Question

I want some tricky elementary integrals. Like tricky tricky and not just regular tricky. Thank you and please.

Answer 1

i cooked up this hard but nice integral when one of our regular users asked a hard integeral that evaluates to her age, I forgot the answer but it should be an integer between 15 and 20 :

\[\large \int_0^{\infty} \left(\frac{1944(e^{ax} - e^{bx})}{\ln((a/b)^{54x})(e^{ax}+1)(e^{bx}+1)}\right) dx = \text{your age }\]

Answer 2

Ok I will give these my full focus after I get a yoohoo.

Answer 3

And thanks.

Answer 4

here is another innocent looking yet very tricky integral :
\[\int\limits_0^{\pi} \dfrac{dt}{1+2\sin^2t}\]

Answer 5

third integral i like :
\[\int{\frac{x^{5}-x}{x^{8}+1}}\:dx\]

Answer 6

@Kainui @ikram002p might have experienced more nice/new integrals :)

Answer 7

if you may allow me tag few others @SithsAndGiggles
@tkhunny @amistre64 @geerky42 @nincompoop @dan815 @johnweldon1993 @iambatman @Zarkon

Answer 8

hey @ganeshie8 that second on I'm doing the sub tan(t/2)=u
which is ending up to be very long.
I'm not really asking for a hint but I guess sort am when I ask is there like a neat little trick to perform to cutout all of that lengthiness. Like just a yes or no question. Don't say anything else for now. Please and thank you. :)

Answer 9

evaluating indefenite integral is a one liner if you divide top and bottom by *something*
the tricky part is in evaluating the limits.. again im sure we will be okay if we are careful :)

Answer 10

Here's a favorite of mine: \[\LARGE \int\limits \frac{dx}{1+x^3}\]

Answer 11

I'm no good with integral, but maybe you can find some tricky integrals here: http://math.stackexchange.com/search?q=tricky+integral

Answer 12

@ganeshie8 I enjoyed that first one!

I gave this one to a calc 2 class a few days ago and no one has given me an answer so far, everyone wants hints but, oh well, lets see if it's that tricky

\[\large \int e^{2x}cos(3x)dx\]

Answer 13

That last one can be done by integration by parts.

Answer 14

I know how to do that integral 3 different ways at least. =P

Answer 15

What are the other two ways to do that one?

Answer 16

that last one is Kainui's favorite one (and mine too) and he somehow manages to work it using only matrices

Answer 17

You can use linear algebra or complex numbers to do it @myininaya

Answer 18

Matrices? Oh may I see? :D lol not very experienced using matrices in integrals!

Answer 19

Sure, so you can look at the derivative of this function and the corresponding sine one as spanning the space of derivatives since they only map onto scalar multiples of each other, so the anti derivative will too, to within a constant of course. So you just invert a 2x2 matrix to get the "integral" matrix! =P

Answer 20

If you have stewart 7e calculus textbook it has pages upon pages with some neat integral problems :P

Answer 21

When I'm not feeling lazy, I can take a pic/ photocopy and send it your way, if you're interested.

Answer 22

I have 6th edition.

Answer 23

Ah ok, cool :).

Answer 24

should have some similar stuff in it
since they never really change the books :p

Answer 25

You're probably right, @ganeshie8 how does one approach your first integral?

Answer 26

@Kainui you probably have some tricky matrix way for the 1/(x^3+1) integral
instead of this partial fraction/trig sub thing ?

Answer 27

\[\LARGE y_1=e^{2x}\cos(3x) \\ \LARGE y_2=e^{2x}\sin(3x)\] Taking the derivative with the product rule we have \[\LARGE y_1'=2y_1 -3 y_2 \\ \LARGE y_2'=3y_1+2 y_2\] So we can stick this in a matrix like this: \[\LARGE \left[\begin{matrix}2 & -3 \\ 3 & 2\end{matrix}\right] \bar y = \bar y'\]

So we invert this matrix to get: \[\LARGE \bar y = \frac{1}{13}\left[\begin{matrix}2 & 3 \\ -3 & 2\end{matrix}\right] \bar y'\] now the vector y' that we're looking to integrate is \[\LARGE \bar y '=\left(\begin{matrix}1 \\ 0\end{matrix}\right)\]

So that's just the first column of the matrix, so the integral appears to be \[\LARGE \frac{1}{13}(2 e^{2x}\cos(3x) - 3e^{2x}\sin(3x))\] but upon checking it looks like I messed up a negative sign somewhere since the answer is really \[\LARGE \frac{1}{13}(2 e^{2x}\cos(3x) + 3e^{2x}\sin(3x))\] but you get the idea haha.

Answer 28

Nope, I don't know a fast way to do that integral with 1/1+ x^3 I'm afraid. It's just deceptively easy looking.

