00:01
We have two parts.
00:03
Part a, use the trigonometric substitution x equals 5 tangent of theta to calculate the indefinite interval from of 1 over x squared plus 25 to the 3 half, and in part be calculated the improper integral of the same function over the real numbers, that is from negative infinity to plus infinity of 1 over x squared plus 25 to the 3 half, or show that that improper integral diverges.
00:36
So let's calculate part a.
00:43
And in part a, doing this change of variable, x equals sign tangent of theta, which is defining the new variable implicitly using x and the new variable being theta.
01:00
So that means that differential of x is five times remember the derivative of tangent of theta is second square of theta and that differential of theta so this is differential of x in terms of differential of theta and doing that we are going to calculate the denominator of the expression we are integrating that is the expression over here x squared plus 25 to the three half so x square plus 24 5 raised to the 3 half means it's the same as x square plus 25 to the one half and that raised to the 3rd.
02:06
It's equivalent.
02:08
And so this is the same as square root of x square plus 25, which is x square plus 25 to the 1 half, and that to the third.
02:18
So this is the equivalent expression to the given expression over here, which is the denominator in the intro we are calculated.
02:30
Okay, so this now is equal to square root of, and now x square is 5 tangent of theta all that square, plus 25, and all that to the third, and that is square root of 5 times tangent square.
02:58
Is 25 tangent square of theta plus 25 to the third and that is we can take a common factor 25 inside square root and we get 25 times tangent square of theta plus 1 and that to the third and that is we separate the square root of the square root of each factor in this product so we get square root of 25 times square root of tangent square of theta plus one, that product to the third, and then we get five times.
03:52
And we remember that using the fundamental trigometric identity, which says the sine square of theta plus cosine square of theta is equal to one, dividing both sides of that identity by cosine square of theta, we get that tangent square of theta plus 1 is second square of theta.
04:11
And when we get the square root of that, we get second of theta.
04:16
And for that, we suppose we are in a range in a domain of theta where that function second of theta is positive to get rid of the absolute value, then normally should be put here.
04:31
And that is 5 to the third times second to the third of theta.
04:40
And so this expression turns into this one when we do this change of variable over here.
04:48
Now we can calculate the indefinite integral we have been given that is the integral of 1 over x square plus 25 to the 3 half is equal differential of x is equal to the integral off and now in the numerator we have one differential of x differential of x is we calculated up here is 5 second square of theta differential of theta and now that divided by and all this pressure in the denominator is just this one here 5 to the third times second cube of theta and we can see we have a lot of simplification here and we have a 5 in numerator and 5 cube in the denominator we get 5 squared in the denominator only at the end and 2nd square in the numerator cancel with second tube the denominator and we end up with 1 over 5 square times second of theta differential of theta and that is 1 over 25 which is a constant and get out of the definite integral times of of 1 over ccquence of 1 over 2nd of theta but remember second of theta is the reciprocal of cosine of theta so we end up with cosine of theta in the interim function sorry cosine which is the reciprocal of second of theta so we get this okay let me arrange that for a moment cosine of theta and the integral of theta is sine of theta so we get sign of theta over 25m because we have ended, we have calculated completely the indefinite integral, we got to add a constant of integration, let's call it c.
07:11
But now we want to express the results in terms of the original variable x.
07:17
We got to do that each time we are calculating an indefinite integral.
07:22
We have to give the answer in terms of the original variable.
07:27
And to do that, typically in this type of exercises when we do a change of a variable of trigonometric nature.
07:39
What we do is to draw a right triangle and put one of the angle, the acute angles, equal to theta to the variable we are defining.
07:51
And we based upon the change of variable we define, that is x is 5 tangents of theta from this change of variable we define, or we that we are using can be deduced a tangent of theta is equal to x over 5, simply by solving this equation for tangent of theta.
08:25
And that means that if we use the definition of tangent of an angle in this right triangle, tangent of theta will be the opposite side to theta, this one here, over the adjacent side.
08:41
To theta that is the one here.
08:44
And because we know that should be equal to x over 5, so it's sufficient to put this opposite side to 5 and the adjacent side to the opposite side to x and the adjacent to 5.
