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Integrals from Brasilian Math Olympiad 2019

Compare the integrals $int_0^fracpi2sin(cos x)dx$ and $int_0^fracpi2cos(sin x)dx$Duo Fresnel-like integrals $(??)$How can I show that these integrals are zeroEvaluate $int_0^n frac2+x^2+sqrtcos nx dx$ or find good upper bound.Analytical solutions to integralsIntegral for the New Year $2019$!Olympiad Algebra Practice Question

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

2

$begingroup$

The following integrals are the Brasilian Mathematical Olympiad in 2019.

Show that $displaystyleint_1^2frace^x(x-1)x(x+e^x),dx=lnleft(frac2+e^22+2eright)$ and $displaystyleint_0^pi/2fracxsin(2x)1+cos(2x)^2,dx=fracpi^216$.

I tried to use traditional methods, but these integrals are very difficult. Has someone any suggestion? Thanks a lot.

integration definite-integrals contest-math

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edited 7 hours ago

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3

2

$begingroup$

The following integrals are the Brasilian Mathematical Olympiad in 2019.

Show that $displaystyleint_1^2frace^x(x-1)x(x+e^x),dx=lnleft(frac2+e^22+2eright)$ and $displaystyleint_0^pi/2fracxsin(2x)1+cos(2x)^2,dx=fracpi^216$.

I tried to use traditional methods, but these integrals are very difficult. Has someone any suggestion? Thanks a lot.

integration definite-integrals contest-math

share|cite|improve this question

edited 7 hours ago

カカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

asked 8 hours ago

CgomesCgomes

67344 silver badges1111 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

The following integrals are the Brasilian Mathematical Olympiad in 2019.

Show that $displaystyleint_1^2frace^x(x-1)x(x+e^x),dx=lnleft(frac2+e^22+2eright)$ and $displaystyleint_0^pi/2fracxsin(2x)1+cos(2x)^2,dx=fracpi^216$.

I tried to use traditional methods, but these integrals are very difficult. Has someone any suggestion? Thanks a lot.

integration definite-integrals contest-math

share|cite|improve this question

edited 7 hours ago

カカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

asked 8 hours ago

CgomesCgomes

67344 silver badges1111 bronze badges

$endgroup$

share|cite|improve this question

edited 7 hours ago

カカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

asked 8 hours ago

CgomesCgomes

67344 silver badges1111 bronze badges

share|cite|improve this question

edited 7 hours ago

カカロット

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asked 8 hours ago

CgomesCgomes

67344 silver badges1111 bronze badges

edited 7 hours ago

カカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

edited 7 hours ago

カカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

asked 8 hours ago

CgomesCgomes

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asked 8 hours ago

CgomesCgomes

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3 Answers
3

active

oldest

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6

$begingroup$

$$int_1^2frace^x(x-1)x(x+e^x)dx=int_1^2 frace^xleft(frac1x-frac1x^2right)1+frace^xxdx=lnleft(1+frace^xxright)bigg|_1^2 =lnleft(frac1+frace^221+eright)$$

---

For the second one, let $fracpi2-x=t$ to get:
$$mathcal J=int_0^pi/2fracxsin(2x)1+cos^2(2x)dx=int_pi/2^0 fracleft(fracpi2-tright)sin(2t)1+cos^2(2t)(-dt)oversett=x=int_0^fracpi2fracleft(fracpi2-xright)sin(2x)1+cos^2(2x)dx$$
Now add them up to get:
$$2mathcal J=fracpi2int_0^fracpi2fracsin(2x)1+cos^2(2x)dxRightarrow mathcal J=-fracpi8arctan(cos(2x))bigg|_0^fracpi2=fracpi^216$$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

カカロットカカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your last integral has the wrong bounds of integration after substitution. Just take out the negative.
  $endgroup$
  – Andrew Chin
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $xto0$ in $fracpi2-x=t$, then what is $tto$?
  $endgroup$
  – Andrew Chin
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndrewChin Are you overlooking $dt=-dx$?
  $endgroup$
  – saulspatz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh that's what I overlooked
  $endgroup$
  – Andrew Chin
  7 hours ago
  

  
  
  
  

add a comment
 | 

0

$begingroup$

Hint: $$int frac 2sin (2x)1+cos^2(2x)dx=-arctan (cos(2x))$$ (you can show this by substitution)

share|cite|improve this answer

answered 7 hours ago

Alessandro CignaAlessandro Cigna

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0

$begingroup$

For the second integral, use $int_a^bf(x) = int_a^bf(a+b-x)$ and add these together to get $I =$$ int_0^fracpi2 frac pi4sin(2x)over1+cos(2x)^2$. Substituting with $cos(2x) = t$ gives the solution easily.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Certainly not a dogCertainly not a dog

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6

$begingroup$

$$int_1^2frace^x(x-1)x(x+e^x)dx=int_1^2 frace^xleft(frac1x-frac1x^2right)1+frace^xxdx=lnleft(1+frace^xxright)bigg|_1^2 =lnleft(frac1+frace^221+eright)$$

---

For the second one, let $fracpi2-x=t$ to get:
$$mathcal J=int_0^pi/2fracxsin(2x)1+cos^2(2x)dx=int_pi/2^0 fracleft(fracpi2-tright)sin(2t)1+cos^2(2t)(-dt)oversett=x=int_0^fracpi2fracleft(fracpi2-xright)sin(2x)1+cos^2(2x)dx$$
Now add them up to get:
$$2mathcal J=fracpi2int_0^fracpi2fracsin(2x)1+cos^2(2x)dxRightarrow mathcal J=-fracpi8arctan(cos(2x))bigg|_0^fracpi2=fracpi^216$$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

カカロットカカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your last integral has the wrong bounds of integration after substitution. Just take out the negative.
  $endgroup$
  – Andrew Chin
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $xto0$ in $fracpi2-x=t$, then what is $tto$?
  $endgroup$
  – Andrew Chin
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndrewChin Are you overlooking $dt=-dx$?
  $endgroup$
  – saulspatz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh that's what I overlooked
  $endgroup$
  – Andrew Chin
  7 hours ago
  

  
  
  
  

add a comment
 | 

6

$begingroup$

$$int_1^2frace^x(x-1)x(x+e^x)dx=int_1^2 frace^xleft(frac1x-frac1x^2right)1+frace^xxdx=lnleft(1+frace^xxright)bigg|_1^2 =lnleft(frac1+frace^221+eright)$$

---

For the second one, let $fracpi2-x=t$ to get:
$$mathcal J=int_0^pi/2fracxsin(2x)1+cos^2(2x)dx=int_pi/2^0 fracleft(fracpi2-tright)sin(2t)1+cos^2(2t)(-dt)oversett=x=int_0^fracpi2fracleft(fracpi2-xright)sin(2x)1+cos^2(2x)dx$$
Now add them up to get:
$$2mathcal J=fracpi2int_0^fracpi2fracsin(2x)1+cos^2(2x)dxRightarrow mathcal J=-fracpi8arctan(cos(2x))bigg|_0^fracpi2=fracpi^216$$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

カカロットカカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your last integral has the wrong bounds of integration after substitution. Just take out the negative.
  $endgroup$
  – Andrew Chin
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $xto0$ in $fracpi2-x=t$, then what is $tto$?
  $endgroup$
  – Andrew Chin
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndrewChin Are you overlooking $dt=-dx$?
  $endgroup$
  – saulspatz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh that's what I overlooked
  $endgroup$
  – Andrew Chin
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

$$int_1^2frace^x(x-1)x(x+e^x)dx=int_1^2 frace^xleft(frac1x-frac1x^2right)1+frace^xxdx=lnleft(1+frace^xxright)bigg|_1^2 =lnleft(frac1+frace^221+eright)$$

---

For the second one, let $fracpi2-x=t$ to get:
$$mathcal J=int_0^pi/2fracxsin(2x)1+cos^2(2x)dx=int_pi/2^0 fracleft(fracpi2-tright)sin(2t)1+cos^2(2t)(-dt)oversett=x=int_0^fracpi2fracleft(fracpi2-xright)sin(2x)1+cos^2(2x)dx$$
Now add them up to get:
$$2mathcal J=fracpi2int_0^fracpi2fracsin(2x)1+cos^2(2x)dxRightarrow mathcal J=-fracpi8arctan(cos(2x))bigg|_0^fracpi2=fracpi^216$$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

カカロットカカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

カカロットカカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

answered 7 hours ago

カカロットカカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

answered 7 hours ago

カカロットカカロット

14.1k11 gold badge2323 silver badges8585 bronze badges

0

$begingroup$

Hint: $$int frac 2sin (2x)1+cos^2(2x)dx=-arctan (cos(2x))$$ (you can show this by substitution)

share|cite|improve this answer

answered 7 hours ago

Alessandro CignaAlessandro Cigna

1,09511 silver badge1414 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

Hint: $$int frac 2sin (2x)1+cos^2(2x)dx=-arctan (cos(2x))$$ (you can show this by substitution)

share|cite|improve this answer

answered 7 hours ago

Alessandro CignaAlessandro Cigna

1,09511 silver badge1414 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Hint: $$int frac 2sin (2x)1+cos^2(2x)dx=-arctan (cos(2x))$$ (you can show this by substitution)

share|cite|improve this answer

answered 7 hours ago

Alessandro CignaAlessandro Cigna

1,09511 silver badge1414 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Alessandro CignaAlessandro Cigna

1,09511 silver badge1414 bronze badges

answered 7 hours ago

Alessandro CignaAlessandro Cigna

1,09511 silver badge1414 bronze badges

answered 7 hours ago

Alessandro CignaAlessandro Cigna

1,09511 silver badge1414 bronze badges

0

$begingroup$

For the second integral, use $int_a^bf(x) = int_a^bf(a+b-x)$ and add these together to get $I =$$ int_0^fracpi2 frac pi4sin(2x)over1+cos(2x)^2$. Substituting with $cos(2x) = t$ gives the solution easily.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Certainly not a dogCertainly not a dog

14099 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

For the second integral, use $int_a^bf(x) = int_a^bf(a+b-x)$ and add these together to get $I =$$ int_0^fracpi2 frac pi4sin(2x)over1+cos(2x)^2$. Substituting with $cos(2x) = t$ gives the solution easily.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Certainly not a dogCertainly not a dog

14099 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

For the second integral, use $int_a^bf(x) = int_a^bf(a+b-x)$ and add these together to get $I =$$ int_0^fracpi2 frac pi4sin(2x)over1+cos(2x)^2$. Substituting with $cos(2x) = t$ gives the solution easily.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Certainly not a dogCertainly not a dog

14099 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Certainly not a dogCertainly not a dog

14099 bronze badges

answered 7 hours ago

Certainly not a dogCertainly not a dog

14099 bronze badges

answered 7 hours ago

Certainly not a dogCertainly not a dog

14099 bronze badges