\(\displaystyle \int (x^{2} + 2x - 1)^{3} 2x dx\)
\(\displaystyle \int(x^2+2x-1)^3=x^6+6x^5+9x^4-4x^3-9x^2+6x-1\) What happened to 2xdx?
\(\displaystyle \int 2x(x^6+6x^5+9x^4-4x^3-9x^2+6x-1)\) What happened to dx?
\(\displaystyle \int 2x^{7} + 12x^{6} + 9x^{5} - 8x^{4} - 18x^{3} + 12x^{2} - 2x\) What happened to dx?
You will notice that when I started manipulating, I did it without reference to integration operators. I was just saying that
f(x) = g(x) to take advantage of this theorem:
\(\displaystyle f(x) = g(x) \implies \int f(x)dx = \int g(x)dx.\)
Integrating term by term:
\(\displaystyle \int \frac{2x^{}8}{8} + \frac{12x^{7}}{7} + \frac{9x^{6}}{6} - \frac{8x^{5}}{5} - \frac{18x^{4}}{4} + \frac{12x^{3}}{3} - \frac{2x^{2}}{2}\)
The integral sign is no longer necessary above because you have ALREADY integrated term by term. Also, being very technical, there is a constant term added for EACH indefinite integral. Now when you add up a bunch of constants, you get a new constant so you can add the big C as the sum of all the constants, but, to be exact, you need the + C at this step. Do you understand why there is a constant associated with each integral?
Reduced Fractions and added \(\displaystyle + C\)
\(\displaystyle \frac{x^{}8}{4} + \frac{12x^{7}}{7} + \frac{3x^{6}}{2} - \frac{8x^{5}}{5} - \frac{9x^{4}}{2} + \frac{4x^{3}}{1} - \frac{x^{2}}{1} + C\)
But when do we use this method of integration as opposed to the other ways?
We use this kind of brute force method only when we cannot find an easier way. A lot of what you are being taught are ways to avoid brute force methods, which are time consuming and very error prone. It is not uncommon in math for there to be several equally valid ways to solve a problem so it is worthwhile looking for alternatives to brute force. So, as Subhotosh Khan said, we NEVER use brute force IF we can find a quicker, less error-prone method.
This post is quite repetitious. Let's consider problem 1: evaluate \(\displaystyle \int (x^2 + 4)^5(2x)dx.\)
We could use the brute force method by raising (x^2 + 4) to the fifth power and multiplying every term of that monstrosity by 2x, giving us an 11th degree polynomial to integrate. I'd be likely to make a mistake so I would have to check my algebra several times. Let's see if there is an easier way. One of the basic integrals is:
\(\displaystyle \int (u^n)du = \dfrac{u^{(n+1)}}{n + 1} + C.\)
In the problem at hand, I see an exponentiated expression and so think to myself: "I wonder if I can use that basic power integral." I do not know whether that approach will work. I showed you an example where it did not work. But trying it certainly seems an economical use of my time and effort. So I make a substitution, either implicitly or explicitly. Once you understand substitutions, it is quicker (but more error prone) to do them implicitly. To understand what you are doing and to reduce the chance of errors, explicit substitution is better. Doing so explicitly, I say
\(\displaystyle Let\ u = (x^2 + 4) \implies (x^2 + 4)^5 = u^5 \implies (x^2 + 4)^5(2x)dx = u^5(2x)dx.\)
Well, that looks closer to what I want, but I have that 2x hanging around, and I cannot integrate (u^5)dx. I need (u^5)du to be able to integrate. OK. No need to despair. The u is a function of x so I can compute du/dx.
\(\displaystyle u = (x^2 + 4) \implies \dfrac{du}{dx} = 2x \implies du = (2x)dx.\)
\(\displaystyle So\ u = (x^2 + 4) \implies \int (x^2 + 4)^5(2x)dx = \int (u^5)du = \dfrac{u^6}{6} + C = \dfrac{(x^2 + 4)^6}{6} + C.\)
That is a whole lot easier than messing with an 11th degree polynomial. Furthermore, instead of memorizing an extremely limited rule about right-hand linear factors, you have used a fairly straight forward logical process that you can use over and over again in many different situations. If you see an expression that is exponentiated, think about using the power rule through an appropriate substitution.
Let's look again at problem 2.
Evaluate \(\displaystyle \int (x^3 - 3x)^{(1/2)}(x^2 - 1)dx.\)
There is probably a brute force way to solve this, but I do not know offhand what it may be and I am reasonably confident that it will be time consuming and error prone. Moreover, there is no right-hand linear factor so the "rule" you thought you saw in problem 1 does not apply. But again there is an exponentiated expression. So it is worth making an effort to see whether the power rule will be useful. Let's try.
\(\displaystyle Let\ u = (x^3 - 3x).\) Now we learned before that, in integration, a substitution from x to u will also require a substitution from dx to du.
\(\displaystyle u = (x^3 - 3x) \implies (x^3 - 3x)^{(1/2)} = u^{(1/2)}\ and\ \dfrac{du}{dx} = 3x^2 - 3 = 3(x^2 - 1) \implies du = 3(x^2 - 1)dx.\)
This is looking promising, but I do not have 3(x^2 - 1)dx. Instead, I have (x^2 - 1)dx = (1/3)du. So I am not quite there BUT
\(\displaystyle u = (x^3 - 3x) \implies \int (x^3 - 3x)^{(1/2)}(x^2 - 1)dx = \int u^{(1/2)}\left(\frac{1}{3}\right)du = \int \left(\frac{1}{3}\right)\left(u^{(1/2)}\right)du.\)
There is a theorem that comes to the rescue: \(\displaystyle a\ is\ a\ constant \implies \int \{a * g(u)\}du = a * \{ \int g(u)du \}.\)
So \(\displaystyle u = (x^3 - 3x) \implies \int (x^3 - 3x)^{(1/2)}(x^2 - 1)dx = \int \left\{\frac{1}{3} * \left(u^{(1/2)}\right) \right\}du = \left(\dfrac{1}{3}\right) * \{ \int \left(u^{(1/2)}\right)du \} = \dfrac{1}{3} * \left(\dfrac{u^{(3/2)}}{\frac{3}{2}} + K\right)= \)
\(\displaystyle \dfrac{1}{3}* \left(\dfrac{2u^{(3/2)}}{3} + K\right) = \dfrac{2u^{(3/2)}}{9} + \dfrac{K}{3} = \dfrac{2u^{(3/2)}}{9} + C.\)
To UNDERSTAND what you are doing, explicit substitution is better. Furthermore, you need to memorize only a handful of theorems and maybe two dozen basic integrals to evaluate the integrals that you are likely to see in a beginning course if you recognize that you are frequently doing substitutions, either implicitly or explicitly.
I apologize for the repetition, but I hope that a somewhat different set of words will help you grasp what I am trying to say.
No, really I plan studying a little more integration, and then harshly reviewing Algebra and Trig over the month of December, and then back into Calculus. I had done some review in Algebra and Trig during Sept and October.