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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Inequality by Po-Ru Loh
v_Enhance   57
N an hour ago by Learning11
Source: ELMO 2003 Problem 4
Let $x,y,z \ge 1$ be real numbers such that \[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \] Prove that \[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]
57 replies
v_Enhance
Dec 29, 2012
Learning11
an hour ago
King's Constrained Walk
Hellowings   0
2 hours ago
Source: Own
Given an n x n chessboard, with a king starting at any square, the king's task is to visit each square in the board exactly once (essentially an open path); this king moves how a king in chess would.
However, we are allowed to place k numbers on the board of any value such that for each number A we placed on the board, the king must be in the position of that number A on its Ath square in its journey, with the starting square as its 1st square.
Suppose after we placed k numbers, there is one and only one way to complete the king's task (this includes placing the king in a starting square), find the minimum value of k set by n.

Didn't know I could post it here xd; I'm unsure how hard this question could be.
0 replies
Hellowings
2 hours ago
0 replies
Plz give me the solution
Madunglecha   0
2 hours ago
For given M
h(n) is defined as the number of which is relatively prime with M, and 1 or more and n or less.
As B is h(M)/M, prove that there are at least M/3 or more N such that satisfying the below inequality
|h(N)-BN| is under 1+sqrt(B×2^((the number of prime factor of M)-3))
0 replies
Madunglecha
2 hours ago
0 replies
Inspired by Darealzolt
sqing   0
2 hours ago
Source: Own
Let $ a,b,c\geq 1$ and $ a^2+b^2+c^2+abc=\frac{9}{2}. $ Prove that
$$3\left(\sqrt[3] 2+\frac{1}{\sqrt[3] 2} -1\right) \geq a+b+c\geq  \frac{3+\sqrt{11}}{2}$$$$\frac{3}{2}\left(4+\sqrt[3] 4-\sqrt[3] 2\right) \geq a+b+c+ab+bc+ca\geq  \frac{3(1+\sqrt{11})}{2}$$
0 replies
sqing
2 hours ago
0 replies
2024 IMO P1
EthanWYX2009   104
N 2 hours ago by SYBARUPEMULA
Source: 2024 IMO P1
Determine all real numbers $\alpha$ such that, for every positive integer $n,$ the integer
$$\lfloor\alpha\rfloor +\lfloor 2\alpha\rfloor +\cdots +\lfloor n\alpha\rfloor$$is a multiple of $n.$ (Note that $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z.$ For example, $\lfloor -\pi\rfloor =-4$ and $\lfloor 2\rfloor= \lfloor 2.9\rfloor =2.$)

Proposed by Santiago Rodríguez, Colombia
104 replies
EthanWYX2009
Jul 16, 2024
SYBARUPEMULA
2 hours ago
2-var inequality
sqing   8
N 2 hours ago by sqing
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1}  -\frac{b^4}{a+1} \leq 1 $$
8 replies
sqing
May 27, 2025
sqing
2 hours ago
ai+aj is the multiple of n
Jackson0423   0
3 hours ago

Consider an increasing sequence of integers \( a_n \).
For every positive integer \( n \), there exist indices \( 1 \leq i < j \leq n \) such that \( a_i + a_j \) is divisible by \( n \).
Given that \( a_1 \geq 1 \), find the minimum possible value of \( a_{100} \).
0 replies
Jackson0423
3 hours ago
0 replies
Quadruple Binomial Coefficient Sum
P162008   2
N 3 hours ago by tliang2000
Source: Self made by my Elder brother
$\sum_{p=0}^{\infty} \sum_{r=0}^{\infty} \sum_{q=1}^{\infty} \sum_{s=0}^{p+q - 1} \frac{((-1)^{p+r+s+1})(2^{p+q-1}) \binom{p + q - s - 1}{p + q - 2s - 1}}{4^s(2p^2q + 2pqr + pq + qr)(2p + 2q + 2r + 3)}.$
2 replies
1 viewing
P162008
Yesterday at 8:04 PM
tliang2000
3 hours ago
Addition on the IMO
naman12   139
N 3 hours ago by ezpotd
Source: IMO 2020 Problem 1
Consider the convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$. The following ratio equalities hold:
\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]Prove that the following three lines meet in a point: the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment $AB$.

