96. Integrals.

by Virgil Nicula, Aug 31, 2010, 10:25 PM

$1.\blacktriangleright$ Ascertain the indefinite integral for the following functions :

$\boxed{\ \begin{array}{ccccccc} \frac{1}{x^4+ax^2+1} & \frac{x^2+x}{(e^x+x+1)^2} & \frac{\ln x}{x^2\sqrt {x^2+1}} & \left(\frac{x}{x\sin x+\cos x}\right)^2 & \frac{x\ln x}{(x^2+1)^2} & \frac{x^2}{(x^2-4)\sin x+4x\cos x} & \frac {1 - \ln x}{x^2 +\ln^2 x}\\\\ \frac {1+\sin x}{1+\sin 2x} & \frac {1}{x^3(x+1)^2} & \frac {1}{\left(x^2+a^2\right)\sqrt{x^2+a^2}} & \frac{2x^3+1}{(x^3-1)\sqrt{x^6-2x^3+x^2+1}} & \frac {\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt {\cos x}} & \frac{1}{x\sqrt{x^n+a}} & \left(\frac{\arctan x}{\arctan x-x}\right)^{2}\end{array}\ }$

Proof.

$\blacktriangleleft I_1=\int \frac{1}{x^4+ax^2+1}\ \mathrm{dx}$ and $J=\int \frac{x^2}{x^4+ax^2+1}\ \mathrm{dx}$ . Thus $I_1+J=\int \frac{\left( x-\frac 1x\right) '}{\left( x-\frac 1x\right)^2+2+a}\ \mathrm{dx}$ and

$J-I_1=\int \frac{\left( x+\frac 1x\right) '}{\left( x+\frac 1x\right)^2+a-2}\ \mathrm{dx}$ . Remark. There are five cases: $a<-2\ ;\ a=-2\ ;\ -2<a<2\ ;\ a=2\ ;\ a>2$ .

$\blacktriangleleft I_2=\int \frac{x^2+x}{(e^x+x+1)^2}\ \mathrm{dx}$ . Denote $A=\int\frac {1}{1+x+e^x}\ \mathrm{dx}$ , $B=\int\frac {x}{1+x+e^x}\ \mathrm{dx}$ , $C=\int\frac{x}{(1+x+e^x)^2}\ \mathrm{dx}$ ,

$D=\int\frac{x^2}{(1+x+e^x)^2}\ \mathrm{dx}$ . Thus, $\left\|\begin{array}{c} B=\int\frac {\left(1+x+e^x\right)-\left(1+e^x\right)}{1+x+e^x}\ \mathrm{dx}=x-\int\frac {\left(1+x+e^x\right)'}{1+x+e^x}\ \mathrm{dx}=x-\ln\left(1+x+e^x\right)+\mathrm C\\\\ C=\int\frac {\left(1+x+e^x\right)-\left(1+e^x\right)}{\left(1+x+e^x\right)^2}\ \mathrm{dx}=\int\frac {1}{1+x+e^x}\ \mathrm{dx}-\int\frac {\left(1+x+e^x\right)'}{\left(1+x+e^x\right)^2}=A+\frac {1}{1+x+e^x}\\\\ D=\int\frac {x\left[\left(1+x+e^x\right)-\left(1+e^x\right)\right]}{\left(1+x+e^x\right)^2}\ \mathrm{dx}=B+\int x\cdot\left(\frac {1}{1+x+e^x}\right)'\ \mathrm{dx}=B+\frac {x}{1+x+e^x}-A\end{array}\right\|$ .

Thus, $I_2=D+C=B-A+\frac {x}{1+x+e^x}+A+\frac {1}{1+x+e^x}$ $\implies$ $\boxed{I_2=x-\ln\left(1+x+e^x\right)+\frac {1+x}{1+x+e^x}+\mathrm C}$ .

