King property for Integrals and some examples
Phil Petrocelli, mymathteacheristerrible.com
May 18, 2019
This is one of the coolest properties of integration, and it can be deployed for some of the most
difficult-looking integration problems. It’s a variant on u-substitution, but appears to do so much
more.
Question.
1) Integrate:
󰁝 π/2
0
sinn x
dx, for any integer n
sinn x + cosn x
2) Integrate:1
󰁝 π/2
1
dx
1 + (tanx)π
󰁝 1
ln(x + 1)
dx
x2 + 1
0
3) Integrate:2
0
Discussion & Exploration.
OK, so the first integral is not really something we see much in elementary Calculus classes. What it really is
asking is for a general formula or answer for any n; perhaps the solution should be expressed in terms of n?
We see that the case where n = 0 is easy, but that the case where n = 1 is difficult. n = 2 is quite nice, while
n = 3 is problematic. I did search for a pattern, to no avail:
• n = 0 (trivial case):
󰁝 π/2
0
1
2
sinn x
dx =
sinn x + cosn x
󰁝 π/2
0
1
dx
2
This integral is covered by blackpenredpen at: https://www.youtube.com/watch?v=9xaGYqiOkPM
This integral is covered by The Religious Atheists at https://www.youtube.com/watch?v=K-M7-gyw5t4,
blackpenredpen at: https://www.youtube.com/watch?v=xGhdatEGsZw,
and Flammable Maths at https://www.youtube.com/watch?v=b4ZuFeInVHE
1
• n = 1 (not fun):
󰁝 π/2
0
• n = 2 (trivial case):
sinn x
dx =
sinn x + cosn x
󰁝 π/2
0
0
0
sinn x
dx =
sinn x + cosn x
• n = 3 (things start to get out of control):
󰁝 π/2
󰁝 π/2
sinn x
dx =
sinn x + cosn x
󰁝 π/2
󰁝 π/2
0
sinx
dx
sinx + cosx
sin2 x dx
0
sin3 x
dx
sin3 x + cos3 x
Kind of headed nowhere.
Then, since we are dealing with sines and cosines, I figured I would look at the oddness/evenness of our integrand, using the techniques found in my article https://mymathteacheristerrible.com/blog/2019/3/23/
crazy-integration-odd-functions-even-functions-calculus-thinking:
f (x) =
sinn x
, n an integer
sinn x + cosn x
Even component:
fe (x) =
󰀆
1󰀅
f (x) + f (−x)
2
󰀗
󰀘
1
sinn x
sinn (−x)
=
+
2 sinn x + cosn x sinn (−x) + cosn (−x)
This gets messy immediately, because, depending on whether n is odd or even, the odd/even trig identities produce
different signs. Really tough to manage.
Odd component: Not even worth it. See above.
Once I saw how messy this whole thing could become, I immediately went to my favorite tool for throwing shapes
around: Desmos (www.demos.com). I created this simple workbook:
https://www.desmos.com/calculator/gdivpe79vd.
Desmos is so awesome with its ability to add a slider for some index, coefficient, or parameter, such as our n in
this problem. Load it up and start changing the value of n with the slider. Give it a try.
Now, let’s take some cases for values of n, as we did before.
• n=0
The graph is a constant. This is our trivial n = 0 case. Clearly, the area under the curve for y = 12 on
x 󰂃 [0, π2 ] is simply:
π 1
π
A=b·h= · =
2 2
4
2
• n=1
The graph is almost a straight line. This is where the provided graph, y = π2 x comes in. The graph of f (x)
is so close to that graph that, if we approximate the area under the curve of f (x), it’s a triangle with base
b = π2 and height h = 1. This gives us the area under the curve for y = π2 x on x 󰂃 [0, π2 ] as simply:
1
1 π
π
A≈ b·h= · ·1=
2
2 2
4
Basically the same area as the n = 0 case? Hm.
• n = 50
See for yourself that the intervening values (n = 2 through n = 49) don’t give solid clues as did our first 2
values. With n = 50, we see that the first half of the curve, on x 󰂃 [0, π4 ] appears to approach y = 0, with a
more and more right-angle-ish jump at x = π4 , while the half on [ π4 , π2 ] is nearly flat at y = 1. We can think
of these halves of the curve as roughly 2 rectangles, where all we would need to do is add their areas:
π
π
A≈0+ ·1=
4
4
A pattern is emerging...
