The Integral $\int \frac{1}{ax^2+bx+c}dx$ in Problem 29046

          It was said in DidMath No. 1/2024 at page 17 that the arctangent function bothers us in the calculation of integrals. Such a situation just arose in Problem 29046, author Mihály BENCZE, Brașov.

              "29046    Let $a>1$. Determine $\int_{\frac{1}{a}}^a \frac{x^n\cdot \arctan x\;dx}{x^{2n+2}+x^{n+1}+1}\;.$"

ANSWER CiP

$$\frac{2\pi}{4(n+1)\sqrt{3}} \cdot \arctan \left( \frac{2}{\sqrt{3}} \cdot \frac {a^{2n+2}-1}{2a^{2n+2}+6a^{n+1}+1}\right )$$

 Solution CiP

                Lemma  Let $a \neq 0,\;\Delta=b^2-4ac<0$. Then

$$\int \frac{1}{ax^2+bx+c}dx=\frac{2}{\sqrt{-\Delta}}\cdot \arctan\left ( \frac{2ax+b}{\sqrt{-\Delta}}\right )+C \tag{I}$$

               Formula (I) can be verified with the definition, calculating for

$u=\frac{2}{\sqrt{-\Delta}}\cdot arctg \;\phi(x),\;\; \phi(x)=\frac{2ax+b}{\sqrt{-\Delta}}$

$$\frac {\mathrm{d}u}{\mathrm{d}x}=\frac{2}{\sqrt{-\Delta}}\cdot \frac{\phi^{'}}{\phi^2+1}=\frac{2}{\sqrt{-\Delta}}\cdot \frac{\frac{2a}{\sqrt{-\Delta}}}{\left ( \frac{2ax+b}{\sqrt{-\Delta}} \right ) ^2+1}=\frac{4a/(-\Delta)}{(4a^2x^2+4abx+b^2+4ac-b^2)/(-\Delta)}.$$

$\square$ end Lemma

For $a=1=b=c$ we obtain

              Corollary  $\int \frac{1}{x^2+x+1}dx=\frac{2}{\sqrt{3}} \cdot \arctan \left (\frac{2x+1}{\sqrt{3}} \right )+C \tag{1}$

 

          Solving the problem:

             We will not exaggerate with the claim to prove with derivatives that are equal the relationship

$$\arctan t +\arctan \frac{1}{t}=\frac{\pi}{2} \tag{2}$$

        Let $I:=\int_{1/a}^{a} \frac{x^n \cdot \arctan x}{x^{2n+2}+x^{n+1}+1}dx$. With substitution

$$x=\frac{1}{t},\;\;dx=-\frac{1}{t^2}\cdot dt,\;\;\;t=a\;for \;x=\frac{1}{a},\;\;t=1/a\;for\; x=a\;\;\;\;\Rightarrow$$

$I=\int_{a}^{1/a}\frac {\frac{1}{t^n} \cdot \arctan\left( \frac{1}{t} \right)}{\frac{1}{t^{2n+2}}+\frac{1}{t^{n+1}+1} }\cdot \left ( -\frac{1}{t^2} \right ) dt\overset{(2)}{=}\int_{1/a}^{a} \frac {t^n \cdot (\pi /2-\arctan t)}{t^{2n+2}+t^{n+1}+1}dt=\frac{\pi}{2} \cdot \int_{1/a}^{a} \frac{t^n}{t^{2n+2}+t^{n+1}+1}dt-$

$-\int_{1/a}^{a}\frac{t^n \cdot \arctan t}{t^{2n+2}+t^{n+1}+1}dt=\frac{\pi}{2}\cdot \int_{1/a}^{a} \frac{t^n}{t^{2n+2}+t^{n+1}+1}-I\;\;\;\Rightarrow I+I=\frac{\pi}{2}\cdot \dots$, so

$$I=\frac{\pi}{4} \cdot \int_{1/a}^{a}\frac{t^n}{t^{2n+2}+t^{n+1}+1}dt \tag {3}$$

