Page 1 of 2

Posted: Fri Oct 07, 2011 10:09 pm UTC
Do any of you have favorite "tricky" integrals?

I'm thinking primarily of single integrals that, in principle, could appear in the second half of a calculus course for first-year undergraduates.

But I'm thinking of integrals that involve a "trick", such as a clever substitution, or maybe taking advantage of some symmetry.

Here's one that I like:

Integral of sqrt( 1 + exp(2x) ) dx.

(Notice, by the way, that this arises "naturally" as the arc length of the graph of the exponential function.)

Others?

Posted: Sat Oct 08, 2011 1:58 am UTC
skullturf wrote:Do any of you have favorite "tricky" integrals?

Something like [imath]\int \frac{1}{1+\sin(x)}\,dx[/imath] is nice. Especially if you don't cover the Weierstrass substitution in class.

Less tricky, but ones that I've always enjoyed are the ones where you "get back the original integral" after an application or two (or three or four) of "integration by parts."

Posted: Sat Oct 08, 2011 5:32 am UTC
The integral of e-x^2 from negative infinity to infinity is fun.

Posted: Sat Oct 08, 2011 5:58 am UTC
lightvector wrote:The integral of e-x^2 from negative infinity to infinity is fun.

I think that would be grossly unfair without some sort of guidance and knowledge of polar coords, but I agree that its a good one.

I would hope that the tan half angle substitution is taught but not tested. Its a nice thing to have in your bag of tricks, but its a bit of a bastard to actually have to work through.

Posted: Sun Oct 09, 2011 5:34 am UTC
The integral of sec x dx is nice. I guess that even the integral of tan x dx might be considered tricky, although writing tan as sin / cos should be a fairly obvious thing to try.

Posted: Sun Oct 09, 2011 3:50 pm UTC
Integral of sin(x)/(sin(x)+cos(x)) from 0 to pi/2

Posted: Sun Oct 09, 2011 10:01 pm UTC
TheWaterBear wrote:Integral of sin(x)/(sin(x)+cos(x)) from 0 to pi/2

Heh, I like that one.

Would you believe: I argue that I can do that in my head, and the answer is
Spoiler:
pi/4
?

Posted: Mon Oct 10, 2011 3:04 am UTC
skullturf wrote:
TheWaterBear wrote:Integral of sin(x)/(sin(x)+cos(x)) from 0 to pi/2

Heh, I like that one.

Would you believe: I argue that I can do that in my head, and the answer is
Spoiler:
pi/4
?

correct

I like it because you *can* carry out the integration, but it has a very simple solution ^^

Posted: Mon Oct 10, 2011 12:11 pm UTC
This kind of integration is pretty simple...a basic use of property of definite integrals

Posted: Mon Oct 10, 2011 10:37 pm UTC
skullturf wrote:
TheWaterBear wrote:Integral of sin(x)/(sin(x)+cos(x)) from 0 to pi/2

Heh, I like that one.

Would you believe: I argue that I can do that in my head, and the answer is
Spoiler:
pi/4
?

That's very nifty. I did it on paper just to make sure I did it right, but it came out in about 3 lines. The temptation do use the half-angle substitution (or something even nastier) is strong, but I have no idea how long that would take (and it would definitely be more prone to error).

Posted: Mon Oct 10, 2011 11:12 pm UTC
Pick any function f(x). For fun, make it really complicated. Ask them to integrate f(x) * f'(x).
Be sure to make liberal use of expanding the integrand. The clever ones may notice that it integrates
to 1/2 * f(x)^2.

I've had integrals of this form pop up a lot in Quantum Mechanics.

Try f(x) = sec(x). You get sec(x) * tan(x) as the integrand. Expand it as sin(x) / cos(x)^3.

Looks tricky, but it's really easy if you know the trig functions and their derivatives. Even easier when
you realize that f(x) * f'(x) integrates to 1/2 * f(x)^2

Fun fact: In my Calc 1 course I took my freshman year, we were asked to differentiate an expression that was
a rational function of twelve different basic functions (exponential, power, or trigonometric). Please never do
that to a student. Ever. It took me the entire front and back of the sheet I was given and I could only do it because
I broke up the differentiation into multiple parts.