• n = 100
Cranking n to 100 shows more of the same as we saw with n = 50; the area appears to be tending toward
A = π4 .
But, what now? Well, time passed. I looked in every Calculus book I had, and even ordered a couple of advanced
integration books. None of them had the trick. I know this is covered in some Electrical Engineering textbooks,
as this function is one of the ways to model a stepping voltage function (you know, the basis of all computing),
but I could not find a book that did a proof nor a derivation of how this should be approached. I put it on the
shelf for a while.
I did notice that the integrand appeared to be a parameterized integrand with parameter n. Still, it did not
lend itself to being solved in any particular way, mostly because the terms were trapped in an integral, and the
parameter was trapped in an exponent.
Fast forward a few weeks to the video I mentioned in footnote 2, above. The trick that appeared to be used was:
󰁝 b
󰁝 b
f (x)dx =
f (a + b − x)dx
a
a
Fair enough, but...? Back to first principles. Let’s apply the Fundamental Theorem of Calculus:
󰁝 b
f (x)dx = F (b) − F (a), where F represents an antiderivative of f
(1)
a
and try to prove the result. Seems easy enough, since a and b are both constants. So:
󰁝 b
󰁝 b
?
f (x)dx =
f (a + b − x)dx
a
a
?
(extra minus signs come from the chain rule!)
?
= −F (a) − (−F (b))
(simplify)
= F (b) − F (a)
?
(simplify)
F (b) − F (a) = −F (a + b − b) − (−F (a + b − a))
F (b) − F (a) = F (b) − F (a)
(2)
(yes. yes, it does. QED.)
(2) warrants a closer look. On the right hand side, our integral is:
󰁝 b
f (a + b − x)dx
a
3
(3)
Let:
u=a+b−x
(a and b are constants)
dx = −du
solve for dx
du = −dx
(differentiate, constants go away)
Performing the substitution, (3) becomes:
−
󰁝 b
f (u)du
(4)
a
So, we can think of this property - called the King property of integration - as a variation on u-substitution. As a
side note, I cannot seem to find any reference to why this is called the King property. If you are reading this and
know why, please email me!
OK, back to the integral at hand:
󰁝 π/2
0
sinn x
dx
sinn x + cosn x
Applying the King property with a = 0 and b = π2 , we have:
󰁝 π/2
0
sinn x
dx =
sinn x + cosn x
󰁝 π/2
=
󰁝 π/2
0
0
󰀃
󰀄
sinn 0 + π2 − x
󰀃
󰀄
󰀃
󰀄 dx
sinn 0 + π2 − x + cosn 0 + π2 − x
󰀃
󰀄
sinn π2 − x
󰀃
󰀄
󰀃
󰀄 dx
sinn π2 − x + cosn π2 − x
The astute observer will notice that something interesting just happened. Recall:
󰀕
󰀖
π
sin
− x = cosx
2
󰀕
󰀖
π
cos
− x = sinx
2
and recall this about notation:
sinn x = (sinx)n
n
cos x = (cosx)
(5)
n
Doing some algebra on (5), we see:
sinn x = (sinx)n
󰀗 󰀕
󰀖󰀘n
π
= cos
−x
2
= cosn
󰀕
π
−x
2
󰀖
Applying this stuff to our integral, it becomes:
󰀃
󰀄
󰁝 π/2
󰁝 π/2
sinn π2 − x
cosn x
󰀃π
󰀄
󰀃π
󰀄 dx =
dx
cosn x + sinn x
sinn 2 − x + cosn 2 − x
0
0
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Doesn’t look much better, I agree. But let’s use a technique that we might have seen with Integration by Parts.
Namely, let:
I=
󰁝 π/2
sinn x
dx
sinn x + cosn x
󰁝 π/2
cosn x
dx
cosn x + sinn x
0
We also know that I is:
I=
0
Adding these two equations, we get:
I +I =
󰁝 π/2
0
2I =
󰁝 π/2 󰀗
0
=
󰁝 π/2
0
cosn x
dx
cosn x + sinn x
󰀘
sinn x
cosn x
+
dx
cosn x + sinn x
cosn x + sinn x
󰁝 π/2
sinn x + cosn x
dx
cosn x + sinn x
󰁝 π/2
dx
0
=
sinn x
dx +
sinn x + cosn x
(apply linearity)
(combine and simplify)
(simplify that fraction! OMG!)
