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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

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0 replies
jlacosta
Nov 1, 2024
0 replies
Easy limit
Snoop76   1
N 21 minutes ago by Levieee
Source: Own
The $y_n$ sequence satisfies the following relation : $ cos(y_n + n y_n)= 2cos(ny_n)$.$ $ Find $\lim_{n\to\infty} cos(y_n)$.
1 reply
Snoop76
4 hours ago
Levieee
21 minutes ago
BA = A^2 + B, then AB=BA?
Cubastic_Boy   1
N an hour ago by AkosS
Let $ n \ge 2, A, B \in \mathbb{R} ^{n \times n}$ 2 matrices such that $ BA = A^2 + B$. Prove that$ AB = BA.$
1 reply
Cubastic_Boy
3 hours ago
AkosS
an hour ago
Law of Large numbers
Ernest532   4
N 5 hours ago by Ernest532
If $(X_n)_{n}$ is a sequence of i.i.d uniformly distributed random variables on $[0,1]$, then $$\frac{n}{X_1+\ldots+X_n}\xrightarrow[n\to\infty]{\text{a.s}}\frac{1}{\mathbb{E}[X_1]}$$
4 replies
Ernest532
Today at 12:04 PM
Ernest532
5 hours ago
Linear Algebra
Sifan.C.Maths   0
5 hours ago
Source: My Issues
$f:\mathbb{R}^4 \rightarrow \mathbb{R}^4$. Prove that there are exist basics of vector space such that the matrix representation of f takes a form:
$$\begin{bmatrix}
a_{11} & a_{12} & a_{13}&a_{14} \\
			a_{21} & a_{22} & a_{23}&a_{24} \\
			0 & 0 & a_{33}&a_{34}\\
			0 & 0 & a_{43}&a_{44}
\end{bmatrix}$$
0 replies
Sifan.C.Maths
5 hours ago
0 replies
No more topics!
Hard integrals
Entrepreneur   27
N Nov 16, 2024 by soryn
Source: Own
$$\color{blue}{\int \sqrt{\cot x}dx}$$$$\color{blue}{\int \tan^2x\sec xdx}$$$$\color{blue}{\int \sec^3xdx}$$
27 replies
Entrepreneur
Aug 12, 2024
soryn
Nov 16, 2024
Hard integrals
G H J
Source: Own
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Entrepreneur
1078 posts
#1 • 2 Y
Y by manlio, w32dedorh30
$$\color{blue}{\int \sqrt{\cot x}dx}$$$$\color{blue}{\int \tan^2x\sec xdx}$$$$\color{blue}{\int \sec^3xdx}$$
This post has been edited 2 times. Last edited by Entrepreneur, Aug 31, 2024, 7:48 PM
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alexheinis
10261 posts
#3
Y by
$\int \sqrt{\cot x}dx=-\int {{\sqrt{p}dp}\over {p^2+1}}=-2\int{{q^2 dq}\over {q^4+1}}$. Now factor the numerator as $(q^2-q\sqrt{2}+1)(q^2+q\sqrt{2}+1)$ and we get ${{-1}\over {2\sqrt{2}}}(\ln ({{q^2+1-q\sqrt{2}}\over {q^2+1+q\sqrt{2}}})+2\arctan (1+q\sqrt{2})-2\arctan (1-q\sqrt{2}))$.

$\int {{\sin^2 x dx}\over {\cos^3x}}=\int {{\sin^2 x d(\sin x)}\over {(1-\sin^2 x)^2}}=\int {{p^2 dp}\over {(1-p^2)^2}}={1\over 4}(\ln ({{1+p}\over {1-p}})+{{2p}\over {1-p^2}}$). After this we can write ${{2p}\over {1-p^2}}={{2\sin x}\over {\cos^2 x}}$ and we can use ${{1+\cos t}\over {1-\cos t}}=\cot^2 (t/2)$. A bit tedious so I'll skip that.