Posted: Tue Oct 11, 2011 12:38 am UTC
ConMan wrote:
skullturf wrote:
TheWaterBear wrote:Integral of sin(x)/(sin(x)+cos(x)) from 0 to pi/2

Heh, I like that one.

Would you believe: I argue that I can do that in my head, and the answer is
Spoiler:
pi/4
?

That's very nifty. I did it on paper just to make sure I did it right, but it came out in about 3 lines. The temptation do use the half-angle substitution (or something even nastier) is strong, but I have no idea how long that would take (and it would definitely be more prone to error).

You should see the step-by-step breakdown for finding the anti-derivative of the function...

http://www.wolframalpha.com/input/?i=in ... %28x%29%29

Posted: Tue Oct 11, 2011 12:50 am UTC
Which is why the tan half angle should be taught but not tested, unless after the substitution you get something really nice, which I have never seen.

Posted: Tue Oct 11, 2011 1:19 am UTC
TheWaterBear wrote:
ConMan wrote:
skullturf wrote:
TheWaterBear wrote:Integral of sin(x)/(sin(x)+cos(x)) from 0 to pi/2

Heh, I like that one.

Would you believe: I argue that I can do that in my head, and the answer is
Spoiler:
pi/4
?

That's very nifty. I did it on paper just to make sure I did it right, but it came out in about 3 lines. The temptation do use the half-angle substitution (or something even nastier) is strong, but I have no idea how long that would take (and it would definitely be more prone to error).

You should see the step-by-step breakdown for finding the anti-derivative of the function...

http://www.wolframalpha.com/input/?i=in ... %28x%29%29

Do you hear that? It's the sound of my brain breaking.

Especially since by looking at the solution I can see that there's actually a pretty simple way to get the indefinite integral using pretty much the same trick as the definite one.

Posted: Thu Oct 13, 2011 2:52 am UTC

Posted: Sun Dec 25, 2011 2:49 pm UTC
One of my favourites is the integral of (x^n)*cos(x). Try to find the general formula.

Spoiler:
Find the recursive formula first, then guess the general formula is a polynomial*cos(x) plus a polynomial*sin(x) then turn the recursive formulae into a differential equation and solve!

Posted: Tue Dec 27, 2011 8:29 am UTC
If you want to be a little evil, you could give this word problem:

There is a silo of radius r in an infinite field of grass. A goat is tied to the side of the silo with a rope of length 2*pi*r. What is the area of grass that the goat can graze? (The vertical dimension can be ignored for the purposes of this problem)

The answer cannot be expressed explicitly in closed form. If I recall correctly, it's approximately 117*r^2. Lots of students will choke on the technical difficulties of applying what they know to this although from what you described I think they should have all the tools necessary to solve it.

Posted: Thu Jan 05, 2012 10:36 am UTC
By the way, what is the solution to Integral of sqrt( 1 + exp(2x) ) dx? I haven't been able to figure it out.

Posted: Thu Jan 05, 2012 4:39 pm UTC
tomtom2357 wrote:By the way, what is the solution to Integral of sqrt( 1 + exp(2x) ) dx? I haven't been able to figure it out.

Even though I'm the one who posed the question, it took me a few tries just now to remember what the "trick" is.

There may be more than one way to evaluate that integral, but I think the following works:

Substitute 1+exp(2x) = u^2

Then 2exp(2x) dx = 2u du

dx = ( u/exp(2x) ) du = ( u/(u^2-1) ) du

That should give you an easier integral.

Posted: Fri Jan 06, 2012 3:26 am UTC
Thanks! Any more tricky integrals?

Posted: Sat Jan 07, 2012 8:02 pm UTC
tomtom2357 wrote:Thanks! Any more tricky integrals?

How about the "tough" secant-tangent integrals, using a hyperbolic substitution? (This usually isn't taught in Calc classes any more.)

Posted: Sun Jan 08, 2012 7:00 am UTC
Any specific integrals, I wouldn't know which ones are *tough*.

Posted: Mon Jan 09, 2012 9:17 am UTC
tomtom2357 wrote:Any specific integrals, I wouldn't know which ones are *tough*.

The cases you don't learn about in Calculus (the power of secant is odd and the power of tangent is even).

Posted: Mon Jan 09, 2012 3:14 pm UTC
Proginoskes wrote:
tomtom2357 wrote:Any specific integrals, I wouldn't know which ones are *tough*.

