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Definite integral problem - sin and its inverse [duplicate]
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Proof of Young's inequality
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The problem is as follows:
let:
$$
0<a<fracpi2 , 0<b<1
$$
show that:
$$
int_0^asin(x)dx+int_0^barcsin(x)dxgeq ab
$$
I've tried to calculate the second integral over y and not over x (over $sin(y)$, didn't got very far.
The solution should be simple, and may involve using double integrals or using the graphs of the sine function and it's inverse in the rectangle that $a$ and $b$ form in $mathbbR^2$.
even an hint would be appreciated.
Thanks
calculus integration definite-integrals
asked Aug 2 at 16:05
juleand
896
marked as duplicate by StubbornAtom, Batominovski, Simply Beautiful Art calculus
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Aug 2 at 16:59
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up vote
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This question already has an answer here:
Proof of Young's inequality
2 answers
The problem is as follows:
let:
$$
0<a<fracpi2 , 0<b<1
$$
show that:
$$
int_0^asin(x)dx+int_0^barcsin(x)dxgeq ab
$$
I've tried to calculate the second integral over y and not over x (over $sin(y)$, didn't got very far.
The solution should be simple, and may involve using double integrals or using the graphs of the sine function and it's inverse in the rectangle that $a$ and $b$ form in $mathbbR^2$.
even an hint would be appreciated.
Thanks
calculus integration definite-integrals
asked Aug 2 at 16:05
juleand
896
marked as duplicate by StubbornAtom, Batominovski, Simply Beautiful Art calculus
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Aug 2 at 16:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
It's more of a general inequality: en.wikipedia.org/wiki/….
– StubbornAtom
Aug 2 at 16:11
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up vote
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down vote
favorite
This question already has an answer here:
Proof of Young's inequality
2 answers
The problem is as follows:
let:
$$
0<a<fracpi2 , 0<b<1
$$
show that:
$$
int_0^asin(x)dx+int_0^barcsin(x)dxgeq ab
$$
I've tried to calculate the second integral over y and not over x (over $sin(y)$, didn't got very far.
The solution should be simple, and may involve using double integrals or using the graphs of the sine function and it's inverse in the rectangle that $a$ and $b$ form in $mathbbR^2$.
even an hint would be appreciated.
Thanks
calculus integration definite-integrals
asked Aug 2 at 16:05
juleand
896
This question already has an answer here:
Proof of Young's inequality
2 answers
The problem is as follows:
let:
$$
0<a<fracpi2 , 0<b<1
$$
show that:
$$
int_0^asin(x)dx+int_0^barcsin(x)dxgeq ab
$$
I've tried to calculate the second integral over y and not over x (over $sin(y)$, didn't got very far.
The solution should be simple, and may involve using double integrals or using the graphs of the sine function and it's inverse in the rectangle that $a$ and $b$ form in $mathbbR^2$.
even an hint would be appreciated.
Thanks
This question already has an answer here:
Proof of Young's inequality
2 answers
calculus integration definite-integrals
asked Aug 2 at 16:05
juleand
896
asked Aug 2 at 16:05
juleand
896
asked Aug 2 at 16:05
juleand
896
asked Aug 2 at 16:05
juleand
896
896
marked as duplicate by StubbornAtom, Batominovski, Simply Beautiful Art calculus
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
It's more of a general inequality: en.wikipedia.org/wiki/….
– StubbornAtom
Aug 2 at 16:11
add a comment |Â
3
It's more of a general inequality: en.wikipedia.org/wiki/….
– StubbornAtom
Aug 2 at 16:11
3
3
It's more of a general inequality: en.wikipedia.org/wiki/….
– StubbornAtom
Aug 2 at 16:11
It's more of a general inequality: en.wikipedia.org/wiki/….
– StubbornAtom
Aug 2 at 16:11
add a comment |Â
1 Answer
1
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0
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The signs and monotonicty here are such that the integrals measure the area below the curve.
The parameter ranges allows the decomposition of the rectangle $[0, b] times [0, arcsin(b)]$ like this:
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b - intlimits_0^arcsin(b) sin(y), dy
$$
For $a le arcsin(b)$ we have
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b
- intlimits_0^a sin(y), dy
- intlimits_a^arcsin(b) sin(y), dy
$$
which means
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(x), dx =
underbracearcsin(b) , b_ge ab
-
underbraceintlimits_a^arcsin(b) sin(x), dx_ge 0
ge a b
$$
We can also decompose $[0, a] times [0, sin(a)]$ like this:
$$
intlimits_0^a sin(x), dx =
a, sin(a) - intlimits_0^sin(a) arcsin(y), dy
$$
This gives an estimation for $b le sin(a)$:
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(y), dy ge
a, b
$$
Finally
$$
b le sin(a) iff arcsin(b) le arcsin(sin(a))= a
$$
so we cover the other case as well.
answered Aug 2 at 16:15
mvw
30.2k22250
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The signs and monotonicty here are such that the integrals measure the area below the curve.