$\int{{dx}\over {\cos^3 x}}=\int {{d(\sin x)}\over {(1-\sin^2 x)^2}}=\int {{dp}\over {(1-p^2)^2}}={1\over 4}(\ln ({{1+p}\over {1-p}})+{{2p}\over {1-p^2}})$.
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Entrepreneur
1078 posts
#4
Y by
$$\color{blue}{\int \frac{dx}{\sqrt{\sin(1+\cos x)}}}$$$$\color{blue}{\int \frac{dx}{\sqrt{\sin(1+\cos^3x)}}}$$
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Entrepreneur
1078 posts
#5
Y by
Entrepreneur wrote:
$$\color{blue}{\int \frac{dx}{\sqrt{\sin(1+\cos x)}}}$$
Substitute $t=\tan \frac x2$ to obtain
\begin{align*}
\mathcal I&= \int \frac{dx}{\sqrt{\sin x(1+\cos x)}}\\
&=\int \frac{dt}{\sqrt{t}}\\
&=2\sqrt t+\mathcal C\\
&=\boxed{2\sqrt{\tan \frac x2}+\mathcal C.}
\end{align*}
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Entrepreneur
1078 posts
#6
Y by
$$\textcolor{blue}{\int \frac{\sqrt{e^x}\cos x}{\sqrt[3]{3\cos x+4\sin x}}dx}$$
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Entrepreneur
1078 posts
#7
Y by
$$\color{blue}{\int_0^{\infty}\sqrt[3]{\frac{\sin^2x}{1+x^4}}dx }$$
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Entrepreneur
1078 posts
#8
Y by
$$\textcolor{blue}{\int \sqrt{\tanh x}dx}$$
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naenaendr
561 posts
#9
Y by
Entrepreneur wrote:
$$\textcolor{blue}{\int \sqrt{\tanh x}dx}$$

nice one!
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Entrepreneur
1078 posts
#10
Y by
$$\textcolor{blue}{\int_0^{\frac{\pi}{2}}\frac{\ln \cos x}{\tan x}\ln\left(\frac{\ln \cos x}{\ln \sin x}\right)dx}$$
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Entrepreneur
1078 posts
#11
Y by
$$\color{blue}{\int_0^{\frac{\pi}{2}}\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$$$$\color{blue}{\int_0^{\frac{\pi}{2}}x\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$$$$\color{blue}{\int_0^{\frac{\pi}{2}}x\cos x\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$$
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Aiden-1089
231 posts
#12
Y by
For $x \in (\frac{\pi}{4}, \frac{\pi}{2})$, we have $\frac{1+\tan{x}}{1-\tan{x}} < 0$, so $\ln \left( \frac{1+\tan{x}}{1-\tan{x}} \right)$ is undefined.
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Entrepreneur
1078 posts
#14
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$$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$$
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Entrepreneur
1078 posts
#15
Y by
$$\color{blue}{\int \sqrt{e^{2x}+e^x+1}dx}$$$$\color{blue}{\int \sqrt{e^{3x}+e^x+1}dx}$$$$\color{blue}{\int \frac{1}{\sqrt{e^{3x}+e^x+1}}dx}$$
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Entrepreneur
1078 posts
#16
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$$\color{blue}{\int_0^{\infty}\frac{x^3\ln x}{(1+x^2)^2}dx}$$
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Entrepreneur
1078 posts
#17
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$$\color{blue}{\int \frac{x\ln x}{(1+x^2)^2}dx}$$
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Entrepreneur
1078 posts
#18
Y by
Entrepreneur wrote:
$$\color{blue}{\int \frac{x\ln x}{(1+x^2)^2}dx}$$

We use $\textcolor{red}{\textit{integration by parts.}}$ Let $u=\ln x, dv=\frac{xdx}{(1+x^2)^2}.$ Then, $du=\frac 1x,v=\frac{-1}{2(1+x^2)}.$ Thus,
$$\mathcal I =uv-\int vdu=\frac{-\ln x}{2(1+x^2)}+\frac 12\color{magenta}{\int\frac{dx}{x(1+x^2)}}.$$Now, for the magenta integral, let $x=\tan \theta.$ Then, $dx=\sec^2\theta d\theta.$ So, $$\color{magenta}{\int\frac{dx}{x(1+x^2)}}=\int\cot \theta d\theta =\ln|\sin \theta|+\mathcal C=\ln\left | \frac{x}{\sqrt{1+x^2}}\right |+\mathcal C.$$Hence, $$\boxed{\int\frac{x\ln x}{(1+x^2)^2}dx=\frac 12\ln\left | \frac{x}{\sqrt{1+x^2}}\right |-\frac{\ln x}{2(1+x^2)}+\mathcal C.}$$
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Entrepreneur
1078 posts
#19
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$$\textcolor{blue}{\int \sqrt[5]{1+e^{\sin x}}dx}$$
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CAS03
135 posts
#20
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Entrepreneur wrote:
$$\color{blue}{\int \sec^3xdx}$$
Haven't seen much on this one and it actually comes together very nicely