The cases you don't learn about in Calculus (the power of secant is odd and the power of tangent is even).

Did I do this right?

∫sec x tan²x dx

= ∫(sec³x - sec x)dx (sum of squares of sine and cosine)

= ∫(sec x)(sec²x)dx - ln|sec x + tan x| (integral of secant, integral of a linear combination)

= sec x tan x - ∫sec x tan²x dx - ln|sec x + tan x| (integration by parts)

So:

∫sec x tan²x dx = sec x tan x - ∫sec x tan²x dx - ln|sec x + tan x|
∫sec x tan²x dx = 1/2 *( sec x tan x - ln|sec x + tan x|) + C

I didn't use hyperbolic functions...

Posted: Tue Jan 10, 2012 4:52 am UTC
$\normalsize\int\frac{1}{sin(x-a)sin(x-b)}dx$

Posted: Tue Jan 10, 2012 7:30 am UTC
gfauxpas wrote:
Proginoskes wrote:
tomtom2357 wrote:Any specific integrals, I wouldn't know which ones are *tough*.

The cases you don't learn about in Calculus (the power of secant is odd and the power of tangent is even).

Did I do this right?

∫sec x tan²x dx

= ∫(sec³x - sec x)dx (sum of squares of sine and cosine)

= ∫(sec x)(sec²x)dx - ln|sec x + tan x| (integral of secant, integral of a linear combination)

= sec x tan x - ∫sec x tan²x dx - ln|sec x + tan x| (integration by parts)

So:

∫sec x tan²x dx = sec x tan x - ∫sec x tan²x dx - ln|sec x + tan x|

Actually, you should have a 2 on the left.

∫sec x tan²x dx = 1/2 *( sec x tan x - ln|sec x + tan x|) + C

I didn't use hyperbolic functions...

No, but you can. And when you do, the integral of [imath]sec x[/imath] turns into the integral of 1.

(For those who don't know this technique, you let [imath]\sec x = \cosh u[/imath] and [imath]\tan x = \sinh u[/imath]. Then
[imath]\sec x \tan x \,dx = \sinh u\,du[/imath], or [imath]dx = {du \over \cosh u}[/imath] and

$\int \sec x \,dx = \int \cosh u \cdot {du \over \cosh u} = \int 1 \,du = u + C = \ln |\sec x + \tan x| + C.$

The last part comes from solving [imath]\sec x = {e^u + e^{-u} \over 2}[/imath], or [imath]2 e^u \sec x = (e^u)^2 + 1[/imath], which is a quadratic equation in [imath]e^u[/imath].

Posted: Tue Jan 10, 2012 9:06 am UTC
Thanks for showing me a new way to do it, that's neat! Why should I have a two on the left? I left " - ∫sec x tan²x dx " on the RHS.

Posted: Tue Jan 10, 2012 11:49 am UTC
Afif_D wrote:$\normalsize\int\frac{1}{sin(x-a)sin(x-b)}dx$

Come on guys. Solve this problem..

Posted: Wed Jan 11, 2012 7:10 am UTC
gfauxpas wrote:Thanks for showing me a new way to do it, that's neat! Why should I have a two on the left? I left " - ∫sec x tan²x dx " on the RHS.

Oops! Didn't see that.

Posted: Wed Jan 11, 2012 12:15 pm UTC

Why dont you guys look at the tricky integral i gave?

Posted: Wed Jan 11, 2012 8:34 pm UTC
Afif_D wrote:$\normalsize\int\frac{1}{sin(x-a)sin(x-b)}dx$

I'm assuming a and b are constants? I cheated and used a list of trig identities

[imath]\int\frac{1}{\sin(x-a)\sin(x-b)}dx[/imath] =

[imath]\int\frac{2dx}{\cos(x-a-x+b)-\cos(x-a+x-b)}[/imath] (Product-to-Sum)

[imath]\int (2 \sec (b-a)) )dx - \int 2(\sec (2x - a - b)dx[/imath](definition of secant, linear combination of integrals)

[imath]2\sec(b-a)x - \ln|\sec(2x-a-b)+\tan(2x-a-b)| + C[/imath]

Is that right?