$\blacktriangleleft I_3=\int\frac{\ln x}{x^2\sqrt {x^2+1}}\ \mathrm{dx}=$ $-\ln x\cdot \frac {\sqrt{1+x^2}}{x}+\int\frac {\sqrt {1+x^2}}{x^2}\ \mathrm{dx}$ because (integrating by parts) $\left\{\begin{array}{ccc} u(x)=\ln x & \implies & u'(x)=\frac 1x\\\\ v'(x)=\frac {1}{x^2\sqrt {1+x^2}} & \implies & v(x)=-\frac{\sqrt {1+x^2}}{x}\end{array}\right\|$ .

Yes, $v(x)=\int\frac {1}{x^2\sqrt{1+x^2}}=$ $\int\frac {1+x^2-x^2}{x^2\sqrt{1+x^2}}=$ $\int\frac {\sqrt{1+x^2}}{x^2}\mathrm{dx}-\ln\left(x+\sqrt {1+x^2}\right)$ . By parts

$\left\{\begin{array}{ccc} f(x)=\sqrt{1+x^2} & \implies & f'(x)=\frac {x}{\sqrt {1+x^2}}\\\\ g'(x)=\frac {1}{x^2} & \implies & g(x)=-\frac 1x\end{array}\right\|$ obtain $\int\frac {\sqrt{1+x^2}}{x^2}\ \mathrm{dx}=$ $-\frac {\sqrt{1+x^2}}{x}+\ln\left(x+\sqrt{1+x^2}\right)$ .

Thus, $v(x)=-\frac {\sqrt{1+x^2}}{x}$ . In conclusion, $\boxed{I_3=-\frac {\ln x\cdot\sqrt{1+x^2}}{x}-\frac {\sqrt{1+x^2}}{x}+\ln\left(x+\sqrt{1+x^2}\right)+\mathrm C}$ .

$\blacktriangleleft I_4=\int\frac{x^2}{(x\sin x+\cos x)^2}\ \mathrm{dx}$ . But $x^2=x\sin x(x\sin x+\cos x)+x\cos x(x\cos x-\sin x)$ . Thus, $I_4=\int\frac{x\sin x}{x\sin x+\cos x}+J$ ,

where $J=\int (x\cos x-\sin x)\cdot\frac {x\cos x}{(x\sin x+\cos x)^2}\ \mathrm{dx}$ . By parts $\left\{\begin{array}{ccc} u(x)=x\cos x-\sin x & \implies & u'(x)=-x\sin x\\\\ v'(x)=\frac {x\cos x}{(x\sin x+\cos x)^2} & \implies & v(x)=-\frac {1}{x\sin x+\cos x}\end{array}\right\|$

obtain $J=\frac {\sin x-x\cos x}{x\sin x+\cos x}-\int\frac {x\sin x}{x\sin x+\cos x}\ \mathrm{dx}$ . Thus, $\boxed{I_4=\frac {\sin x-x\cos x}{x\sin x+\cos x}+\mathrm C}$ .

$\blacktriangleleft I_5=\frac{x\ln x}{(x^2+1)^2}=$ $-\frac 12\cdot\frac {1}{1+x^2}\cdot\ln x+\frac 12\cdot\int\frac {1}{x(1+x^2)}\ \mathrm {dx}$ because (by parts) $\left\{\begin{array}{ccc} f(x)=\ln x & \implies & f'(x)=\frac 1x\\\ g'(x)=\frac {x}{\left(1+x^2\right)^2} & \implies & g(x)=-\frac {1}{2\left(1+x^2\right)}\end{array}\right\|$ .

$\frac {1}{x\left(1+x^2\right)}=\frac 1x-\frac {x}{1+x^2}\implies$ $\int\frac {1}{x\left(1+x^2\right)}\ \mathrm{dx}=$ $\ln |x|-\frac 12\cdot \ln\left(1+x^2\right)+C$ .

Thus, $\boxed{I_4=-\frac {\ln x}{2\left(1+x^2\right)}+\frac 12\cdot\ln |x|-\frac 14\cdot\ln\left(1+x^2\right)+C}$ .