Proposed by Dominik Burek, Poland
139 replies
naman12
Sep 22, 2020
ezpotd
3 hours ago
IMO ShortList 1998, number theory problem 5
orl   66
N 4 hours ago by lksb
Source: IMO ShortList 1998, number theory problem 5
Determine all positive integers $n$ for which there exists an integer $m$ such that ${2^{n}-1}$ is a divisor of ${m^{2}+9}$.
66 replies
orl
Oct 22, 2004
lksb
4 hours ago
2020 EGMO P5: P is the incentre of CDE
alifenix-   50
N 4 hours ago by EpicBird08
Source: 2020 EGMO P5
Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$. The circumcircle $\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$.

Prove $P$ is the incentre of triangle $CDE$.
50 replies
alifenix-
Apr 18, 2020
EpicBird08
4 hours ago
2023 Putnam A1
giginori   29
N 4 hours ago by kidsbian
For a positive integer $n$, let $f_n(x)=\cos (x) \cos (2 x) \cos (3 x) \cdots \cos (n x)$. Find the smallest $n$ such that $\left|f_n^{\prime \prime}(0)\right|>2023$.
29 replies
giginori
Dec 3, 2023
kidsbian
4 hours ago
A MATHEMATICA E BONITA
P162008   0
Yesterday at 7:54 PM
Source: Self made by my Elder brother
Let $K = \sum_{i=0}^{\infty} \sum_{j=0}^{\infty}\sum_{m=0}^{\infty}\sum_{l=0}^{\infty} \frac{1}{(i+j+m+l)!}$ where $i,j,k$ and $l \in W.$

Now, consider the ratio $Z$ defined as
$Z = \frac{\sum_{r=0}^{\lfloor k \rfloor} \sum_{i=0}^{\lfloor k \rfloor} (-1)^r \binom{\lfloor k \rfloor}{r}(\lfloor k \rfloor - r)^i}{\sum_{r=0}^{\lfloor k \rfloor + 1}(-1)^r\binom{\lfloor k \rfloor + 1}{r}(\lfloor k \rfloor + 1 - r)^{\lfloor k \rfloor + 1}}.$

The summation function $S(n)$ is given by
$S(n) = \sum_{j=1}^{n} \left(\binom{n}{j} (j!)\left(\sum_{b=0}^{j} \frac{(-1)^b}{b!}\right)\right)$

Let $p$ denotes the number of points of intersection between the curves
$x^2 + y^2 - \tan(e^x) - \frac{|x|}{\sin y} = 0, (x\sin (a))^y + (y - x\cos(a))^x = |a|.$

Define $A(m)$ as
$A(m) = p\left(\sum_{k=0}^{m} \binom{2m + 1}{k} ((2m + 1) - 2k) (-1)^k\right).$

The value of $X$ is
$X = \lim_{n \to \infty} \frac{\sqrt{n}}{e^n} \text{exp} \left(\int_{0}^{\infty} \lfloor ne^{-x} \rfloor \text{dx}\right).$

And, $8Y =$ Number of subsets of $\left(1,2,3,\cdots,100\right)$ whose sum of elements is divisible by $5.$

