Mixing
Annual MIT Integration Bee - Year Change in Integral
$begingroup$
Problem $18$ in the Annual MIT Integration Bee cleverly uses the year $2017$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
There is a solution on MSE at MIT Integration Bee 2017 problem:$int_0^{pi/2}frac 1 {1+tan^{2017} x} , dx$ : Need hints
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Can a general form be found to meet that criteria?
integration
asked Nov 13 '17 at 16:03
MooMoo
5,61131020
$endgroup$
add a comment |
$begingroup$
Problem $18$ in the Annual MIT Integration Bee cleverly uses the year $2017$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
There is a solution on MSE at MIT Integration Bee 2017 problem:$int_0^{pi/2}frac 1 {1+tan^{2017} x} , dx$ : Need hints
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Can a general form be found to meet that criteria?
integration
asked Nov 13 '17 at 16:03
MooMoo
5,61131020
$endgroup$
3
$begingroup$
$frac{1}{1+a}+frac{1}{1+frac{1}{a}}=1$ that is the heart of the "miracle".
$endgroup$
– FDP
Nov 13 '17 at 20:43
add a comment |
$begingroup$
Problem $18$ in the Annual MIT Integration Bee cleverly uses the year $2017$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
There is a solution on MSE at MIT Integration Bee 2017 problem:$int_0^{pi/2}frac 1 {1+tan^{2017} x} , dx$ : Need hints
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Can a general form be found to meet that criteria?
integration
asked Nov 13 '17 at 16:03
MooMoo
5,61131020
$endgroup$
Problem $18$ in the Annual MIT Integration Bee cleverly uses the year $2017$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
There is a solution on MSE at MIT Integration Bee 2017 problem:$int_0^{pi/2}frac 1 {1+tan^{2017} x} , dx$ : Need hints
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Can a general form be found to meet that criteria?
integration
integration
asked Nov 13 '17 at 16:03
MooMoo
5,61131020
asked Nov 13 '17 at 16:03
MooMoo
5,61131020
edited Nov 13 '17 at 16:09
Moo
asked Nov 13 '17 at 16:03
MooMoo
5,61131020
asked Nov 13 '17 at 16:03
MooMoo
5,61131020
asked Nov 13 '17 at 16:03
MooMoo
5,61131020
5,61131020
3
$begingroup$
$frac{1}{1+a}+frac{1}{1+frac{1}{a}}=1$ that is the heart of the "miracle".
$endgroup$
– FDP
Nov 13 '17 at 20:43
add a comment |
3
$begingroup$
$frac{1}{1+a}+frac{1}{1+frac{1}{a}}=1$ that is the heart of the "miracle".
$endgroup$
– FDP
Nov 13 '17 at 20:43
3
3
$begingroup$
$frac{1}{1+a}+frac{1}{1+frac{1}{a}}=1$ that is the heart of the "miracle".
$endgroup$
– FDP
Nov 13 '17 at 20:43
$begingroup$
$frac{1}{1+a}+frac{1}{1+frac{1}{a}}=1$ that is the heart of the "miracle".
$endgroup$
– FDP
Nov 13 '17 at 20:43
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you follow through the linked solution the fact that the exponent is $2017$ is not used at all. For any real $a$ we have $int_0^{pi/2} frac 1{1+tan^a(x)}dx=frac pi 4$. It does depend on the limits of integration being $0$ and $frac pi 2$ so the reflection works.
answered Nov 13 '17 at 16:10
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
I should have looked at that a lot closer! Thanks.
$endgroup$
– Moo
Nov 13 '17 at 16:53
add a comment |
$begingroup$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
This uses the transformation
$$I=int_a^bf(x) dx = int_a^bf(a+b-x) dx$$
Then, when they are added,
$$2I = int_a^b1 dx=b-a$$
$$I=frac{b-a}{2} text{, as long as } a+b =pi/2$$
As another example
$$int_{pi/6}^{pi/3} frac 1 {1+tan^{2017} x} , dx = pi/12$$
Note that the year, or the power to which $tan(x)$ is raised to, doesn't matter.
answered Nov 13 '17 at 16:38
John LouJohn Lou
2,261816
$endgroup$
add a comment |
$begingroup$
Substitute the following definition of $tan x$ as the quotient of $sin x$ and $cos x$ into the integral. $$tan x=dfrac{sin x}{cos x}implies I=int_{0}^{pi/2} dfrac{cos^{2017} x}{sin^{2017} x + cos^{2017} x} dx tag1$$ Now use the reflective property of Integrals stated as follows $$int_a^b f(x)dx =int_a^b f(a+b-x) dx tag2$$ to obtain the equivalent of $I$ as $$I=int_0^{pi/2} dfrac{sin^{2017} x}{sin^{2017} x +cos^{2017} x}dx tag3$$ Add Equations $(1)$ and $(3)$ to get $2I=int_{0}^{pi/2} dx$ which gives you the value of $I$ as $frac{pi}{4}$.