Separate:
\[\int \sec^3 x \, dx = \int \sec x \cdot \sec^2 x \, dx.\]Integration by parts:
\[\begin{array}{cc}
u=\sec x & dv = \sec^2 x \, dx \\
du = \sec x \tan x \, dx & v = \tan x
\end{array}\]Substituting:
\begin{align*}
\int \sec^3x \, dx &= \sec x \tan x - \int \sec x \tan^2 \, dx \\
\int \sec^3x \, dx &= \sec x \tan x - \int \sec x (\sec^2 x -1) \, dx \\
2\int \sec^3 x \, dx &= \sec x \tan x + \int \sec x \, dx
\end{align*}So final answer:
\[\boxed{\int \sec^3x \, dx = \frac 12 \left( \sec x \tan x + \log |\sec x + \tan x | + C \right)}.\]
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CAS03
135 posts
#21
Y by
Thanks for sending these, they are pretty cool and fun especially as I literally just learned integration by parts this week in class lol

That one at least was good practice :)
This post has been edited 1 time. Last edited by CAS03, Sep 9, 2024, 6:56 PM
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Entrepreneur
1078 posts
#22
Y by
Click to reveal hidden text My pleasure :)
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Entrepreneur
1078 posts
#23
Y by
$$\color{blue}{\int \frac{\tan x}{\sqrt{1+x^2}}dx}$$$$\color{blue}{\int \frac{\arctan x}{\sqrt{1+x^2}}dx}$$$$\color{blue}{\int \frac{\tanh x}{\sqrt{1+x^2}}dx}$$
This post has been edited 1 time. Last edited by Entrepreneur, Sep 13, 2024, 6:51 PM
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Entrepreneur
1078 posts
#24
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$$\textcolor{blue}{\int \frac{\cot^{-1}\sqrt{1+2x}}{x\sqrt{1+2x}}dx}$$$$\textcolor{blue}{\int \frac{\cot^{-1}x}{x^2-1}dx}$$
This post has been edited 1 time. Last edited by Entrepreneur, Sep 14, 2024, 12:14 PM
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Entrepreneur
1078 posts
#25
Y by
Entrepreneur wrote:
$$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$$

$G$ is the Catalan's constant.
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Entrepreneur
1078 posts
#26
Y by
Entrepreneur wrote:
$$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$$

We have
\begin{align*}
\mathcal I&= \int_0^1\frac{\ln x}{1+x^2}dx\\
&=\int_0^1\ln x\sum_{k=0}^\infty(-1)^kx^{2k}dx\\
&=\sum_{k=0}^\infty\int_0^1x^{2k}\ln xdx\\
&=\boxed{\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+1)^2} =-G.}
\end{align*}$\color{green}{\cal{QED.}}$
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Entrepreneur
1078 posts
#27
Y by
Entrepreneur wrote:
Entrepreneur wrote:
$$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$$

We have
\begin{align*}
\mathcal I&= \int_0^1\frac{\ln x}{1+x^2}dx\\
&=\int_0^1\ln x\sum_{k=0}^\infty(-1)^kx^{2k}dx\\
&=\sum_{k=0}^\infty\int_0^1x^{2k}\ln xdx\\
&=\boxed{\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+1)^2} =-G.}
\end{align*}$\color{green}{\cal{QED.}}$

Generalized Result:
$$\color{blue}{\int_0^1\frac{\ln x}{x^a\pm1}dx=\sum_{k=0}^\infty\frac{(\mp 1)^{k+1}}{(ak+1)^2}.}$$
Can this be generalized any further?
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Entrepreneur
1078 posts
#28
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$$\textcolor{blue}{\int x^n\ln^axdx=\frac{x^n\ln^ax}{n+1}-\frac{x^n}{n+1}\sum_{k=1}^{a}\frac{\ln^{a-k}x}{(n+1)^k}+\cal C.}$$Is this result true?
This post has been edited 1 time. Last edited by Entrepreneur, Nov 16, 2024, 10:01 AM
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Entrepreneur
1078 posts
#29
Y by
$$\color{blue}{\int_0^\frac{\pi}{4}\ln(\cos x)dx}$$$$\color{blue}{\int_0^\frac{\pi}{4}\ln^2(\cos x)dx}$$
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soryn
5189 posts
#30
Y by
For the first integral,denote I the desired integral,and J the integral with ln(sinx). After calculatios,obtain the system with the equations : I+J=-π/2ln2 and I-J=G, where G is the Catalan's constant.
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