$\blacktriangleleft$ Let $x\in (0,\pi )$ and $I=\int {\frac{{x^2 }}{{(x^2 - 4)\sin x + 4x\cos x}}}\mathrm{dx} =$ $\int {\frac{{\frac{{x^2 }}{{(x^2 + 4)}}}}{{\frac{{(x^2 - 4)}}{{(x^2 + 4)}}\sin x + \frac{{4x}}{{(x^2 + 4)}}\cos x}}}\ \mathrm{dx}$ . Use the substitution $\boxed{x = 2\cot\frac t2}$ , i.e.

$t=2\arctan\frac 2x$ . Therefore, $\left\|\begin{array}{c} \mathrm{dx} = -\csc ^2 \frac t2\\\\ \frac{{(x^2 - 4)}}{{(x^2 + 4)}} = \cos t\\\\ \frac{{4x}}{{(x^2 + 4)}} = \sin t\end{array}\right\|$ . Thus $I = - \int \frac{\cos^2\frac t2\cdot \csc ^2 \frac t2}{\cos t \sin x + \sin t\cos x}\ \mathrm{dt}=$ $- \int \frac{\cot^2 \frac t2}{\sin\left( t + 2\cot\frac t2\right)}\ \mathrm{dt}$ . Use the

substitution $\boxed{y = t + 2\cot\frac t2}$ , where $\mathrm{dy}= - \cot^2 \frac t2\ \mathrm{dt}$ . In conclusion, $I=- \int \frac{\cot^2 \frac t2}{\sin\left( t + 2\cot\frac t2\right)}\ \mathrm{dt}= \int \csc y\ \mathrm{dy}=$ $\ln\left|\tan\frac y2\right|=$

$\ln\left|\tan\left(\frac t2+\cot\frac t2\right)\right|=$ $\ln\left|\tan\left(\arctan\frac 2x+\frac x2\right)\right|=$ $\ln\left|\frac {\frac 2x+\tan\frac x2}{1-\frac 2x\cdot\tan\frac x2}\right|$ $\implies$ $\boxed {I=\ln\left|\frac {2+x\cdot\tan\frac x2}{x-2\cdot\tan\frac x2}\right|+C}$ .

$\blacktriangleleft\ I_7=\int\frac {1 - \ln x}{x^2 +\ln^2 x}\ \mathrm{dx}=$ $\int\frac {\frac {1-\ln x}{x^2}}{1+\left(\frac {\ln x}{x}\right)^2}\ \mathrm{dx}=$ $\int\frac {\left(\frac {\ln x}{x}\right)'}{1+\left(\frac {\ln x}{x}\right)^2}\ \mathrm{dx}=$ $\arctan\frac {\ln x}{x}+\mathbb C$ .

$\blacktriangleleft\ I_8=\int\frac {1+\sin x}{1+\sin 2x}\ \mathrm{dx}$ . Denote $W=\int\frac {1}{1+\sin 2x}\ \mathrm {dx}$ , $U=\int \frac {\sin x}{1+\sin 2x}\ \mathrm{dx}$ , $V=\int\frac {\cos x}{1+\sin 2x}\ \mathrm {dx}$ . Thus, $I=W+U$ .

$W=\int \frac {\sin^2x+\cos^2x}{(\sin x+\cos x)^2}\ \mathrm {dx}=$ $\int\frac {1+\tan^2x}{(1+\tan x)^2}\ \mathrm{dx}=$ $\int\frac {(1+\tan x)'}{(1+\tan x)^2}\ \mathrm {dx}=-\frac {1}{1+\tan x}+\mathrm C\implies \boxed {\ W=-\frac {\cos x}{\sin x+\cos x}+\mathrm C\ }$ .

$V-U=\int\frac {\cos x-\sin x}{1+\sin 2x}\ \mathrm {dx}=$ $\int\frac {(\sin x+\cos x)'}{(\sin x+\cos x)^2}\ \mathrm {dx}=$ $-\frac {1}{\sin x+\cos x}+\mathrm C\implies \boxed {\ V-U=-\frac {1}{\sin x+\cos x}+\mathrm C\ }$ .

