Integral Calculus Bee: Antidifferentiating & Finding Primative Functions

[phoenixthoth]

And, yes, I mean what you think it means: antidifferentiating. Finding a primative. Whatever...

At my school, we're conducting an integration bee not unlike similar bees done elsewhere.

The purpose of this thread is two-fold:
1. To compile a list of non-routine integrals
2. To discuss how these integrals are done

Here is the list I came up with. Please add to the list. I'm not looking for 300 integrals no one at a community college can solve (and, yes, by "solve", I mean to antidifferentiate). They should stump a significant percentage of those who would get an A in Integral Calculus though not 100%.

A few of these are downright easy but they can stump the woefully inexperienced.

One or two of them are potentially very difficult if you're not clever enough.

1. Log[Sqrt[x]]/Sqrt[x]
2. Sin[Sqrt[x]]
3. (2^x)(Sqrt[1-4^x])
4. Log[Sqrt[x]]
5. Exp[x] Sin[x]
6. Exp[2 x]/Sqrt[Exp[x]+1]
7. 1/(Exp[x]+1)
8. 1/(2+Tan[x])
9. 1/(x^4 + 64)
10. Sqrt[Tan[x]]
11. 1/(1-(Sin[x])^2)
12. 1/(3+5Sin[x])
13. (1+x^2)^(-3)
14. (x^2)/(1+x^2)
15. (1+x)/Sqrt[-4+2x+x^2]
16. Log[(x+1)/(x-1)]/x^2
17. Cos[2x]/(8+Sin[2x]^2)
18. Sec[x]^2/Sqrt[1-4Tan[x]^2]
19. ArcTan[x]
20. Log[1+x]/x^2.

Thanks for your input!

 

[marlon]

What is the most difficult one according to you ?

i will solve it

marlon

 

[mruncleramos]

Most are standard textbooks problems.

 

[phoenixthoth]

I would guess that #9 would stump the most people but I really have no idea.

You may find #10 and #2 fun as well.

 

[Crosson]

Those are nice, but the tricks don't really get fun until definite integration:

\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}

\int_{-\infty}^{\infty}\frac{sin(x)}{x} dx = \pi

The first one is a fairly trivial change of variable trick, and the second is found by contour integration. In addition to having these down pat, an integrator champ should be able to pull all the tricks: differentiating under the integral sign, direct integration of differential eq corresponding to the function, etc.

 

[marlon]

9 is probably the most easy one

2 is easy with a good substitution and ten is nice

marlon

 

[quantumdude]

I like these:

<br /> \int e^{ax} sin(bx)dx<br />

<br /> \int e^{ax} cos(bx)dx<br />

<br /> \int sec^n(x) dx<br />

(n is odd and positive in the last one).

They all have one thing in common: when integrating by parts, you have to recognize that you eventually get the same integral you started with, and you have to add a multiple of it to both sides to finish the problem.

 

[phoenixthoth]

You think 9 is easier than 4, 5, 14, 15, and 19? Hmm... Is there an easy way to do #9 that I don't know about? What's the easy way to solve #9?

 

[marlon]

9 is equal to

\int \frac{dx}{(x^2 + 8)^2 -16 x^2}

then use a²-b² = (a-b)(a+b) in the denominator
then integration by partial fractions

marlon

 

[marlon]

9 is easy because it is quite straightforeward this is the easiest way out

marlon, though i admit it requires some calculation

 

[phoenixthoth]

Yeah that's how I'd do it but I wouldn't call factoring x^4+64 "easy." Sure, compared to the proof of FLT it's easy but... ;)

 

[marlon]

how did you do 10 ?

marlon

 

[marlon]

i got an answer by using the substitution t²=tanx and then apply partial fractions

marlon

 

[phoenixthoth]

how did you do 10 ?

marlon

I'm no integral bee champion so this is probably not the best way to do it:

u=Tan[x] turns it into
Sqrt/(1+u^2).

Re-write:
u/(Sqrt(1+u^2)).

Now let v=Sqrt to get
2v^2 / (1+v^4) which is integrable by partial fractions as the denominator splits into (v^2+Sqrt[2]v+1)(v^2-Sqrt[2]v+1).

 

[marlon]

any other questions ?

marlon

 

[dextercioby]

\int e^{tan x} \ dx =... ?

\int \sqrt{\sin x} \ dx =... ? Piece of cake for champions like Marlon

 

[marlon]

The integral of sqrt(sinx) is not exactly solvable. You will need elliptic integrals or when you use the substitution t=sinx, you will get a beta-function...

marlon

http://mathforum.org/kb/message.jspa?messageID=474805&tstart=0

 

[marlon]

The integral exp(tanx) is not that difficult, if i am right. Write the exp(tanx) as a series :
sum over n of ((tanx)^n)/n!

Now the integral sign and summation sign can be interchanged so what we really need to integrate is (tan(x))^n...

i suppose that by writing tan as sin/cos, you can construct a recursion relation. This relation can be constructed by integration by parts...

marlon

 

[motai]

holy... geez talk about scaring a Calc 1 student... I don't even know how to approach some of those integrals... geez. The only thing that comes close that I can solve is probably applying an arctangent rule to no. 9 from my knowledge, but as marlon said I probably have to review partial fractions.

ack.

 

[dextercioby]

Marlon,the \int e^{\tan x} \ dx admits an exact (nonseries) solution among special functions.

Daniel.

 

[marlon]

Marlon,the \int e^{\tan x} \ dx admits an exact (nonseries) solution among special functions.

Daniel.

Which is ?

marlon

 

[dextercioby]

\int e^{\tan x} \ dx=\frac{1}{2}ie^{i}\func{Ei}\left( 1,-\tan x+i\right) -\frac{1}{2}ie^{-i}\func{Ei}\left( 1,-\tan x-i\right) +C

Daniel.

 

[marlon]

What a clear 'exact' solution ?? !

I really think my approach is more intuitive and you don't need them 'electronic backups'

marlon

 

[dextercioby]

Then do it.

Daniel.

 

[marlon]

Then do it.

Daniel.

Well read the content of my posts...

marlon

 

[dextercioby]

It's so simple to talk,to give advice and not to calculate... I think that,regarding intuition,looking for this integral in G & R is simpler that computing a recursion formula...

Daniel.

 

[El zaouk]

try to solve integral of x^3-2x^2+2x+3/(x^2+2x+2)^2
,, waiting for your answers guys ! have fun

 

[Opus_723]

I'm taking my first integral Calc class, and thought I was getting pretty good. These things are nasty though. Thanks for teaching me some humility . I'm curious though how you learn to do these. I've looked ahead in my book, and I really don't think we're going to learn any methods in this class that suddenly make these integrals possible.