08:58
Because when we calculate the tangent of theta in the right triangle using the definition of tangent of theta, that will be the opposite leg to theta that is x over the adjacent leg to theta that is 5.
09:11
Get just this, which is the change of variable we made.
09:18
And now this other side here is the hypotenuse of the triangle, and we use the pythagorean theorem to get this square root of the sum of the squares of the legs that is x squared plus five square that is 25.
09:34
And having this triangle, we can calculate the function we want of theta that is sine of theta, through our metric function of sine of theta and that by using definition of sine of theta in the right triangle here.
09:51
That is the opposite leg to theta, that is x over the hypotenuse square root of x2 plus 25.
10:04
And that's the way we do these calculations.
10:11
And now we can go back here and set this is equal to.
10:16
And we put instead of sinus theta we put this expression so we get x over 25 times square root of x squared plus 25 plus a constant of integration that is the interval of 1 over x square plus 25 to the 3 half differential of x in proper the indefinite integral here is equal to x over 25 times square root of x squared plus 25 plus constant of integration c.
11:14
This is a result of part a which will be using in part b.
11:23
Now in part b we want to talk about the convergence or divergence of the intro from negative infinity to plus infinity of 1 over x squared plus 25 to the 3 half.
11:41
That is the same function we are integrating in part a.
11:47
And by definition, this is equal to the sum of these two improper integrals, provided they both converge.
11:56
The negative infinity to 0, which we can select another value.
12:01
It doesn't have to be 0 necessarily, but i'm going to show 0.
12:05
You can choose another value if you want it's going to be the same of 1 over x squared plus 25 to the 3 half differential of x plus the interval from 0 to plus infinity of 1 over square root plus sorry x square i meant x square plus 25 to the 3 half differential of x that is we are putting the improper integral here in terms of two other improper integrals and provided both improper intervals here converges, this one will also converge.
12:59
And we are going to prove another thing here is that this improper integral converges to the same value that this improper integral converges if, of course, convergence occur, that is, we only get to prove that, for example, these improperly into converges because in that case, this will also converge to the same value that converges the other one.
13:29
To see that, we are going to, first we are going to state what i'm saying here.
13:37
I'm saying that if the integral from zero to plus infinity of 1 over x squared plus 25 to this, three half that is this improper interval here converges to a number v let's say a real number then the other integral let me get rid of this error for a moment then the other interval from negative infinity to zero of one over x square plus 25 to the three half also to the same number.
14:46
That is, it converges and converges to the same number that this other improper integral.
14:53
That's why if this expression is true, this implication is true, then we only need to verify that this improper integral here converges.
15:03
So let's see this is true this statement here.
15:10
And for that, let's say that the integral from negative infinity to 0 of 1 over x squared plus 25 to the 3 half is defined as the limit when n goes to negative infinity of the integral from n to 0 of 1 over x square plus 25 to the 3 half differential of x as definition of the improper integral here.
15:50
And now we can do a change of variable here in this definite integral we have around here, and this change of variable we will be using is u equal negative x.
16:04
And from here, differential of u is negative differential of x.
16:09
And from here differential of x is negative differential of u.
16:13
The other thing we got to verify is which are the new limits of integration.
16:18
That is x equal n which is the low limit here means that u is equal to using this expression here you got to be negative n and if x is zero which is the upper limit of the integral we have here then u is also equal to zero but if we put zero in this definition around here u is equal to zero also so we get that and now we can let me arrange this this way to know that this is what justify what i'm going to write at this moment here this is the limit when n goes to plus to negative infinity sorry of the integral from the new limits are negative n and zero then we have one over x square but if we solve this expression here for x x is negative u so we get negative u or that square plus 25 to the three half and here we have let me put that over here not to get clover here so it's negative differential of you because of this year differential of x is negative differential of view so this negative okay let me arrange this a little bit to be clear here negative differential of view and this negative here can be get out of the integral so this is limit when n goes to negative infinity of the integral negative out, this negative get out of the interval and we get inter for negative n to zero of differential of u over.
18:43
Now negative u, all that square is the same as u square because we are squaring also the negative sign plus 25 to the three half.
18:56
And when we have a negative in front of an integral, in front of a definite integral, we can change, we get rid of the sign and get and interchange the limits of integration...