$V+U=\int\frac {\sin x+\cos x}{(\sin x+\cos x)^2}\ \mathrm{dx}=\int\frac {1}{\sin x+\cos x}\ \mathrm {dx}=\int\frac {\left(x-\frac {\pi}{4}\right)'}{\sqrt 2\cdot \cos\left(x-\frac {\pi}{4}\right)}\ \mathrm{dx}$ . Since $\int\frac {1}{\cos t}\ \mathrm{dt}=\frac 12\cdot\ln\left|\frac {1+\sin t}{1-\sin t}\right|+\mathrm C$

obtain $\boxed {\ V+U=\frac {\sqrt 2}{4}\cdot\ln\left|\frac {1+\sin \left(x-\frac {\pi}{4}\right)}{1-\sin\left(x-\frac {\pi}{4}\right) }\right|+\mathrm C\ }$ a.s.o.

$\blacktriangleleft$ For $\{m,n\}\subset\mathbb N^*$ we have $I_{m,n}\equiv\int\frac {1}{x^m(x+1)^n}\ \mathrm {dx}=\int\frac {(x+1)-x}{x^m(x+1)^n}\ \mathrm {dx}\implies$ $\boxed{I_{m,n}=I_{m,n-1}-I_{m-1,n}}$ .

For $m\ge 2$ we have $I_{m,0}=\int\frac {1}{x^m}\ \mathrm {dx}=-\frac {1}{(m-1)x^{m-1}}+\mathrm C$ and $I_{0,m}=\int\frac {1}{(x+1)^m}\ \mathrm {dx}=-\frac {1}{(m-1)(x+1)^{m-1}}+\mathrm C$ .

In conclusion, $I_{3,2}=I_{3,1}-I_{2,2}=$ $\left[I_{3,0}-I_{2,1}\right]-\left[I_{2,1}-I_{1,2}\right]=I_{3,0}-2\cdot I_{2,1}+I_{1,2}=$

$I_{3,0}-2\cdot \left[I_{2,0}-I_{1,1}\right]+\left[I_{1,1}-I_{0,2}\right]=$ $I_{3,0}-2\cdot I_{2,0}+3\cdot I_{1,1}-I_{0,2}=$ $I_{3,0}-2\cdot I_{2,0}+3\cdot\left[I_{1,0}-I_{0,1}\right]-I_{0,2}$ $\implies$

$I_{3,2}=I_{3,0}-2\cdot I_{2,0}+3\cdot\left[I_{1,0}-I_{0,1}\right]-I_{0,2}$ $\implies$ $\boxed{\ I_{3,2}=-\frac {1}{2x^2}+\frac 2x+3\ln\left|\frac {x}{x+1}\right|+\frac {1}{x+1}+\mathrm C\ }$ .

$\blacktriangleleft\ J=\frac{1}{2}\cdot\int(x^{2}+a^{2})^{-\frac{3}{2}}(x^{2}+a^{2})'\ dx$ $\Longrightarrow$ $\boxed{\ J=-\frac{1}{\sqrt{x^{2}+a^{2}}}+C\ }$ and $I_9=\frac{1}{a^{2}}\int\frac{(x^{2}+a^{2})-x^{2}}{(x^{2}+a^{2})\sqrt{x^{2}+a^{2}}}\ dx=$

$\frac{1}{a^{2}}\left(\ln\left|x+\sqrt{x^{2}+a^{2}}\right|-K\right)\ ,$ where $K=\int\frac{x^{2}}{(x^{2}+a^{2})\sqrt{x^{2}+a^{2}}}\ dx$ . Thus, $\boxed{\ \begin{array}{ccc}f(x)=x & \Longrightarrow & f'(x)=1\\\\ g'(x)=\frac{x}{(x^{2}+a^{2})\sqrt{x^{2}+a^{2}}}& \Longrightarrow & g(x)=J=-\frac{1}{\sqrt{x^{2}+a^{2}}}\end{array}\ }$ $\Longrightarrow$

