to the power of 2. It is logical to say that the expression 2x + 1 is t to the power of 6. 6 is the smallest common multiple of 3 and 2 exponents of the roots, right?
So let’s see what happens here. The root to the power of 3, this is squared, it is to the power of 4, and here, too, everything is extracted, fine! You just need to understand what dx is. So, we express x, x is one second t to the power of 6 minus one second, as a result dx is equal, we find the derivative six t to the power of 5, 3 t to the power of 5 dt. That’s it, let’s move on to the replacement. In the numerator of 3 t to the power of 5, we replaced dt by dx. Now be careful, the root to the power of 3 – this will be t squared and another square, t to the power of 4. t to the power of 4 minus the root to the power of 2 is t to the power of 3, fine.
Let’s simplify it. To do this, we take 3 out of the integral, t to the power of 3 is a common factor, we reduce by t to the power of 3, t squared remains in the numerator, in the denominator after we have reduced t to the power of 3, t – 1, dt remains. You can divide by a column, but it is easier to perform the conversion for an incorrect fraction obtained. We divide it slowly, then we write t + 1 plus 1 divided by t-1, well, then we got a completely simple integral, we use the table of integrals, write 3, one second t squared plus t, the last terms give us the logarithm of the denominator modulus, plus c, and then, of course, we need to go back to the variable x. t squared is the root to the power of 3. Yes, we must extract the root to the power of 3.
So, we write: three second roots to the power of 3 of two x plus one, minus, what is t – this is the root to the power of 6, there will be 3 roots to the power of 3 2x + 1, so, well, and plus 3 natural logarithms, instead of t again we write the root to the power of 6 of two x plus one, so, minus one and plus c. So, as you can see, the problem is solved.
Here, under the root sign is a fractional linear function. There’s only one of them, and that’s probably not bad. What are we doing? What replacement? We denote, our task is to get rid of irrationality, we denote the root expression by t squared, so the root is t, the replacement is already there. The problem now is dx. To do this, we must express x from this expression. So, x minus one is xt squared, x is transferred to the left side minus xt squared equals 1, x is 1 minus t squared, this is 1.
So, x is 1 divided by 1 minus t squared, then dx, let’s find dx, it is the derivative, let’s write it like this, 1 minus t squared to the power of minus one, prime dt, it is easier for us to calculate the integral, isn’t it, the derivative? So, we got minus, it turned out to be minus the second power, one minus t squared, squared and the derivative of the internal function minus 2t, so, minus 2t dt, so, as a result, we get 2t divided by 1 minus t squared, dt. So, we go to the integral, this is t, and here 2t, 1 minus t squared, squared, dt. Let’s remove this example because it turns out too difficult, we will get bogged down, that is, I think it will be easier to use example 2. I clean up. As if everything is correct, everything is fine, but it turned out to be cumbersome.
Well, now. We will not just consider it. The next method, the second one, how to get rid of irrationality, turn to trigonometry. This applies to those cases when under the integral you see these roots: x squared and a squared, in different combinations. In this case, we rely on trigonometry formulas. Look, here it is, the diagram, how it can be used, formulas to further apply the trigonometry formulas. So, let’s take an example. Look, to use one of the formulas, I’ll take x – this is two sine t that appears under the root, let’s carefully write out: 4 minus x squared, this will be equal to 4 minus 4 sine t squared, we take 2out before the sign of the root, the basic trigonometric unit appears under the root, which gives us the cosine t.
So, dx is a derivative 2 cosine t and do not forget the multiplier dt. We turn to the integral again. The first factor gives us 2 cos t and dx 2 also gives the cosine of t, and dt. So, we take 4 out of the integral, under the integral, cosine t squared dt. So, we got the integral of the trigonometric function, and then, you see, the problem is whether we come to a fractional-rational function using substitution, or to a class of integrals that we have an idea of how to solve them, we know the methods. This is where we apply the downgrade formula. So, this is 1 plus the cosine of 2 t in half, and here we reduce by 2, we get this expression. Then we find the integral of each term, the integral of the cosine will be the sine, so, plus.
So, well. So, we get 2 t the integral of cosine is sine with a coefficient of one second, we get plus the sine of t plus c, and then remember what t is, t is associated with x through this formula, so we have to express here, we have x in half, it is the sine of t and t obtained is the arc sine of x in half, so it will be 2 arc sine of x in half, it is certainly possible to express the sine of 2 t, but we are not going to do it through x, so plus sine of two x arctinus in half, let’s leave it like this. And another approach, how to get rid of irrationality, how to remove the roots in the integrand.
There are three Euler substitutions. So, look, so, they relate to the case when under the sign of the integral there is a rational function with respect to x, and the roots of the square trinomial. The first ones, they are not associated with the roots, the first two and the third substitution are associated with the root of the square trinomial. So, let’s consider an example. So, what do we do? We denote the root t minus x. What is my guide when choosing an Euler substitution? Well, what we have, look, with this replacement, we have x plus the root, it is t, the denominator immediately becomes the variable t, the whole problem is how to find dx. What do we do for this? This expression is squared, converted, squaring the right and left parts, t squared minus 2 tx plus x squared, so x squared on the right and left are eliminated and x, in general, is easily expressed, let’s see, 2 tx is t squared minus 1, and therefore, x is t squared minus 1 divided by 2t.
So, here the main task is to find x and find dx, so we find the derivative of the fraction, the derivative of the numerator 2t is multiplied by 2t, this is 4t squared minus the derivative of the denominator, we multiply 2 by the numerator 2t squared plus 2, so the denominator is squared, 4t squared dt. So, what do we get? 2t squared plus 2, it is clear that you can reduce by 2, it is t squared plus 1 divided by 2 t squared dt, that’s what we have here, so, go to the integral. Dx is the expression t squared plus 1, 2 t squared, and in the denominator we still have t. Well, I’ll write it like this, dt. Let’s take one second out of the integral, leaving t squared plus 1 divided by t in the cube. Divide the numerator by the denominator, 1 divided by t, plus, lets write t to the power of minus three, so that we can integrate more conveniently, dt.
So, one second, the first term gives the logarithm of the module t, the second term we adds 1, we have minus 2 and divide by minus 2. So, it became minus one fourth of t to the power of minus 2, Yes, plus c. All that’s left is to go back to the x variable. So, t is x plus the root of x squared plus 1. So, we write the answer, one second is the natural logarithm of the modulus of this expression, the answer is a little cumbersome, yes, well, we can’t do anything about it. So, minus one-fourth, yes, and the denominator will still be the square of this expression x plus the root of x squared plus 1 squared, well, plus c. That’s it. The problem has been solved. So, here we have x squared, right? So, the problem is solved.