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Sure, but here's some more neatness to this integral, you need not even change the limits of integration each year to get the ${pi}/{4}$. In fact one might as well prove that the integral denoted $I_n=int_{a}^{b} frac{1}{1+tan^n x} dx$ is independent of $n$.
In fact while setting an integral like this you might wanna experiment with the limits of integration to get even more elegant results. But for the reflective property i.e. equation $(2)$ to hold, the following needs to hold true: $a+b=pi/2$
Exercise
- Evaluate $int_{a}^{b} frac{1}{1+tan^n x} dx$ in terms of $a$ and $b$ where $a+b=pi/2$.
- (Bonus) Judge whether $int_{pi/2-pi^{e}}^{pi^e} frac{1}{1+tan^n x} dx$ is negative or positive.
answered Jan 15 at 7:19
Paras KhoslaParas Khosla
632213
$endgroup$
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
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$begingroup$
If you follow through the linked solution the fact that the exponent is $2017$ is not used at all. For any real $a$ we have $int_0^{pi/2} frac 1{1+tan^a(x)}dx=frac pi 4$. It does depend on the limits of integration being $0$ and $frac pi 2$ so the reflection works.
answered Nov 13 '17 at 16:10
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
I should have looked at that a lot closer! Thanks.
$endgroup$
– Moo
Nov 13 '17 at 16:53
add a comment |
$begingroup$
If you follow through the linked solution the fact that the exponent is $2017$ is not used at all. For any real $a$ we have $int_0^{pi/2} frac 1{1+tan^a(x)}dx=frac pi 4$. It does depend on the limits of integration being $0$ and $frac pi 2$ so the reflection works.
answered Nov 13 '17 at 16:10
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
I should have looked at that a lot closer! Thanks.
$endgroup$
– Moo
Nov 13 '17 at 16:53
add a comment |
$begingroup$
If you follow through the linked solution the fact that the exponent is $2017$ is not used at all. For any real $a$ we have $int_0^{pi/2} frac 1{1+tan^a(x)}dx=frac pi 4$. It does depend on the limits of integration being $0$ and $frac pi 2$ so the reflection works.
answered Nov 13 '17 at 16:10
Ross MillikanRoss Millikan
297k23198371
$endgroup$
If you follow through the linked solution the fact that the exponent is $2017$ is not used at all. For any real $a$ we have $int_0^{pi/2} frac 1{1+tan^a(x)}dx=frac pi 4$. It does depend on the limits of integration being $0$ and $frac pi 2$ so the reflection works.
answered Nov 13 '17 at 16:10
Ross MillikanRoss Millikan
297k23198371
answered Nov 13 '17 at 16:10
Ross MillikanRoss Millikan
297k23198371
answered Nov 13 '17 at 16:10
Ross MillikanRoss Millikan
297k23198371
answered Nov 13 '17 at 16:10
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
I should have looked at that a lot closer! Thanks.
$endgroup$
– Moo
Nov 13 '17 at 16:53
add a comment |
$begingroup$
I should have looked at that a lot closer! Thanks.
$endgroup$
– Moo
Nov 13 '17 at 16:53
$begingroup$
I should have looked at that a lot closer! Thanks.
$endgroup$
– Moo
Nov 13 '17 at 16:53
$begingroup$
I should have looked at that a lot closer! Thanks.
$endgroup$
– Moo
Nov 13 '17 at 16:53
add a comment |
$begingroup$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
This uses the transformation
$$I=int_a^bf(x) dx = int_a^bf(a+b-x) dx$$
Then, when they are added,
$$2I = int_a^b1 dx=b-a$$
$$I=frac{b-a}{2} text{, as long as } a+b =pi/2$$
As another example
$$int_{pi/6}^{pi/3} frac 1 {1+tan^{2017} x} , dx = pi/12$$
Note that the year, or the power to which $tan(x)$ is raised to, doesn't matter.
answered Nov 13 '17 at 16:38
John LouJohn Lou
2,261816
$endgroup$
add a comment |
$begingroup$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
This uses the transformation
$$I=int_a^bf(x) dx = int_a^bf(a+b-x) dx$$
Then, when they are added,
$$2I = int_a^b1 dx=b-a$$
$$I=frac{b-a}{2} text{, as long as } a+b =pi/2$$
As another example
$$int_{pi/6}^{pi/3} frac 1 {1+tan^{2017} x} , dx = pi/12$$
Note that the year, or the power to which $tan(x)$ is raised to, doesn't matter.