$K=-\frac{x}{\sqrt{x^{2}+a^{2}}}+\int\frac{1}{\sqrt{x^{2}+a^{2}}}\ dx$ $\Longrightarrow$ $\boxed{\ K=-\frac{1}{\sqrt{x^{2}+a^{2}}}+\ln\left|x+\sqrt{x^{2}+a^{2}}\right|\ }$ and $\boxed{\ I_9=\frac{x}{a^{2}\sqrt{x^{2}+a^{2}}}+C\ }\ .$

$\blacktriangleleft$ Let $x>1$ and $F=\int \frac{2x+\frac{1}{x^2}}{\left(x^2-\frac 1x\right)\sqrt{x^4-2x+1+\frac{1}{x^2}}}\ dx=$ $\int \frac{\left(x^2-\frac 1x\right)'}{\left(x^2-\frac 1x\right)\sqrt{\left( x^2-\frac 1x\right)^2+1}}\ dx=$ $G\left(x^2-\frac 1x\right)+C$ ,

where $G=\int \frac{1}{t\sqrt{1+t^2}}\ dt=-\int \frac{\left(\frac 1t\right)'}{\sqrt{1+\left(\frac 1t\right)^2}}\ dt=$ $-H\left(\frac 1t\right)+C$ , where $H=\int \frac{1}{\sqrt{1+y^2}}\ dy=\ln \left(y+\sqrt{1+y^2}\right)+C$ .

Therefore, $G(t)=\ln\left(\sqrt{1+t^2}-1\right)-\ln t+C$ $\Longrightarrow$ $\boxed{F(x)=\ln \frac{\sqrt{x^6-2x^3+x^2+1}-x}{x^3-1}+C}$ .

$\blacktriangleleft\ I_{12}\equiv\int \frac {\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt {\cos x}}\ \mathrm{dx}=$ $\int \frac {\sqrt{\tan x}}{1+\sqrt{\tan x}}\ \mathrm{dx}=$

$\int \frac {\sqrt{\tan x}}{1+\sqrt{\tan x}}\cdot\frac {2\sqrt{\tan x}}{1+\tan^2x}\cdot\left(\sqrt{\tan x}\right)'\ \mathrm{dx}=$ $F(\tan x)$ , where $F(y)=\int\frac {2y^2}{(1+y)(1+y^4)}\ \mathrm{dx}$ a.s.o.

$\blacktriangleleft\ I_{13}\equiv \int \frac{1}{x\sqrt{x^n+a}}\ \mathrm{dx},\ n\in N^*,\ a\ne 0$ . Thus, $\boxed {y\equiv \phi (x)=x^n}\ ,\ \phi '(x)=nx^{n-1}$ $\Longrightarrow$

$I_{13}=\frac 1n\int \frac{1}{x^n\sqrt{x^n+a}} (x^n)'\ \mathrm{dx}=$ $\frac 1n Y\ ,\ Y\equiv \int \frac{1}{y\sqrt{y+a}}\ \mathrm{dy}$ and $\boxed {z\equiv \psi (y)=\sqrt{y+a}}\ ,\ \psi '(y)=\frac{1}{2\sqrt{y+a}}$ $\Longrightarrow$

$Y=2\int \frac{1}{\left(\sqrt{y+a}\right)^2-a}\left(\sqrt{y+a}\right)'\ \mathrm{dy}=2Z\ ,\ Z\equiv \int \frac{1}{z^2-a} \ \mathrm{dz}$ . If $a<0$ , then $Z=\frac{1}{\sqrt{-a}} \arctan \frac{z}{\sqrt{-a}}+C$ $\Longrightarrow$

$I_{13}=\frac{2}{n\sqrt{-a}}\arctan \sqrt{\frac{x^n+a}{-a}}+C$ . If $a>0$ , then $Z=\frac{1}{2\sqrt a} \ln \left|\frac{z-\sqrt a}{z+\sqrt a}\right| +C$ $\Longrightarrow$ $I_{13}=\frac{1}{n\sqrt a}\ln \frac{\sqrt{x^n +a}-\sqrt a}{\sqrt{x^n+a}+\sqrt a}+C$ .