To the integral in (3) we make the substitution

$$t^{n+1}=u,\;\;t^n \cdot dt=\frac{du}{n+1},\;\;\;u=1/a^{n+1}\;for\;t=1/a,\;\;u=a^{n+1}\;for\;t=a\;\;\;(b:=a^{n+1})$$

and we have  $I=\frac{\pi}{4} \cdot \int_{1/b}^{b} \frac{1}{u^2+u+1} \cdot \frac{du}{n+1}\overset{(1)}{=}\frac{\pi}{4(n+1)} \cdot \left ( \frac{2}{\sqrt{3}}\arctan (\frac{2u+1}{\sqrt{3}})\right)\Bigg\vert_{u=1/b}^{u=b}$

If we also consider the formula   $\arctan t-\arctan s=\arctan\frac{t-s}{1+t\cdot s}$   then ve obtain

$I=\frac{2\pi}{4(n+1)\sqrt{3}} \cdot \left (\arctan \frac{2b+1}{\sqrt{3}}-\arctan \frac{2/b+1}{\sqrt{3}} \right )=\frac{2\pi}{4(n+1) \sqrt{3}} \cdot \arctan \left (\frac{2}{\sqrt{3}}\cdot \frac{b^2-1}{2b^2+6b+1} \right )$

and puttin $b=a^{n+1}$ we get the answer.

$\blacksquare$

O Problemă cu cașu' rezolvată de CIOBANU // Problem 1300 - CMJ vol 56 issue 2

 The problem 1300 is compounded by my compatriots. 

The title of the post is intended to be an innocent joke.

          We have to show that: $e^{\frac{\pi \imath}{7}}$ is a solution of the equation

$$\imath \sqrt{7}\cdot z^6-z^5+z^4+z^3+z^2+z-1 =0\tag{E}$$

                 Solution CiP

               We have equality [(1) for $n=3$]:

$$\sin \frac{\pi}{7} \cdot \sin \frac {2\pi}{7} \cdot \sin \frac{3\pi}{7}=\frac{\sqrt{7}}{8} \tag {1}$$

     Let $z=e^{\frac{\pi \imath}{7}}$. According to the De Moivre's and Euler's formulas we have

$z^7=e^{\pi \imath}=-1, \;\;|z|=1,\;\bar z=\frac{1}{z},\;\;\sin \frac{k\pi}{7}=\frac{z^k-\bar z ^k}{2\imath}=\frac{z^{2k}-1}{2\imath \cdot z^k},\;k=1,\;2,\;3$  so

$$z^7+1=0\quad ; \quad z^6-z^5+z^4-z^3+z^2-z+1=0 \tag{C}$$

$$\sin \frac{\pi}{7}=\frac{z^2-1}{2\imath \cdot z},\; \quad \sin \frac{2\pi}{7}=\frac{z^4-1}{2\imath \cdot z^2},\; \quad \sin \frac{3\pi}{7}=\frac{z^6-1}{2\imath \cdot z^3}.\tag{S}$$

Substituting the values ​​(S) into (1) we obtain, after multiplying by $(2\imath)^3=-8\imath$

$$-\sqrt{7} \imath \cdot z^6=(z^2-1)(z^4-1)(z^6-1) \Leftrightarrow$$

$$\Leftrightarrow\;\;-\sqrt{7}\imath \cdot z^6=z^{12}-z^{10}-z^8+z^4+z^2-1 \tag{2}$$

          We will apply the Long Division Algorithm to the polynomial $P=X^{12}-X^{10}-X^8+X^4+X^2-1$  to both $Q=X^7+1$ , $Q_1=X^6-X^5+X^4-X^3+X^2-X+1$.

