Math.

Math is math, I've learned so much of it in the past semester it's mind boggling. I remember when I left high school looking at factoring a quadratic was scary, now I factor cubics without even blinking. I used to look at binomial expansions as impossible, now I can do them in seconds. I used to think I would never be good in algebra, but now I've made A's in everyone of my calculus classes. I've mastered trigonometry, algebra, and calculus - and it seems like a skill everyone should possess. I'm already learning the tools of the trade, that possessing skills equates to higher earned wages. I'm going to be applying for a math tutor at Jester learning center on campus next semester with my teacher recommendation. I'll be able to tutor calculus 1 and 2 students which to me is supremely easy, although calculus 2 is harder than calculus 3 in my opinion.

See the thing about integrals is that it's a process: you know that every integral they give you can be integrated using elementary functions and the ones you can't you can express as infinite series.

The type of integrals they give you at this school are nasty, but they can be done. The process which I want to teach the young ones is simple and mathematical, and worked for me when I was in their shoes.

First: expect no easy one step integrals... the integrals they give will always contain at least 2 to three steps of changing the integrand before you can make a valid integration.

so my steps are as follows:

Go through a process of thought with each integral.
1.) look at the integrand and see if there is a possible u-substitution (simple one) that strikes you immediately.

2.) If the u-substitution fails or is a weak one i.e. it produces several u's instead of one u, try integration by parts to see if that is viable.

3.) if your integrand isn't expressed or can't implicitly be expressed as a product of functions you have to move on to trig substitution, this is generally used when there is a square root sign on the denominator - however, it can also be used if the function is on the numerator but it would then involve a further step of integration by parts.

4.) if the function is an expression of 3 to 4 term polynomial on both numerator and denominator it almost always is an integration by partial fractions problem which is the easiest you're going to get at this school. First use polynomial long division where applicable and then split the fraction into a sum of individual fractions using algebra.

5.) if all these steps fail or you're a little bit off from being able to accomplish a step, then you know a more complex abstract substitution is involved, or completing the square is involved - after a while you can usually gain a feel for what tricky abstract thing they require...

for example:

1
-------------------------------
sqrt (x^2 - 4x -5)

The progression of steps to solve this integral:

1.) Does a u-substitution help? Well... I could try a variety of u-substitutions, but my du term will always contain a linear term that can't be factored.

2.) integration by parts? You would just get caught, because this is obviously not a product of two functions.

3.) partial fractions seems tempting but the sqrt on bottom locks up your quadratic.

So you know something special has to take place, well the only special thing that could happen in this problem is completing the square, take the linear term 4x - take the coefficient, 4 divide it by two and then square the result = 4 ... so now add 4 and subtract 4 on your denominator (so you don't change the equation)

x^2 - 4x -5 + 4 - 4

always group the positive added in your quadratic, because only if the second term in your quadratic is positive can both your signs be the same in the factored result which is what you want.

(x^2 - 4x + 4) - 9 (combining the -5 with the -4 and the + 4 with the quadratic)

(x-2)^2 - 9

so this is factored

sqrt (x-2)^2 - 9

This looks like a sexy trig substitution. Once you do enough of these you instantly relate the identity

sec^2 - 1 = tan^2

so now make the "X" substitution x-3 = 3 sec x note: x substitution is way different than u substitution:

so your dx now becomes = 3 sec x tan x directly

anyways we'll concern ourselves with that later continue on to your denominator:

sqrt (3 sec x) ^2 - 9

distribute the ^2 inside the parentheses since all you have is products

sqrt (9 sec^2 x - 9 )

factor out the 9

sqrt (9(sec^x - 1 ) = = sqrt (9 (tan^2) ) == 3 tan x !! good good

now since your dx = 3 sec x tan x that is multiplied so it goes on the numerator

3 sec x tan x
-------------
3 tan x

cancel out the threes and tan x

== sec x

so now you've algebraically manipulated your integrand into an a function that can be integrated

however, the integral of sec x is not trivial. Again, it involves some dirty algebra: since sec x = 1/cos x if you make a u-substitution cos x you're still left with a -sin x factor that isn't expressable in terms of u

so:

you're going to have to add some things to this function:

sec x * sec x + tan x
... --------------- <--- this fraction is simply just 1
sec x + tan x

then you get


sec^2 x + tan x sec x
----------------------
sec x + tan x

you should immediately see this form as du/u since your numerator is exactly the derivative of your denominator

u = tan x + sec x

du = sec ^2 + sec x tan x dx

du
----------- = dx
sec^2 + sec x tan x

This cancels out your numerator so now you're left with

1
----------
u

the integral of that is trivial

Ln (u)

put u back into terms of x

Ln( sec x + tan x ) + C

and you're done! Now make a mailbox in your brain and whenever you see a function of 1/ sqrt (quadratic) follow these steps for the solution as this applies to almost all circumstance of this function.

I can't wait to tutor the newbies!