Your favorite "tricky" integrals?
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Your favorite "tricky" integrals?
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 firstyear 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?
I'm thinking primarily of single integrals that, in principle, could appear in the second half of a calculus course for firstyear 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?
Re: Your favorite "tricky" integrals?
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."

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Re: Your favorite "tricky" integrals?
The integral of e^{x^2} from negative infinity to infinity is fun.
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Re: Your favorite "tricky" integrals?
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.
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Re: Your favorite "tricky" integrals?
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.

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Re: Your favorite "tricky" integrals?
Integral of sin(x)/(sin(x)+cos(x)) from 0 to pi/2
Re: Your favorite "tricky" integrals?
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:

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Re: Your favorite "tricky" integrals?
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:
correct
I like it because you *can* carry out the integration, but it has a very simple solution ^^
Re: Your favorite "tricky" integrals?
This kind of integration is pretty simple...a basic use of property of definite integrals
Re: Your favorite "tricky" integrals?
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:
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 halfangle 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).
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Re: Your favorite "tricky" integrals?
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.
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.
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Re: Your favorite "tricky" integrals?
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:
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 halfangle 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 stepbystep breakdown for finding the antiderivative of the function...
http://www.wolframalpha.com/input/?i=in ... %28x%29%29
 jestingrabbit
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Re: Your favorite "tricky" integrals?
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.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Your favorite "tricky" integrals?
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:
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 halfangle 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 stepbystep breakdown for finding the antiderivative 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.
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Re: Your favorite "tricky" integrals?
One of my favourites is the integral of (x^n)*cos(x). Try to find the general formula.
Spoiler:
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Re: Your favorite "tricky" integrals?
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.
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.

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Re: Your favorite "tricky" integrals?
By the way, what is the solution to Integral of sqrt( 1 + exp(2x) ) dx? I haven't been able to figure it out.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Re: Your favorite "tricky" integrals?
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^21) ) du
That should give you an easier integral.

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Re: Your favorite "tricky" integrals?
Thanks! Any more tricky integrals?
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Re: Your favorite "tricky" integrals?
tomtom2357 wrote:Thanks! Any more tricky integrals?
How about the "tough" secanttangent integrals, using a hyperbolic substitution? (This usually isn't taught in Calc classes any more.)

