Post #4

This week in Calculus, we learned partial fractions and how to use a table to integrate easier. I think partial fractions are pretty easy, but that's being said with fingers crossed! So, let me try to explain what i don't understand. How do you know what part of the equation to use to find the right equation in the table?

So, let me explain partial fractions.

1. How do you know its a partial fraction?
Well duhhh sillies, its going to have a fraction with some quadratics

2. So, then what do you do? You need to break the fraction up.

3. You break up the fraction, by factoring the bottom and rewriting it as multiple new fractions.

4. Take the separate denominators with a numerator of A, B, C, or D.

5. Next, you create a common denominator and set that equal to the numerator of the initial fraction.

6. You then pick convient values to solve for the variables of the equation.

7. After you find the values, you plug everything back in to the fractions you created

8. Integrate

9. And hopefully box off the correct answerrrr!

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.

Post #2

Kay. Since I've yet to touch on L'Hopital's Rule, I shall do so now.

The most important thing to learn about l'Hôpital's rule is when it should not be used:

Definitely do NOT use it when the limits of the two parts are not both 0, or both infinity. In this case the rule is likely to give a wrong answer!

Example:

limx->0+ (cos x)/x

is positive infinity, because the numerator approaches 1 while the denominator approaches 0. If we incorrectly apply l'Hôpital's rule, we get

limx->0+ (- sin x)/1 = 0.

So you DO use L'Hopital's Rule when you get an indeterminate in the first place...this is inf/inf, 0/0, etc.

Okay, for Trig SUB!!!!! I'm getting pretty good at this, so bear with me....

My trick is: Everytime I see a trig function to an odd power, I take out an even...After this I use an appropriate identity. It's really not all that hard...I have my notecards somewhere...just ask me for them..


OkAY!!!! for things you can comment on....

Does anyone know how to:

1. Divide stuff? like x^2 + x+ 7 all over x-8. the other day BRob tried to do a problem like that, and I failedddd miserably. Easier way??

2. Chasing the Rabbit. One time I ended up with chasing the rabbit, but the answer was something super easy. any hints as to when you should use by parts i.e. chasing the rabbit?

alright. night.

blog 2

alright alright, so this week we had our first real test. it was kindaaaa hard, but i think i did well :) we are still working on integration .. by parts, trig sub, wallis formula, all that good stuff.

so, first i'll tell you what wallice's formula is (by the way idk how to spell it so ignore that)...
if you have an integration problem of sin or cos raised to a power.. this is when you use this
S cos^5(x)
alright, so if your degree is ODD, you do (2/3)(4/5)..(n-1/n) and simply multiply them together. so your answer would be 8/15.
S sin^8(x)
if your degree is EVEN, you do (1/2)(3/4)...(n-1/n) (pi/2). then multiply
so your answer would be (1/2)(3/4)(5/6)(7/8)(pi/2). i don't feel like multiplying it out haha.

HELP:
alright, trig sub. it's pretty much a bunch of formulas telling you what to substitute in and when to do it when you are integrating trig functions.
i know how to do these.. i just tend to mess up cuz i don't memorize when i have to do what. & i also didn't bring my book home to remember to put anything about it on here.. :x so.. could someone maybe go over a few formulas for me?

when to do synthetic division:
when your top function degree is larger than the top.

so say you had S (x^2 + 2x +5)/(x-6)
6 would go in your box, then 1, 2, 5...
i think. if i'm wrong someone please let me know! also, i get kinda lost after that.. i stop and don't remember what to do next, the only thing i remember is that i have to put my remainder over the bottom of the fraction at the end... help please

Post #2

Hello my Calculus BC friends,

TRIG SUBSTITUTION!

Some basic integrals:
S sinu du = -cos u + C
S cosu du = sin u + C
S tan u du = -ln|cos u| + C
S cot u du = ln|sin u| + C
S secu du = ln|sec u + tan u| + C
S cscu du = -ln|csc u + cot u| + C
S sec^2 u du = tan u + C
S csc^2 u du = -cot u + C

Some identities:
sin^2x + cos^2x = 1 .
sin^2x = (1 - cos 2x)/2
cos^2x = (1 + cos 2x)/2

*What I try to do: usually try to take out some kind of squared, then change the to an identity, distribute in, and substitute.

*ALL the Rules:
SIN & COS guidelines:
1. If the power of the sine is odd and positive, save one sine
factor and convert the remaining factors to cosines. Then, expand
and integrate.
2. If the power of the cosine is odd and positive, save one cosine
factor and convert the remaining factors to sines. Then, expand
and integrate.
3. If the powers of both sine and cosine are even and
non negative, make repeated use of the half-angle identities for
sin^2x and cos^2x to convert the integrand to odd powers of the
cosine. Then proceed as in guideline 2.

SEC & TAN guidelines:
1. If the power of the secant is even and positive, save a secantsquared
factor and convert the remaining factors to tangents.
Then expand and integrate.
2. If the power of the tangent is odd and positive, save a secanttangent
factor and convert the remaining factors to secants. Then
expand and integrate.
3. If there are no secant factors and the power of the tangent is
even and positive, convert a tangent-squared factor to a secantsquared
factor, then expand and repeat if necessary.
4. If the integral is of the form S secmx dx, where m is odd and
positive, use integration by parts.
5. If none of the first four guidelines applies, try converting to

Wallis formula:
Only works with sin and cos when going from 0 to pi/2. n is the exponent
when n is ODD: (2/3)(4/5)(6/7)...(n -1)/n
EVEN: (1/2)(3/4)(5/6)...((n-1)/n)(pi/2)

HERE IS WHAT YOU CAN COMMENT ON:
Now I understand everything, but I somehow cannot always work the problems. Does anyone have some kind of trick on how to know when you look at a problem and know you have to either substitute, by part it, or trig sub? Also, do you know something that can help me remember how to do trig sub? (like the steps explained easier or a trick to remember or the steps you follow EVERY time?)

Post #2

Okay, so after what felt like the longest week ever, it's now time to do the blog. This week in Calculus BC I was kinda discouraged by trig sub because its something i really don't understand. I only do the problems right when i have the formulas in front of me...and after studying them for a week straight, and still getting mixed up on them, i'm finding it almost hopeless.

Hopefully someone can show me their study techniques?

But lets go over a few things...

For trig sub, something i always get wrong is WHEN to actually do the method..so, i believe it is when you can't basically bi-part something? correct?

Also, you should never bi-part or trig sub things when you only have the trig function and its derivative/ integral..just saying. I do it all the time and it is definately the hard way.

I really wish there is something i can explain that i know how to do, but there really isn't..

I guess i'll explain Wallis Formula.

So, you do this when you have sin or cos and the degree is EVEN:

1. Start with 1/2 and multiply the chronological numbers until you get to the exponent.
2. Then multiply by pi/two
3. Add +c
4. Box or circle your answer

When the degree is ODD:

1. Start with 1/2 and multiply the chronological numbers unitl you get to the exponent.
2. Put a + c
3. Box or circle your answer


So, i really feel like a baby and hopefully someone can help me..
i really just need all the helpful hints and basic problems explained to me.
i'm not quite sure why my brain hasn't kicked into school mode yet..