answered Nov 13 '17 at 16:38
John LouJohn Lou
2,261816
$endgroup$
add a comment |
$begingroup$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
This uses the transformation
$$I=int_a^bf(x) dx = int_a^bf(a+b-x) dx$$
Then, when they are added,
$$2I = int_a^b1 dx=b-a$$
$$I=frac{b-a}{2} text{, as long as } a+b =pi/2$$
As another example
$$int_{pi/6}^{pi/3} frac 1 {1+tan^{2017} x} , dx = pi/12$$
Note that the year, or the power to which $tan(x)$ is raised to, doesn't matter.
answered Nov 13 '17 at 16:38
John LouJohn Lou
2,261816
$endgroup$
$$int_0^{pi/2} frac 1 {1+tan^{2017} x} , dx = dfrac{pi}{4}$$
This uses the transformation
$$I=int_a^bf(x) dx = int_a^bf(a+b-x) dx$$
Then, when they are added,
$$2I = int_a^b1 dx=b-a$$
$$I=frac{b-a}{2} text{, as long as } a+b =pi/2$$
As another example
$$int_{pi/6}^{pi/3} frac 1 {1+tan^{2017} x} , dx = pi/12$$
Note that the year, or the power to which $tan(x)$ is raised to, doesn't matter.
answered Nov 13 '17 at 16:38
John LouJohn Lou
2,261816
answered Nov 13 '17 at 16:38
John LouJohn Lou
2,261816
answered Nov 13 '17 at 16:38
John LouJohn Lou
2,261816
answered Nov 13 '17 at 16:38
John LouJohn Lou
2,261816
2,261816
add a comment |
add a comment |
$begingroup$
Substitute the following definition of $tan x$ as the quotient of $sin x$ and $cos x$ into the integral. $$tan x=dfrac{sin x}{cos x}implies I=int_{0}^{pi/2} dfrac{cos^{2017} x}{sin^{2017} x + cos^{2017} x} dx tag1$$ Now use the reflective property of Integrals stated as follows $$int_a^b f(x)dx =int_a^b f(a+b-x) dx tag2$$ to obtain the equivalent of $I$ as $$I=int_0^{pi/2} dfrac{sin^{2017} x}{sin^{2017} x +cos^{2017} x}dx tag3$$ Add Equations $(1)$ and $(3)$ to get $2I=int_{0}^{pi/2} dx$ which gives you the value of $I$ as $frac{pi}{4}$.
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Sure, but here's some more neatness to this integral, you need not even change the limits of integration each year to get the ${pi}/{4}$. In fact one might as well prove that the integral denoted $I_n=int_{a}^{b} frac{1}{1+tan^n x} dx$ is independent of $n$.
In fact while setting an integral like this you might wanna experiment with the limits of integration to get even more elegant results. But for the reflective property i.e. equation $(2)$ to hold, the following needs to hold true: $a+b=pi/2$
Exercise
- Evaluate $int_{a}^{b} frac{1}{1+tan^n x} dx$ in terms of $a$ and $b$ where $a+b=pi/2$.
- (Bonus) Judge whether $int_{pi/2-pi^{e}}^{pi^e} frac{1}{1+tan^n x} dx$ is negative or positive.
answered Jan 15 at 7:19
Paras KhoslaParas Khosla
632213
$endgroup$
add a comment |
$begingroup$
Substitute the following definition of $tan x$ as the quotient of $sin x$ and $cos x$ into the integral. $$tan x=dfrac{sin x}{cos x}implies I=int_{0}^{pi/2} dfrac{cos^{2017} x}{sin^{2017} x + cos^{2017} x} dx tag1$$ Now use the reflective property of Integrals stated as follows $$int_a^b f(x)dx =int_a^b f(a+b-x) dx tag2$$ to obtain the equivalent of $I$ as $$I=int_0^{pi/2} dfrac{sin^{2017} x}{sin^{2017} x +cos^{2017} x}dx tag3$$ Add Equations $(1)$ and $(3)$ to get $2I=int_{0}^{pi/2} dx$ which gives you the value of $I$ as $frac{pi}{4}$.
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Sure, but here's some more neatness to this integral, you need not even change the limits of integration each year to get the ${pi}/{4}$. In fact one might as well prove that the integral denoted $I_n=int_{a}^{b} frac{1}{1+tan^n x} dx$ is independent of $n$.