Finally, compute the value of $\frac{1}{Z} +S(4) + 1 + e^{A(20)} + X\sqrt{8\pi} + Y.$
0 replies
P162008
Yesterday at 7:54 PM
0 replies
Ultra-hyper saddle with logarithmic weight
randomperson1021   0
Yesterday at 5:22 PM
Fix integers \(k\ge 3\) and \(1<r<k\), a parameter \(\lambda>0\), and a real log-exponent \(\beta\in\mathbb R\). For every real \(a\) define
$$
F_{a,\beta}^{(k,r)}(x)
  \;:=\;
  \sum_{n\ge 1}
       n^{\,a}\,(\log n)^{\beta}\,e^{\lambda n^{r}}\,x^{\,n^{k}},
  \qquad 0\le x<1.
$$
Put
$$
\Lambda_{k,r,\lambda}
   \;:=\;
   \lambda\!\left(1-\frac{r}{k}\right)
   \left(\frac{\lambda r}{k}\right)^{\!\frac{r}{\,k-r\,}},
   \qquad
   \gamma=\frac{r}{k-r}.
$$
(1) Show that there exists a real constant \(c=c(k,r)\) (independent of \(\lambda\) and of \(\beta\)) such that
$$
\lim_{x\to 1^{-}}
      F_{a,\beta}^{(k,r)}(x)\,
      e^{-\Lambda_{k,r,\lambda}\,(1-x)^{-\gamma}}
      \;=\;
      \begin{cases}
          0, & a<c,\\[6pt]
          \infty, & a>c.
      \end{cases}
$$
(2) Determine this critical value \(c\) explicitly and verify that it coincides with the classical case \(r=1\), namely \(c=-\tfrac12\).

(3) Evaluate the finite, non-zero limit that occurs at the borderline \(a=c\) (your answer may depend on \(k,r,\lambda\) but not on \(\beta\)).
0 replies
randomperson1021
Yesterday at 5:22 PM
0 replies
The ratio between two integral
Butterfly   5
N Mar 31, 2025 by vanstraelen



Prove $\frac{I_1}{I_2}=\sqrt{2}$ where $I_1=\int_{0}^{\frac{\sqrt{3}-1}{\sqrt{2}}} \frac{1-x^2}{\sqrt{x^8-14x^4+1}}dx$ and $I_2=\int_{0}^{\sqrt{2}-1} \frac{1+x^2}{\sqrt{x^8+14x^4+1}}dx.$
5 replies
Butterfly
Mar 28, 2025
vanstraelen
Mar 31, 2025
The ratio between two integral
G H J
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Butterfly
581 posts
#1
Y by
Prove $\frac{I_1}{I_2}=\sqrt{2}$ where $I_1=\int_{0}^{\frac{\sqrt{3}-1}{\sqrt{2}}} \frac{1-x^2}{\sqrt{x^8-14x^4+1}}dx$ and $I_2=\int_{0}^{\sqrt{2}-1} \frac{1+x^2}{\sqrt{x^8+14x^4+1}}dx.$
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paxtonw
35 posts
#3
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We are comparing the integrals:

I_1 = \int_0^{\frac{\sqrt{3} - 1}{\sqrt{2}}} \frac{1 - x^2}{\sqrt{x^8 - 14x^4 + 1}} , dx

I_2 = \int_0^{\sqrt{2} - 1} \frac{1 + x^2}{\sqrt{x^8 + 14x^4 + 1}} , dx

Let x = \frac{1}{t} \Rightarrow dx = -\frac{1}{t^2} dt

This changes I_1 into:

I_1 = \int_{\frac{\sqrt{2}}{\sqrt{3} - 1}}^{\infty} \frac{1 - t^2}{\sqrt{t^8 - 14t^4 + 1}} , dt

So now I_1 covers the interval [ \frac{\sqrt{2}}{\sqrt{3} - 1}, \infty )

The integrand is:

f(x) = \frac{1 + x^2}{\sqrt{x^8 + 14x^4 + 1}}

It satisfies:

f(\frac{1}{x}) = f(x) \cdot \frac{1}{x^2}

and

dx \to -\frac{1}{x^2} dx

So:

f(\frac{1}{x}) dx = -f(x) dx

The \frac{1}{x^2} cancels out due to the Jacobian, making the integral symmetric.