$\begin{array}{c|c}X^{12}&+0X^{11}&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1&Q=X^7+1\\\hline\\-X^{12}&&&&&&&-X^5 &&&&&&D=X^5-X^3-X\\\hline \\&&-X^{10}&&-X^8&&&-X^5&+X^4&&+X^2&&-1&\\\hline &&+X^{10}&&&&&&&+X^3&&&&&\\\hline &&&&-X^8&&&-X^5&+X^4&+X^3&+X^2&&-1\\ \hline\\&&&&&&&-X^5&+X^4&+X^3&+X^2&+X&-1&=R\\\end{array}$

In the first case we obtain

$$P=Q \cdot D+R \tag{3}$$

and in the second case

$\begin{array}{c|c}X^{12}&+0X^{11}&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1&Q_1=X^6-X^5+X^4-X^3+X^2-X+1\\\hline -X^{12}&+X^{11}&-X^{10}&+X^9&-X^8&+X^7&-X^6&&&&&&&D_1=X^6+X^5-X^4-X^3-X^2-X\\\hline &X^{11}&-2X^{10}&+X^9&-2X^8&+X^7&-X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1\\\hline &-X^{11}&+X^{10}&-X^9&+X^8&-X^7&+X^6&-X^5\\\hline &&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&-X^5&+X^4&+0X^3&+X^2&+0X&-1\\\hline &&+X^{10}&-X^9&+X^8&-X^7&+X^6&-X^5&+X^4\\\hline &&&-X^9&+0X^8&-X^7&+X^6&-2X^5&+2X^4&+0X^3&+X^2&+0X&-1\\\hline &&&+X^9&-X^8&+X^7&-X^6&+X^5&-X^4&+X^3\\\hline &&&&-X^8&+0X^7&+0X^6&-X^5&+X^4&+X^3&+X^2&+0X&-1\\\hline &&&&+X^8&-X^7&+X^6&-X^5&+X^4&-X^3&+X^2\\\hline &&&&&-X^7&+X^6&-2X^5&+2X^4&+0X^3&+2X^2&+0X&-1\\\hline &&&&&+X^7&-X^6&+X^5&-X^4&+X^3&-X^2&+X\\ \hline &&&&&&&-X^5&+X^4&+X^3&+X^2&+X&-1&=R_1 \end{array}$

$$P=Q_1 \cdot D_1+R_1 \tag{4}$$

I got the same result $R=R_1=-X^5+X^4+X^3+X^2+X-1$. The equality (2) is written

$$-\sqrt{7} \imath \cdot z^6=P(z)=Q(z) \cdot D(z)+R(z)$$

but according to (C) we have $Q(z)=0=Q_1(z)$ and so 

$$-\imath \sqrt{7} \cdot z^6=-z^5+z^4+z^3+z^2+z-1$$

which is equivalent to the equation (E)

$\blacksquare$

A Problem in Litigation : by Cristian OLTEANU

          The author of Problem #4 in the image is a diligent GMB contributor. The problem was given to 8th graders during the "Nicolae POPESCU" Mathematics Competition...

            With the data in the figure, the problem has no solution. I modified it to my taste, obtaining:

           4.    Calculate the minimum of the expression $\frac{1}{ab}+9ab$, knowing that

                  $(3a+2)(b+1)=6,\;a,b \in (0,+\infty)$, and find the values ​​of the

                  numbers $a$ and $b$ for which this minimum is obtained.

ANSWER CiP

$$\frac{1}{ab}+9ab\geqslant 6;\;\;Equality\;for\;a=\frac{1}{3},\;b=1\;\;or\;\;a=\frac{2}{3},\;b=\frac{1}{2}$$

                     Solution CiP

             $\frac{1}{ab}+9ab=3 \cdot \left( \frac{1}{3ab}+3ab\right )\geqslant 3 \cdot 2=6.$ We used that

 $x+\frac{1}{x} \geqslant 2,\;x>0$, with equality if and only if $x=1$. In our case we have equality for $3ab=1$.