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Re: Your favorite "tricky" integrals?
Any specific integrals, I wouldn't know which ones are *tough*.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
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Re: Your favorite "tricky" integrals?
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).
Re: Your favorite "tricky" integrals?
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  lnsec x + tan x (integral of secant, integral of a linear combination)
= sec x tan x  ∫sec x tan²x dx  lnsec x + tan x (integration by parts)
So:
∫sec x tan²x dx = sec x tan x  ∫sec x tan²x dx  lnsec x + tan x
∫sec x tan²x dx = 1/2 *( sec x tan x  lnsec x + tan x) + C
I didn't use hyperbolic functions...
Re: Your favorite "tricky" integrals?
[math]\normalsize\int\frac{1}{sin(xa)sin(xb)}dx[/math]
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Re: Your favorite "tricky" integrals?
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  lnsec x + tan x (integral of secant, integral of a linear combination)
= sec x tan x  ∫sec x tan²x dx  lnsec x + tan x (integration by parts)
So:
∫sec x tan²x dx = sec x tan x  ∫sec x tan²x dx  lnsec x + tan x
Actually, you should have a 2 on the left.
∫sec x tan²x dx = 1/2 *( sec x tan x  lnsec 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
[math]\int \sec x \,dx = \int \cosh u \cdot {du \over \cosh u} = \int 1 \,du = u + C = \ln \sec x + \tan x + C.[/math]
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].
Re: Your favorite "tricky" integrals?
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.
Last edited by gfauxpas on Tue Jan 10, 2012 2:19 pm UTC, edited 1 time in total.
Re: Your favorite "tricky" integrals?
Afif_D wrote:[math]\normalsize\int\frac{1}{sin(xa)sin(xb)}dx[/math]
Come on guys. Solve this problem..
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Re: Your favorite "tricky" integrals?
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.
Re: Your favorite "tricky" integrals?
Why dont you guys look at the tricky integral i gave?
Re: Your favorite "tricky" integrals?
Afif_D wrote:[math]\normalsize\int\frac{1}{sin(xa)sin(xb)}dx[/math]
I'm assuming a and b are constants? I cheated and used a list of trig identities
[imath]\int\frac{1}{\sin(xa)\sin(xb)}dx[/imath] =
[imath]\int\frac{2dx}{\cos(xax+b)\cos(xa+xb)}[/imath] (ProducttoSum)
[imath]\int (2 \sec (ba)) )dx  \int 2(\sec (2x  a  b)dx[/imath](definition of secant, linear combination of integrals)
[imath]2\sec(ba)x  \ln\sec(2xab)+\tan(2xab) + C[/imath]
Is that right?
Re: Your favorite "tricky" integrals?
gfauxpas wrote:[imath]\int\frac{2dx}{\cos(xax+b)\cos(xa+xb)}[/imath] (ProducttoSum)
[imath]\int (2 \sec (ba)) )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].
Re: Your favorite "tricky" integrals?
jaap wrote:gfauxpas wrote:[imath]\int\frac{2dx}{\cos(xax+b)\cos(xa+xb)}[/imath] (ProducttoSum)
[imath]\int (2 \sec (ba)) )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
Re: Your favorite "tricky" integrals?
One I like:
[math]\frac{1}{6} \int_0^t\ e^{x}(tx)^3 dx[/math]
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!)
[math]\frac{1}{6} \int_0^t\ e^{x}(tx)^3 dx[/math]
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!)
Re: Your favorite "tricky" integrals?
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.
Re: Your favorite "tricky" integrals?
Afif_D wrote:[math]\normalsize\int\frac{1}{sin(xa)sin(xb)}dx[/math]
To solve this there is only one short good way .
Multiply denominator and numerator by
[math]\normalsize\sin(ab)[/math]
which is a constant
Then leave the denominator intact.
Now change the numerator as
[math]\huge\sin((xb)(xa))[/math]
Now expand this thing in terms of sine and cosine functions. It will make it reduce to some familiar forms. Do it and see.
Re: Your favorite "tricky" integrals?
vilidice wrote:One I like:
[math]\frac{1}{6} \int_0^t\ e^{x}(tx)^3 dx[/math]
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??
Re: Your favorite "tricky" integrals?
using the two facts: tx = 0 at x=t, and the binomial formula to get:
[math]\frac{1}{6} \int_0^t e^x (t^33t^2x+3tx^2x^3)dx[/math]
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:
[math]1+t+\frac{t^2}{2}+\frac{t^3}{6} e^t[/math]
OR
by applying the formula for iterated integration:
[math]\frac{1}{(n1)!}\int_a^tf(x)(tx)^{n1}dx[/math] with f(x) = exp(x), a=0, and n=4[/math] to get that the integral is going to be the 4thiteration integral of the function e^x (but I don't think that's covered in most cal1 classes.)
OR (most likely)
If given the summation definition for exp(1), and exp(x) (my cal1 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 [(tx)^k]/k! it's the first k terms of the series, minus exp(x), collectively multiplied by (1)^k+1.
[math]\frac{1}{6} \int_0^t e^x (t^33t^2x+3tx^2x^3)dx[/math]
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:
[math]1+t+\frac{t^2}{2}+\frac{t^3}{6} e^t[/math]
OR
by applying the formula for iterated integration:
[math]\frac{1}{(n1)!}\int_a^tf(x)(tx)^{n1}dx[/math] with f(x) = exp(x), a=0, and n=4[/math] to get that the integral is going to be the 4thiteration integral of the function e^x (but I don't think that's covered in most cal1 classes.)
OR (most likely)
If given the summation definition for exp(1), and exp(x) (my cal1 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 [(tx)^k]/k! it's the first k terms of the series, minus exp(x), collectively multiplied by (1)^k+1.
Re: Your favorite "tricky" integrals?
This one is cute. Find the area bounded by the positive xaxis and the line [math]e^{x} + e^{y} = 1[/math]
Re: Your favorite "tricky" integrals?
dissonant wrote:This one is cute. Find the area bounded by the positive xaxis and the line [math]e^{x} + e^{y} = 1[/math]
Are you sure the area is bound and finite?
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