The parameter ranges allows the decomposition of the rectangle $[0, b] times [0, arcsin(b)]$ like this:
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b - intlimits_0^arcsin(b) sin(y), dy
$$
For $a le arcsin(b)$ we have
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b
- intlimits_0^a sin(y), dy
- intlimits_a^arcsin(b) sin(y), dy
$$
which means
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(x), dx =
underbracearcsin(b) , b_ge ab
-
underbraceintlimits_a^arcsin(b) sin(x), dx_ge 0
ge a b
$$
We can also decompose $[0, a] times [0, sin(a)]$ like this:
$$
intlimits_0^a sin(x), dx =
a, sin(a) - intlimits_0^sin(a) arcsin(y), dy
$$
This gives an estimation for $b le sin(a)$:
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(y), dy ge
a, b
$$
Finally
$$
b le sin(a) iff arcsin(b) le arcsin(sin(a))= a
$$
so we cover the other case as well.
answered Aug 2 at 16:15
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
The signs and monotonicty here are such that the integrals measure the area below the curve.
The parameter ranges allows the decomposition of the rectangle $[0, b] times [0, arcsin(b)]$ like this:
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b - intlimits_0^arcsin(b) sin(y), dy
$$
For $a le arcsin(b)$ we have
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b
- intlimits_0^a sin(y), dy
- intlimits_a^arcsin(b) sin(y), dy
$$
which means
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(x), dx =
underbracearcsin(b) , b_ge ab
-
underbraceintlimits_a^arcsin(b) sin(x), dx_ge 0
ge a b
$$
We can also decompose $[0, a] times [0, sin(a)]$ like this:
$$
intlimits_0^a sin(x), dx =
a, sin(a) - intlimits_0^sin(a) arcsin(y), dy
$$
This gives an estimation for $b le sin(a)$:
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(y), dy ge
a, b
$$
Finally
$$
b le sin(a) iff arcsin(b) le arcsin(sin(a))= a
$$
so we cover the other case as well.
answered Aug 2 at 16:15
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The signs and monotonicty here are such that the integrals measure the area below the curve.
The parameter ranges allows the decomposition of the rectangle $[0, b] times [0, arcsin(b)]$ like this:
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b - intlimits_0^arcsin(b) sin(y), dy
$$
For $a le arcsin(b)$ we have
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b
- intlimits_0^a sin(y), dy
- intlimits_a^arcsin(b) sin(y), dy
$$
which means
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(x), dx =
underbracearcsin(b) , b_ge ab
-
underbraceintlimits_a^arcsin(b) sin(x), dx_ge 0
ge a b
$$
We can also decompose $[0, a] times [0, sin(a)]$ like this:
$$
intlimits_0^a sin(x), dx =
a, sin(a) - intlimits_0^sin(a) arcsin(y), dy
$$
This gives an estimation for $b le sin(a)$:
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(y), dy ge
a, b
$$
Finally
$$
b le sin(a) iff arcsin(b) le arcsin(sin(a))= a
$$
so we cover the other case as well.
answered Aug 2 at 16:15
mvw
30.2k22250
The signs and monotonicty here are such that the integrals measure the area below the curve.
The parameter ranges allows the decomposition of the rectangle $[0, b] times [0, arcsin(b)]$ like this:
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b - intlimits_0^arcsin(b) sin(y), dy
$$
For $a le arcsin(b)$ we have
$$
intlimits_0^b arcsin(x), dx =
arcsin(b), b
- intlimits_0^a sin(y), dy
- intlimits_a^arcsin(b) sin(y), dy
$$
which means
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(x), dx =
underbracearcsin(b) , b_ge ab
-
underbraceintlimits_a^arcsin(b) sin(x), dx_ge 0
ge a b
$$
We can also decompose $[0, a] times [0, sin(a)]$ like this:
$$
intlimits_0^a sin(x), dx =
a, sin(a) - intlimits_0^sin(a) arcsin(y), dy
$$
This gives an estimation for $b le sin(a)$:
$$
intlimits_0^a sin(x), dx + intlimits_0^b arcsin(y), dy ge
a, b
$$
Finally
$$
b le sin(a) iff arcsin(b) le arcsin(sin(a))= a
$$
so we cover the other case as well.
answered Aug 2 at 16:15
mvw
30.2k22250
edited Aug 2 at 16:51
answered Aug 2 at 16:15
mvw
30.2k22250
answered Aug 2 at 16:15
mvw
30.2k22250
answered Aug 2 at 16:15
mvw
30.2k22250
30.2k22250
add a comment |Â
add a comment |Â
3
It's more of a general inequality: en.wikipedia.org/wiki/….
– StubbornAtom
Aug 2 at 16:11