0
=x
󰀘π/2
0
2I =
π
2
I=
π
4
(evaluate and simplify)
I mean, seriously. Applying the King property just totally slayed the nth power situation in the original integral!
This is like, the coolest thing I’ve seen in a while. This integral drove me nuts for a few weeks while it simmered
on the back burner, then suddenly ... DONE. Wow.
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Two more examples.
1) Integrate:
󰁝 π/2
0
1
dx
1 + (tanx)π
This is one where the King property can be applied right away, so:
󰁝 π/2
󰁝 π/2
1
1
󰀅
󰀆π dx
dx =
π
1 + (tanx)
1 + tan(0 + π2 − x)
0
0
=
󰁝 π/2
0
=
󰁝 π/2
0
=
󰁝 π/2
0
1
dx
1 + (cotx)π
1
󰀄π
1
tanx
dx
(apply arithmetic inverse identity)
(tanx)π
dx
tanx)π + 1
(simplify the complex fraction)
1+
󰀃
󰀃
(apply cofunction identity)
So, we know that our integral, I, can be thought of in two ways:
󰁝 π/2
1
I=
dx
1 + (tanx)π
0
I=
󰁝 π/2
0
󰀃
(tanx)π
dx
tanx)π + 1
Adding both equations and solving, we get:
󰁝 π/2
󰁝 π/2
1
(tanx)π
󰀃
2I =
dx
+
dx
1 + (tanx)π
tanx)π + 1
0
0
=
󰁝 π/2 󰀗
0
=
󰁝 π/2
1 + (tanx)π
dx
1 + (tanx)π
󰁝 π/2
dx
0
=
󰀘
1
(tanx)π
󰀃
+
dx
1 + (tanx)π
tanx)π + 1
(apply linearity)
(combine fractions w. common denominator)
(simplify fraction!)
0
2I =
π
2
(evaluate and simplify)
I=
π
4
(Done.)
6
2) Integrate:
󰁝 1
0
ln(x + 1)
dx
x2 + 1
Straight away, we should make the usual substitution of u = tanθ to get started, so our integral becomes:
󰁝 π/4
ln(tanθ + 1)dθ =
0
󰁝 π/4
0
=
󰁝 π/4
0
=
󰁝 π/4
0
=
󰀘
tan( π4 ) − tanθ
󰀃
ln
+ 1 dθ
1 + tan π4 ) · tanθ
ln(tanθ + 1)dθ =
󰁝 π/4
ln(tanθ + 1)dθ =
0
2
󰁝 π/4
󰀕
(combine fraction using common denom)
󰀕
󰀖
2
dθ
1 + tanθ
ln 2 dθ −
󰁝 π/4
ln(1 + tanθ)dθ
󰁝 π/4
ln 2 dθ −
󰁝 π/4
ln(1 + tanθ)dθ
󰁝 π/4
ln 2 dθ
(apply log property of division)
0
0
(evaluate and simplify)
0
π
= ln 2
4
0
(simplify)
ln
= ln 2 · θ
󰁝 π/4
󰀗
󰁝 π/4
0
0
(angle difference identity for tan)
󰀖
1 − tanθ + 1 + tanθ
ln
dθ
1 + tanθ
0
󰁝 π/4
󰀗
󰀘
1 − tanθ
ln
+ 1 dθ
1 + tanθ
0
=
(do the trig sub, then apply King property)
󰁝 π/4
0
=
󰀗
󰀘
󰀃π
󰀄
ln tan
− θ + 1 dθ
4
ln(tanθ + 1)dθ =
󰀘π/4
0
π
ln 2
8
(Done.)
Done.
7
Reporting errors and giving feedback
I am so pleased that you have downloaded this study guide and have considered the techniques herein. To that
end, I am the only writer and the only editor of these things, so if you find an error in the text or calculations,
please email me and tell me about it! I am committed to prompt changes when something is inaccurate. I also
really appreciate it when someone takes a moment to tell me how I’m doing with these sorts of things, so please
do so, if you feel inclined.
My email address is: phil.petrocelli@gmail.com.
Please visit https://mymathteacheristerrible.com for other study guides. Please tell others about it.
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