Post #4

Well this week we talked about partial fractions and B integration tables. So I guess that is what I will be covering in this blog.

Partial Fractions:
problem: S 1/x^2+5x+6

first: factor the bottom
1/(x+3)(x+2)

second: set up the fractions, start off with a over the first, set equal to problem
A/x+3 + B/x+2 = 1/x^2+5x+6

third: multiply to get denominates equal, which gives you
A(x+2) + B(x+3) = 1

fourth: find convenient values for x to solve for A and B
A(x+2) + B(x+3)

*choose -2 to plug into A b/c it will give you zero
A(-2+2) + B(-2+3)=1
A(0) + B(1) =1
B=1
*choose -3 for B /c it will give you zero
A(-3+2) + B(-3+ -3)=1
A(-1) + B(0)=1
A=-1

fifth: plug A and B back into the fractions A/x+3 + B/x+2
S -1/x+3 + 1/x+2

*1/x gives you a ln
=-ln(x+3) + ln(x+2) +C
*simplify
ln(x+2/x+3) +C


Note: if you were given S fx^2+20x+6/x(x+1)^2, your fractions would be, it is a rule
A/x + B/x+1 + C/(x+1)^2


B integration tables:
This is from A21 in the book. It has a listing of all kinds of integrals and what they equal. You basically have to figure out which integral applies to the problem. If you need a u, a, or something you have to figure that out from the problem. All there is left is to plug in and simplify. The only tricky part is make sure you pick the right integral to use.

My Question:
Can anyone tell me which integral I would use for these two:
S 1/squareroot x(1-cos squareroot x)
or
S x^7lnx
*for this one there is a u^n formula nd a 1/u formula? but what do I use?

week THREE

This week was ALL about Trig. Sub.

Trig. Sub. is used anytime there is a square root in the form of sqrt(a^2 - x^2), sqrt (x^2 - a^2), or sqrt(x^2 + a^2) in an integration problem.

If you see sqrt(a^2 - x^2) then x=asin(theta) and sqrt(a^2 - x^2)=acos(theta).

If you see sqrt (x^2 - a^2) then x=asec(theta) and sqrt (x^2 - a^2)= atan(theta).

If you see sqrt(x^2 + a^2) then x=atan(theta) and sqrt(x^2 + a^2)=asec(theta).

After figuring out which one you have to use, you then find dx by taking the derivative of x and if you have an (x) variable in the problem, then solve for that using the x equation.

After integrating, you have to switch the theta versions of the problem back to x.

Some issues I'm having with this is when you actually have to integrate.
So Ryan, here's some integration formulas to remember:
Ssec = lnsec + tan
Ssec^2x = tan
Ssec^3x = (1/2)sectan + (1/2)lnsec + tan
Scsc*cot = -csc
Stan = -lncos
Scot = lnsin

And some trigonometry formulas to remember:
sin^2x + cos^2x = 1
1 + tan^2x = sec^2x
1 + cot^2x = csc^2x
sin2x = 2sinxcos

Ryan

Post 3

Okay. So I would like to state a couple of things that Abbey and I learned while doing that partner thing:

1. When, after doing trig sub, you get either SIN^2 or COS^2, that would be when you use the following Reduction Formulas:

cos^2 = (1+cos2x)/2
sin^2 = (1-cos2x)/2

The way I remember that is, sin is negative because it's negative that its not sine. and cosine is cosine, so it's positive.

2. So instead of doing by parts like the textbook says to do when you get S of sec to an odd power, you should just memorize s=S sec^3(x) because it shows up quite often. I think it's something like 1/3secxtanx + 1/3ln(secx + tanx) + C

3. When doing trig sub, you have to be able to see basic trig concepts, so to speak. Say I have S 1/sec^2(x). One of the problems Abbey had was realizing that that was the same as cos^2(x). That's just something you have to watch out for.

4. Also, I'm pretty sure everyone could brush up on some things...ie trig formulas.

5. Memorize your triangles...I'm thinking.

6. In normal trig integration. You have to realize that sometimes...say you have tan and sec in an integral...if you can some how get a du (ie sec^2 or sectan) out of it, thats the way you want to go because by doing so, you just have to do normal substitution integration again. Got it? good.

Okay, for stuff you can comment on...umm...

Could someone explain the process of doing definite integrals for me? I'm kinda sketchy on the part where you plug the original values in?? it's all kind of a blur. Anyways..thanks!!

Post #3

Alrighty here we go...

First, I would like to list some identities that I need to learn:

sinx= 1/2 - 1/2cos2x
cosx= 1/2 + 1/2cos2x

cos^2x + sin^2x = 1
1 + tan^2x = sec^2x
1 + cot^2x = csc^2x

tanx = sinx/cosx
cotx = cosx/sinx

sin(2x) = 2sinxcosx

Well, we have been doing trig sub like all week. So let me try to explain it the best way I can since I haven't gotten the hang of it just yet.

*Say you have the square root of x^2/squarerootof 25-x^2 dx

1. You see what box you need to use (which I need to memorize). There are three different cases: (a is the number, u is the x)
square root of a^2 - u^2 --->asin(t)
square root of a^2 + u^2 --->atan(t)
square root of u^2 - a^2 --->asec(t)
This example would follow a^2 - u^2

2. Then find your x and dx:
x is from the three different cases so x= 5sin(t)
dx is the derivative so dx= 5cos(t)

3. Next you have the square root:
squarerootof 25-x^2 --->5cos(t)
*those are from your chart thing too, but Mal Pal said that it is usually like the opposite of the x

4. and don't forget to account for the x^2:
so take your x, x=5sin(t) and square it which gives you x^2=25sin^2(t)

5. Now plug everything in!
so you should get
25sin^2(t)5cos(t)/5cos(t)
*simplify: the 5cost cancel leaving 25sin^2(t)

Now I'm pretty sure you use the power reduction formulas here right?
But this is where I get stuck..how do I use 1/2 - 1/2cos2x for 25sin^2(t)?

6. I know you integrate after that.

7. form the triangle and use that to plug in to the trig functions
*don't forget SOHCAHTOA

Post # 3

Hello Blogmates.

This week in Calculus we learned trig sub and then we reviewed integration. Integration is something i really need help with becasue I can't seem to understand when to integrate regularly, bi part, substitute, or trig sub. Hints?

So, in order to help myself, i'll list a few formulas we should not forget.

S sinx = -cos x + c
S -sinx = cos x + c
S cosx = sin x + c
S tanx = ln /cosx/ + c
S secx = ln /secx + tanx/ + c
S cscx = - ln / secx + cotx/ + c
S cotx = ln/sinx/ + c

Next, it is very important that you learn these POWER REDUCTION FORMULAS:

cosx = 1/2 + 1/2cos2x
sinx = 1/2 - 1/2cox2x

And this PYTHAGOREAN IDENTITY:

cos^2x + sin^2x = 1

Now, i think the things i need help with the most is with choosing which integration method to do..hopefully that atleast comes with time?

The last thing i want to explain is a nice summary of trig sub.

So, first, you choose what box your problem is and find the information necessary; including: x, dx, sqrt, and whatever else is in the problem.

Second, you plug all of that back into the problem and hopefully cancel some things.

Third, integrate.

Fourth, form the triangle by the information given in the selected box.

Fifth, find the trig functions in the problem's answer by using SOHCAHTOA and the triangle.

Lastly, pray you got it right.

Now, a quote from my favorite youtube video, glozell,

PEACE AND BLESSINGS. PEACE AND BLESSINGS.