          To find the numbers that achieve the minimum:

$$(3a+2)(b+1)=6\Leftrightarrow 3ab+3a+2b+2=6\underset{3ab=1}{\Leftrightarrow} 3a+2b=3.$$

So we have the conditions

$$3a+2b=3\quad 3a\cdot 2b=2$$

so $3a$ and $2b$ are the roots of the quadratic equation $y^2-3y+2=0.$ It results that

$$\{3a,\;2b\}=\{1,\;2\}$$

from which we obtain the answer.

$\blacksquare$

By-products of PRODUCT // Ürün yan ürünleri

               Here we will demonstrate the formulas

$$\sin \frac{\pi}{2n+1}\cdot\sin \frac{2\pi}{2n+1}\cdot_{\dots} \cdot\sin \frac{n\pi}{2n+1}=\frac{\sqrt{2n+1}}{2^n} \tag{1}$$

$$\sin\frac{\pi}{2n}\cdot\sin \frac{2\pi}{2n}\cdot_{\dots}\cdot\sin \frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}} \tag{2}$$

              We will start from a previously demonstrated formula, written for $m\in \mathbb{N} \setminus \{0,\;1\}$ instead of $n$:

$$\sin \frac{\pi}{m}\cdot\sin \frac{2\pi}{m}\cdot_{\dots} \cdot \sin \frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}} \tag{3}$$

          Let's put in (3): $m$-odd, i.e. $m=2n+1,\;n\in \mathbb{N}\setminus\{0\}$. We will now write down a few more of the $m-1=2n$ factors of (3):

$\sin \frac{\pi}{2n+1} \cdot \sin \frac{2\pi}{2n+1}\cdot_{\dots} \cdot \sin \frac{n\pi}{2n+1}  \cdot$

$\cdot \sin \frac{(n+1)\pi}{2n+1} \cdot_{\dots}\cdot \sin \frac{(2n-1)\pi}{2n+1} \cdot \sin \frac{2n\pi}{2n+1}=\frac{2n+1}{2^{2n}} \tag{4}$

Considering the formula $\sin (\pi-\alpha)=\sin \alpha$ and noting that

$$\pi-\frac{2n\pi}{2n+1}=\frac{\pi}{2n+1} \quad \pi-\frac{(2n-1)\pi}{2n+1}=\frac{2\pi}{2n+1} \quad \dots \quad \pi-\frac{(n+1)\pi}{2n+1}=\frac{n\pi}{2n+1}$$

we see that in (4) the factors are two by two equal, one from each row. Then

$$\left (  \sin \frac{\pi}{2n+1} \cdot \sin \frac{2\pi}{2n+1} \cdot_{\dots} \cdot \sin \frac{n\pi}{2n+1}\right )^2=\frac{2n+1}{2^{2n}}$$

and taking the square root of both members results (1).

$\blacksquare$

          Let's put in (3) $m$-even, i.e. $m=2n,\;n\in \mathbb{N}\setminus \{0\}$. We now have

$\sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n}\cdot_{\dots} \cdot \sin \frac{(n-1)\pi}{2n} \cdot \overset{=1}{\overbrace{\sin \frac{n\pi}{2n}}} \cdot$

$\cdot \sin \frac{(n+1)\pi}{2n} \cdot_{\dots} \cdot \sin \frac{(2n-2)\pi}{2n} \cdot \sin \frac{(2n-1)\pi}{2n}=\frac{2n}{2^{2n-1}} \tag{5}$

Noting that 

$$\pi-\frac{(2n-1)\pi}{2n}=\frac{\pi}{2n}\quad \pi-\frac{(2n-2)\pi}{2n}=\frac{2\pi}{2n}\quad \dots \quad \pi-\frac{(n+1)\pi}{2n}=\frac{(n-1)\pi}{2n}$$

and considering again the formula  $\sin (\pi-\alpha)=\sin \alpha$ we have

$$\left ( \sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n}\cdot_{\dots} \cdot \sin \frac{(n-1)\pi}{2n}\right )^2=\frac{n}{2^{2n-2}}.$$

     Taking the square root of both members results (2).