Posted: Wed Jan 11, 2012 8:44 pm UTC
gfauxpas wrote:[imath]\int\frac{2dx}{\cos(x-a-x+b)-\cos(x-a+x-b)}[/imath] (Product-to-Sum)

[imath]\int (2 \sec (b-a)) )dx - \int 2(\sec (2x - a - b)dx[/imath](definition of secant, linear combination of integrals)

You don't have a linear combination here: [imath]\frac{1}{r+s}[/imath] is not equal to [imath]\frac{1}{r} + \frac{1}{s}[/imath].

Posted: Wed Jan 11, 2012 9:21 pm UTC
jaap wrote:
gfauxpas wrote:[imath]\int\frac{2dx}{\cos(x-a-x+b)-\cos(x-a+x-b)}[/imath] (Product-to-Sum)

[imath]\int (2 \sec (b-a)) )dx - \int 2(\sec (2x - a - b)dx[/imath](definition of secant, linear combination of integrals)

You don't have a linear combination here: [imath]\frac{1}{r+s}[/imath] is not equal to [imath]\frac{1}{r} + \frac{1}{s}[/imath].

Oh.. for some reason I thought 1/(cos x + cos y) = sec x + sec y. My b. How embarrasing

Posted: Wed Jan 11, 2012 10:53 pm UTC
One I like:

$\frac{1}{6} \int_0^t\ e^{x}(t-x)^3 dx$

a lot of people get hung up on using integration by parts, when it's actually a whole lot easier.

Edit: supposed to be 1/6 not 1/4 (1/3!)

Posted: Wed Jan 11, 2012 11:05 pm UTC
I particularly like ∫(e^x)cos(x)dx, while it's really simple to do by applying parts twice in a row, you can also try to integrate (e^x)sin(x) by using parts with u equal to sin(x) and dv/dx equal to e^x, then integrate it with u equal to e^x and dv/dx equal to sin(x), and with a little rearranging you'll stumble upon the integral of (e^x)cos(x). Not "tricky" in a complicated way, but it seems a nice trick.

Posted: Thu Jan 12, 2012 4:49 am UTC
Afif_D wrote:$\normalsize\int\frac{1}{sin(x-a)sin(x-b)}dx$

To solve this there is only one short good way .
Multiply denominator and numerator by
$\normalsize\sin(a-b)$
which is a constant
Then leave the denominator intact.
Now change the numerator as
$\huge\sin((x-b)-(x-a))$

Now expand this thing in terms of sine and cosine functions. It will make it reduce to some familiar forms. Do it and see.

Posted: Thu Jan 12, 2012 4:53 am UTC
vilidice wrote:One I like:

$\frac{1}{6} \int_0^t\ e^{x}(t-x)^3 dx$

a lot of people get hung up on using integration by parts, when it's actually a whole lot easier.

Edit: supposed to be 1/6 not 1/4 (1/3!)

So how do you do this without using parts??

Posted: Thu Jan 12, 2012 6:02 am UTC
using the two facts: t-x = 0 at x=t, and the binomial formula to get:

$\frac{1}{6} \int_0^t e^x (t^3-3t^2x+3tx^2-x^3)dx$

at the lower limit x=0 all terms except the -[imath]\int_0^te^xt^3 dx[/imath] terms go to 0, so they can be disregarded, same for the sign changes for the interior bits, leading to all that going to 0, and this one factor of exp(t)("easily" seen By using the rules for integrating a polynomial) the result falls out:

$1+t+\frac{t^2}{2}+\frac{t^3}{6} -e^t$

OR

by applying the formula for iterated integration:

$\frac{1}{(n-1)!}\int_a^tf(x)(t-x)^{n-1}dx$ with f(x) = exp(x), a=0, and n=4[/math] to get that the integral is going to be the 4th-iteration integral of the function e^x (but I don't think that's covered in most cal-1 classes.)

OR (most likely)

If given the summation definition for exp(1), and exp(x) (my cal-1 class was, I don't know if it's standard to do that anymore), you can substitute that in, with the factored polynomial, and evaluate it out as the first few terms of the series plus the remainder, and upon evaluation the above falls out. It actually can be generalized to an expression for that, where for the integral with [(t-x)^k]/k! it's the first k terms of the series, minus exp(x), collectively multiplied by (-1)^k+1.

This one is cute. Find the area bounded by the positive x-axis and the line $e^{-x} + e^{-y} = 1$
dissonant wrote:This one is cute. Find the area bounded by the positive x-axis and the line $e^{-x} + e^{-y} = 1$