$\blacktriangleleft\ I_{14}=\int\left(\frac{\arctan x}{\arctan x-x}\right)^{2}\ \mathrm{dx} =$ $\int\frac{\left(x^2+1\right)\arctan^2x}{x^2}\cdot$ $\frac {x^2}{\left(x^2+1\right)(\arctan x-x)^2}\ \mathrm{dx}$ . Integrate by parts :

$\left\{\begin{array}{ccc} u(x)=\frac{\left(x^2+1\right)\arctan^2x}{x^2} & \implies & u'(x)=-\frac {2\arctan x(\arctan x-x)}{x^3}\\\\ v'(x)=\frac {x^2}{\left(x^2+1\right)(\arctan x-x)^2} & \implies & v(x)=\frac {1}{\arctan x-x}\end{array}\right\|\implies$ $\boxed{I_{14}=\frac{\left(x^2+1\right)\arctan^2x}{x^2(\arctan -x)}+2J}$ ,

where $J=\int\frac {\arctan x}{x^3}\ \mathrm{dx}$ . Integrate by parts : $\left\{\begin{array}{ccc} u(x)=\arctan x & \implies & u'(x)=\frac {1}{x^2+1}\\\\ v'(x)=\frac {1}{x^3} & \implies & v(x)=-\frac {1}{2x^2}\end{array}\right\|\implies$ $\boxed{J=-\frac {\arctan x}{2x^2}+\frac 12\cdot K}$ ,

where $K=\int\frac {1}{x^2\left(x^2+1\right)}\ \mathrm{dx}=$ $\int\left(\frac {1}{x^2}-\frac {1}{x^2+1}\right)\ \mathrm{dx}\implies$ $\boxed{K=-\frac 1x-\arctan x+\mathbb C}$ . In conclusion, obtain by return way that

$J=-\frac {\arctan x}{2x^2}-\frac {1}{2x}-\frac 12\cdot\arctan x+\mathbb C\implies$ $\boxed{I_{14}=\frac {\left(x^2+1\right)\arctan^2x}{x^2(\arctan x-x)}-\frac {\arctan x}{x^2}-\frac 1x-\arctan x+\mathbb C}$ .

$2.\blacktriangleright$ Prove that $\int_{0}^{\frac{\pi}{2}}\sin x\sqrt{\sin 2x}\ dx\ =\sqrt 2\int^{\frac{\pi}{4}}_0 \cos x\sqrt{\cos 2x}\ dx=\ \int_{-1}^{1}\sqrt{1-x^2}\ dx$ ;

Proof. Let $I=\int_{0}^{\frac{\pi}{2}}\sin x\sqrt{\sin 2x}\ \mathrm{dx}$ . So $I=\int_{0}^{\frac{\pi}{2}}\cos x\sqrt{\sin 2x}\ \mathrm{dx}$ . Adding we get $2I=\int_{0}^{\frac{\pi}{2}}(\sin x+\cos x)\sqrt{\sin 2x}\ \mathrm{dx}$ .

Now substitute $t=\sin x-\cos x$ , where $\mathrm{dt}= (\cos x+\sin x)\mathrm{dx}$ . When $x=0$ , then $t= -1$ and when $x=\frac{\pi}2$ , then $t=1$ .

Use $\sin 2x=1-(\sin x-\cos x)^2$ . But this will give $2I=\int_{-1}^{1}\sqrt{1-x^2}\ \mathrm{dx}$ , i.e. $I=\frac {\pi}{4}$ .

Otherwise. $A\equiv \int^{\frac{\pi}{2}}_0 \sin x\sqrt{\sin 2x}\ \mathrm{dx}=$ $\int^{\frac{\pi}{2}}_0 \sin \left( \frac{\pi}{2}-x\right)\sqrt{\sin 2\left(\frac{\pi}{2}-x\right)}\ \mathrm{dx}=$ $\int^{\frac{\pi}{2}}_0 \cos x\sqrt{\sin 2x}\ \mathrm{dx}\Longrightarrow$