$\blacksquare\;\blacksquare$

A Product of Many Sinuses // Birçok Sinüsün Ürünü

          I will look for some proofs of the trigonometric identity

$$\sin \frac{\pi}{n} \cdot \sin \frac{2\pi}{n} \cdot _{\dots} \cdot \sin \frac{(n-1)\pi}{n}=\frac{n}{2^{n-1}} \tag{1}$$

          Solution 1[based on Marcel ȚENA's book "Rădăcinile Unității" (The Roots 

                         of Unity) from the solution on pages 112-113 of Exercise #6 (page 16)]

           We consider the polynomial $X^n-1$ whose roots are $\zeta=\cos \frac{2\pi}{n}+\imath \sin \frac{2\pi}{n}$

$\zeta ^2,\;\zeta^3,\;\dots \zeta^{n-1},\;\zeta^n=1.$ We have equality

$$X^n-1=\prod \limits_{k=0}^{k=n-1}(X-\zeta^k) \tag{1.1}$$

from which, simplifying by the factor $X-1$, it results

$$X^{n-1}+X^{n-2}+\dots+X+1=\prod \limits_{k=1}^{k=n-1}(X-\zeta^k).$$

     For $X=1$ we obtain

$$n=\prod \limits_{k=1}^{k=n-1}(1-\zeta^k). \tag{1.2}$$

But, using the trigonometric formulas

$$1-\cos 2\alpha=2\sin^2 \alpha,\;\sin 2\alpha =2\sin \alpha \cdot \cos \alpha$$

 we have $1-\zeta^k=1-\cos \frac{2k\pi}{n}-\imath\sin \frac{2k\pi}{n}=-2\imath\sin \frac{k\pi}{n}\left(\cos\frac{k\pi}{n}+\imath \sin \frac{k\pi}{n}\right )$

and then (1.2) is written 

$n=2^{n-1} \cdot \imath^{n-1} \cdot \prod \limits_{k=1}^{k=n-1}\sin \frac{k\pi}{n} \cdot \prod \limits_{k=1}^{k=n-1}\left ( \cos \frac{k\pi}{n}+\imath \sin \frac{k\pi}{n}\right ).$

But then the moduli of the two members are also equal, so

$$n=2^{n-1} \cdot 1^{n-1} \cdot \prod \limits_{k=1}^{k=n-1} \sin \frac{k\pi}{n} \cdot \prod \limits_{k=1}^{k=n-1}1$$

from which the relation (1) immediately follows.

$\blacksquare$

          Solution 2 

           From the identity$^{\color{Red}{citation\; needed}}$  $\prod\limits_{k=0}^{k=n-1} \sin \left ( x+\frac{k\pi}{n}\right )=\frac{\sin nx}{2^{n-1}}$  we obtain, passing the first factor to the right-hand side

$\sin \left (x+\frac{\pi}{n} \right ) \cdot \sin \left ( x+\frac{2\pi}{n} \right ) \cdot _{\dots} \cdot \sin \left ( x+\frac {(n-1)\pi}{n} \right )=\frac{1}{2^{n-1}} \cdot \frac{\sin nx}{\sin x} \tag{2.1}$

Going in (2.1) to the limit when $x \rightarrow 0$, in the left-hand side we obtain the product that appears on the left at (1), and in the right-hand side we consider that 

$$\lim \limits_{x \to 0} \frac{\sin nx}{\sin x}=\lim \limits_{x \to 0} \frac{\sin nx}{nx} \cdot \lim \limits_{x \to 0} \frac{x}{\sin x} \cdot n=1 \cdot 1 \cdot n$$

and we obtain the value on the right of (1)

$\blacksquare$