Therefore:

I_1 = \sqrt{2} I_2

Final result:

\frac{I_1}{I_2} = \sqrt{2}

Someone please format this for me.. I am a new user
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vanstraelen
9063 posts
#4
Y by
Latex-version of "paxtonw"

We are comparing the integrals:

$I_1 = \int_0^{\frac{\sqrt{3} - 1}{\sqrt{2}}} \frac{1 - x^2}{\sqrt{x^8 - 14x^4 + 1}}\ dx$.
$I_2 = \int_0^{\sqrt{2} - 1} \frac{1 + x^2}{\sqrt{x^8 + 14x^4 + 1}}\ dx$.

Let $x = \frac{1}{t} \Rightarrow dx = -\frac{1}{t^2} dt$.
This changes $I_1$ into:
$I_1 = \int_{\frac{\sqrt{2}}{\sqrt{3} - 1}}^{\infty} \frac{1 - t^2}{\sqrt{t^8 - 14t^4 + 1}}\ dt$.
So now $I_1$ covers the interval $[ \frac{\sqrt{2}}{\sqrt{3} - 1}, \infty )$.

The integrand is:
$f(x) = \frac{1 + x^2}{\sqrt{x^8 + 14x^4 + 1}}$.
It satisfies:
$f(\frac{1}{x}) = f(x) \cdot \frac{1}{x^2}$.
and
$dx \to -\frac{1}{x^2} dx$.

So:
$f(\frac{1}{x}) dx = -f(x) dx$.
The $\frac{1}{x^2}$ cancels out due to the Jacobian, making the integral symmetric.

Therefore:
$I_1 = \sqrt{2} I_2$.
Final result:
$\frac{I_1}{I_2} = \sqrt{2}$.
I think, this is wrong.
It satisfies:
$f(\frac{1}{x}) = f(x) \cdot \frac{1}{x^2}$.
This post has been edited 2 times. Last edited by vanstraelen, Mar 29, 2025, 4:29 PM
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Butterfly
581 posts
#5
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see https://www.zhihu.com/question/1888722957578258415/answer/1888924402114068602
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paxtonw
35 posts
#6
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You’re right. Here’s a different proof that may be correct.

Let x = tan(theta). Then both upper limits become theta = pi/8, so both integrals go from 0 to pi/8.

For I1:
The integrand becomes:
(1 - tan²(theta)) * sec²(theta) / sqrt(tan⁸(theta) - 14tan⁴(theta) + 1)

Using trig identities:
• 1 - tan²(theta) = cos(2theta) / cos²(theta)
• sec²(theta) = 1 / cos²(theta)

So the numerator becomes:
cos(2theta) / cos⁴(theta)

For I2:
The integrand becomes:
(1 + tan²(theta)) * sec²(theta) / sqrt(tan⁸(theta) + 14tan⁴(theta) + 1)

Since 1 + tan²(theta) = sec²(theta), and sec²(theta) * sec²(theta) = 1 / cos⁴(theta),
the numerator is just:
1 / cos⁴(theta)

So:
I1 has cos(2theta) in the numerator,
I2 does not.

Over the interval from 0 to pi/8, the integral of I1 ends up being exactly sqrt(2) times the value of I2.

Conclusion:
I1 / I2 = sqrt(2).

Please point out any apparent flaws that I’m not noticing here.. if the proof works for you then cheers!
This post has been edited 1 time. Last edited by paxtonw, Mar 30, 2025, 4:41 PM
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vanstraelen
9063 posts
#7
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Full solution, following the website of @Butterfly

$I_{1}=\int_{0}^{\frac{\sqrt{3}-1}{\sqrt{2}}} \frac{1-x^{2}}{\sqrt{x^{8}-14x^{4}+1}}\ dx=\int_{0}^{\frac{\sqrt{3}-1}{\sqrt{2}}} \frac{\frac{1}{x^{2}}-1}{\sqrt{x^{4}-14+\frac{1}{x^{4}}}}\ dx$.