Answer 29

That is neat approach for the integral of e^(2x)cos(3x)

Answer 30

It's really pretty quick and painless since there's a trick for inverting 2x2 matrices. It's always nice to see some new alternative too. The more math I do the more I realize how useful linear algebra is haha.

Answer 31

I wish I can give all of you a medal for all your fancy integrals to look at.

Answer 32

Here's another fun trick: \[\LARGE \int\limits \frac{dx}{1-x^2}\] but now let \[\LARGE x= i \tan \theta \\ \LARGE dx = i \sec^2 \theta d \theta\] we get \[\LARGE \int\limits \frac{ i \sec^2 \theta d \theta}{1+\tan^2 \theta}=i \int\limits d \theta= i \theta\]\[\LARGE \int\limits \frac{dx}{1-x^2}=i \tan^{-1} (-ix)\]

Answer 33

@ganeshie8 for some reason I want to write that first integral you wrote as two separate fractions
e^(ax)-e^(bx)=e^(ax)+1-e^(bx)-1
but I don't think that is helping me

Answer 34

Wow! i thought of the same thing for first step @myininaya :
\[\large \large \int \dfrac{1}{x(e^{bx}+1)} - \int \dfrac{1}{x(e^{bx}+1)} = F(b) - F(a) \]
\]

Answer 35

**** \[\large \int \dfrac{1}{x(e^{bx}+1)} - \int \dfrac{1}{x(e^{ax}+1)} = F(b) - F(a)\]

Answer 36

yeah times that constant 36

Answer 37

Ahh yes and some log stuff also i think..

Answer 38

oh yeah
over ln(a/b)

Answer 39

Did you just give me some weird hint with the F(b)-F(a) thing?

Answer 40

I didn't really mean your hint was weird if it was a hint.

Answer 41

or even if it wasn't a hint

Answer 42

The F(b)-F(a) confuses me though because why would those integrals be constants

Answer 43

differentiate F(b) with respect to b,
( notice that we can commute differentiation and integration : \(\large \dfrac{d}{db} \int dx = \int \dfrac{d}{db}~dx\))

do the same thing for F(a)..

thats one method ^^
there is another quick method also

Answer 44

** spoiler **
solution using that long way worked by @ShailKumar
http://assets.openstudy.com/updates/attachments/53d488bce4b0f2f7e5e971e2-shailkumar-1406447288904-diffunderintegralsign.jpg

Answer 45

That's crazy. (as in crazy neat)

Answer 46

here is a more neat+short solution :)
http://openstudy.com/users/vishweshshrimali5#/updates/53d488bce4b0f2f7e5e971e2

Answer 47

Have fun :D

Answer 48

Proving the following integral identity is fairly tricky if you haven't done it before:
\[\int_{-\infty}^\infty e^{-x^2}~dx=\sqrt{\pi}\]

Answer 49

well , using complex makes any integral solvable and able to solve hmm
but i like the one that @sithangiggles mentioned

Answer 50

\[\int\limits_{0}^{2\pi}\frac{1}{1+2\sin^2(x)}dx=4\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{2-\cos(2x)} dx \\ \tan(x)=t \\ \sec^2(x) dx=dt \\ dx=\cos^2(x) dt \\ dx=\frac{dt}{t^2+1} \\ 4 \int\limits_{0}^{\infty} \frac{1}{t^2+1}\frac{1}{2- \frac{1-t^2}{1+t^2}} dt \\ =4\int\limits_{0}^{\infty}\frac{1}{2(t^2+1)-(1-t^2)} dt \\ =4\int\limits_{0}^{\infty}\frac{1}{3t^2+1} dt \\ =4 \int\limits_{0}^{\infty}\frac{1}{(\sqrt{3}t)^2+1} dt \\ \lim_{a \rightarrow \infty} \frac{4}{\sqrt{3}} \arctan(\sqrt{3}x)|_0^{a} \\ \lim_{a \rightarrow \infty}\frac{4}{\sqrt{3}} \arctan(\sqrt{3}a)=\frac{4}{\sqrt{3}} \cdot \frac{\pi}{2}=\frac{2 \pi }{\sqrt{3}}\]

Answer 51

my age might change by the time i solve it though have u considered that possibility!

Answer 52

Ahh very nicely worked @myininaya :) I ran into weird problems when I first attempted it.. here are few more alternative methods :
http://math.stackexchange.com/questions/958920/evaluate-int-limits-0-pi-fracdx12-sin2x

Answer 53

@dan815 that integral gives your age correctly upto an uncertainty of 100 years.. so we are fine technically speaking ;p good point though :)

Answer 54

:3

Answer 55

@ganeshie8 y'=(x^5-x)/(x^8+1) has got me stumped

Answer 56

And yeah that one I seen the way I first tried to work out and someone totally finished it that way but I wasn't about to because it had ugly factorization

Answer 57

And I manipulated that one substitution tan(x/2)=t since I had a double angle

Answer 58

nice thread on integrals