In fact while setting an integral like this you might wanna experiment with the limits of integration to get even more elegant results. But for the reflective property i.e. equation $(2)$ to hold, the following needs to hold true: $a+b=pi/2$
Exercise
- Evaluate $int_{a}^{b} frac{1}{1+tan^n x} dx$ in terms of $a$ and $b$ where $a+b=pi/2$.
- (Bonus) Judge whether $int_{pi/2-pi^{e}}^{pi^e} frac{1}{1+tan^n x} dx$ is negative or positive.
answered Jan 15 at 7:19
Paras KhoslaParas Khosla
632213
$endgroup$
add a comment |
$begingroup$
Substitute the following definition of $tan x$ as the quotient of $sin x$ and $cos x$ into the integral. $$tan x=dfrac{sin x}{cos x}implies I=int_{0}^{pi/2} dfrac{cos^{2017} x}{sin^{2017} x + cos^{2017} x} dx tag1$$ Now use the reflective property of Integrals stated as follows $$int_a^b f(x)dx =int_a^b f(a+b-x) dx tag2$$ to obtain the equivalent of $I$ as $$I=int_0^{pi/2} dfrac{sin^{2017} x}{sin^{2017} x +cos^{2017} x}dx tag3$$ Add Equations $(1)$ and $(3)$ to get $2I=int_{0}^{pi/2} dx$ which gives you the value of $I$ as $frac{pi}{4}$.
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Sure, but here's some more neatness to this integral, you need not even change the limits of integration each year to get the ${pi}/{4}$. In fact one might as well prove that the integral denoted $I_n=int_{a}^{b} frac{1}{1+tan^n x} dx$ is independent of $n$.
In fact while setting an integral like this you might wanna experiment with the limits of integration to get even more elegant results. But for the reflective property i.e. equation $(2)$ to hold, the following needs to hold true: $a+b=pi/2$
Exercise
- Evaluate $int_{a}^{b} frac{1}{1+tan^n x} dx$ in terms of $a$ and $b$ where $a+b=pi/2$.
- (Bonus) Judge whether $int_{pi/2-pi^{e}}^{pi^e} frac{1}{1+tan^n x} dx$ is negative or positive.
answered Jan 15 at 7:19
Paras KhoslaParas Khosla
632213
$endgroup$
Substitute the following definition of $tan x$ as the quotient of $sin x$ and $cos x$ into the integral. $$tan x=dfrac{sin x}{cos x}implies I=int_{0}^{pi/2} dfrac{cos^{2017} x}{sin^{2017} x + cos^{2017} x} dx tag1$$ Now use the reflective property of Integrals stated as follows $$int_a^b f(x)dx =int_a^b f(a+b-x) dx tag2$$ to obtain the equivalent of $I$ as $$I=int_0^{pi/2} dfrac{sin^{2017} x}{sin^{2017} x +cos^{2017} x}dx tag3$$ Add Equations $(1)$ and $(3)$ to get $2I=int_{0}^{pi/2} dx$ which gives you the value of $I$ as $frac{pi}{4}$.
My question is if it is possible to always get such a clean solution by modifying the year to the current year (each year) and by changing the limits of integration?
Sure, but here's some more neatness to this integral, you need not even change the limits of integration each year to get the ${pi}/{4}$. In fact one might as well prove that the integral denoted $I_n=int_{a}^{b} frac{1}{1+tan^n x} dx$ is independent of $n$.
In fact while setting an integral like this you might wanna experiment with the limits of integration to get even more elegant results. But for the reflective property i.e. equation $(2)$ to hold, the following needs to hold true: $a+b=pi/2$
Exercise
- Evaluate $int_{a}^{b} frac{1}{1+tan^n x} dx$ in terms of $a$ and $b$ where $a+b=pi/2$.
- (Bonus) Judge whether $int_{pi/2-pi^{e}}^{pi^e} frac{1}{1+tan^n x} dx$ is negative or positive.
answered Jan 15 at 7:19
Paras KhoslaParas Khosla
632213
edited Jan 15 at 14:19
answered Jan 15 at 7:19
Paras KhoslaParas Khosla
632213
answered Jan 15 at 7:19
Paras KhoslaParas Khosla
632213
answered Jan 15 at 7:19
Paras KhoslaParas Khosla
632213
632213
add a comment |
add a comment |
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$begingroup$
$frac{1}{1+a}+frac{1}{1+frac{1}{a}}=1$ that is the heart of the "miracle".
$endgroup$
– FDP
Nov 13 '17 at 20:43