First substitution: $x+\frac{1}{x}=t$, then $(1-\frac{1}{x^{2}})\ dx=dt$ and $t^{4}=x^{4}+\frac{1}{x^{4}}+4(x^{2}+\frac{1}{x^{2}})+6$,
$I_{1}=-\int_{\infty}^{\sqrt{6}} \frac{1}{\sqrt{t^{4}-4t^{2}-12}}\ dt=\int_{\sqrt{6}}^{\infty} \frac{1}{\sqrt{t^{4}-4t^{2}-12}}\ dt$.

Second substitution: $t=\frac{\sqrt{6}}{\cos u}$, then $dt=\frac{\sqrt{6}\sin u}{\cos^{2}u}\ du$ and
$I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\frac{36}{\cos^{4}u}-\frac{24}{\cos^{2}u}-12}} \cdot \frac{\sqrt{6}\sin u}{\cos^{2}u}\ du$.
$I_{1}=\sqrt{6}\int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sqrt{36-24\cos^{2}u-12\cos^{4}u}}\ du$.
$I_{1}=\frac{\sqrt{6}}{\sqrt{12}} \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sqrt{3-4\cos^{2}u-\cos^{4}u}}\ du$.
$I_{1}=\frac{\sqrt{6}}{\sqrt{12}} \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sqrt{3-4(1-\sin^{2}u)-(1-\sin^{2}u)^{2}}}\ du$.
$I_{1}=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sqrt{4\sin^{2}u-\sin^{4}u}}\ du$.
$I_{1}=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{4-\sin^{2}u}}\ du$.
$I_{2}=\int_{0}^{\sqrt{2}-1} \frac{1+x^{2}}{\sqrt{x^{8}+14x^{4}+1}}\ dx=\int_{0}^{\sqrt{2}-1} \frac{\frac{1}{x^{2}}+1}{\sqrt{x^{4}+14+\frac{1}{x^{4}}}}\ dx$.

First substitution: $x-\frac{1}{x}=-2t$, then $(1+\frac{1}{x^{2}})\ dx=-2\ dt$ and $16t^{4}=x^{4}+\frac{1}{x^{4}}-4(x^{2}+\frac{1}{x^{2}})+6$,
$I_{2}=\int_{\infty}^{1} \frac{-2}{\sqrt{16t^{4}+16t^{2}+16}}\ dt$.
$I_{2}=\frac{1}{2} \int_{1}^{\infty} \frac{1}{\sqrt{t^{4}+t^{2}+1}}\ dt$.

Second substitution: $t=\cot \frac{u}{2}$, then $dt=-\frac{1}{2\sin^{2}\frac{u}{2}}\ du$ and
$I_{2}=\frac{1}{2} \int_{\frac{\pi}{2}}^{0} \frac{1}{\sqrt{\cot^{4}\frac{u}{2}+\cot^{2}\frac{u}{2}+1}} \cdot \frac{-1}{2\sin^{2}\frac{u}{2}}\ du$.
$I_{2}=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos^{4}\frac{u}{2}+\cos^{2}\frac{u}{2}\sin^{2}\frac{u}{2}+\sin^{4}\frac{u}{2}}}\ du$.
$I_{2}=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{(\cos^{2}\frac{u}{2}+\sin^{2}\frac{u}{2})^{2}-\sin^{2}\frac{u}{2}\cos^{2}\frac{u}{2}}}\ du$.
$I_{2}=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^{2}\frac{u}{2}\cos^{2}\frac{u}{2}}}\ du$.
$I_{2}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{4-4\sin^{2}\frac{u}{2}\cos^{2}\frac{u}{2}}}\ du$.
$I_{2}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{4-\sin^{2}u}}\ du$.
This post has been edited 1 time. Last edited by vanstraelen, Apr 1, 2025, 6:44 AM
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