$2A= \int^{\frac{\pi}{2}}_0 (\cos x+\sin x)\sqrt{\sin 2x}\ \mathrm{dx}=$ $\int^{\frac{\pi}{2}}_0 \sqrt{1-(\sin x-\cos x)^2}\cdot (\sin x-\cos x)'\ \mathrm{dx}=$ $\int^1_{-1} \sqrt{1-x^2}\ \mathrm{dx}=$

$2\int^1_0 \sqrt{1-x^2}\ \mathrm{dx}=$ $\frac{\pi}{2}\Longrightarrow A=\frac{\pi}{4}$ . Thus, $CC\equiv \int^{\frac{\pi}{4}}_0 \cos x\sqrt{\cos 2x}\ \mathrm{dx}=$ $\int^{\frac{\pi}{4}}_0 (1-2\sin^2x)^{\frac 12} (\sin x)'\ \mathrm{dx}=$

$\int^{\frac{\sqrt 2}{2}}_0 \sqrt{1-2x^2}\ \mathrm{dx}=$ $\frac{1}{\sqrt 2}\int^{\frac{\sqrt 2}{2}}_0 \sqrt{1-(x\sqrt 2 )^2}\cdot (x\sqrt 2)'\ \mathrm{dx}=$ $\frac{1}{\sqrt 2}\int^1_0 \sqrt{1-x^2}\ \mathrm{dx}=\frac{\pi \sqrt 2}2\ .$

Remark Ascertain elegantly the following definite integrals : $CC\equiv \int^{\frac{\pi}{4}}_0 \cos x\sqrt{\cos 2x}\ \mathrm{dx}\ ,$ $CS\equiv \int^{\frac{\pi}{4}}_0 \cos x\sqrt{\sin 2x}\ \mathrm{dx}\ ,$

$SS\equiv \int^{\frac{\pi}{4}}_0 \sin x\sqrt{\sin 2x}\ \mathrm{dx}\ ,$ $SC\equiv \int^{\frac{\pi}{4}}_0 \sin x\sqrt{\cos 2x}\ \mathrm{dx}$ . Indication. $CS+SS=\ldots\ ,\ CS-SS=\ \ldots$ a.s.o.

$3.\blacktriangleright$ Let $f\ :\ (0,\infty)\rightarrow R$ be a function, where $f(\lambda)=\int_{0}^{2\pi}\frac{\sin x}{x+\lambda}\ \mathrm{dx}$ . Prove that $f(\lambda)>0$ , $\lim_{\lambda\rightarrow\infty}f(\lambda)=0$ and $\lim_{\lambda\rightarrow\infty}{\lambda}^2f(\lambda)=2\pi$ .

Proof. Integrate by parts, using $(1-\cos x)$ as the antiderivative of $\sin x\ :\ f(\lambda)=$ $\int_0^{2\pi} \frac{\sin x}{x+\lambda}\ \mathrm{dx}=$ $\left|\frac{1-\cos x}{x+\lambda}\right|_0^{2\pi}+ \int_0^{2\pi} \frac{1-\cos x}{(x+\lambda)^2}\ \mathrm{dx}=$

$\int_0^{2\pi} \frac{1-\cos x}{(x+\lambda)^2}\ \mathrm{dx}$ . Since the integrand in this version is positive on $(0,2\pi),$ it is clear that $f(\lambda)>0$ and $\lambda^2f(\lambda)=$ $\int_0^{2\pi} \frac{1-\cos x}{\left(1+\frac{x}{\lambda}\right)^2}\ \mathrm{dx}$ . As $\lambda\to\infty$

the integrand converges uniformly to $(1-\cos x)$ . Hence $\lim_{\lambda\to\infty}\lambda^2f(\lambda)=\int_0^{2\pi}(1-\cos x)\ \mathrm{dx}=2\pi$ . After the fact, the existence of this limit implies that

$f(\lambda)\to 0$ as $\lambda\to\infty$ although we could have proved that directly in several other ways. It is also clear that $f(\lambda)$ is a decreasing function for $\lambda>0$ and that

$\lim_{\lambda\searrow 0}f(\lambda)$ exists and equals $\int_0^{2\pi}\frac{\sin x}x\ \mathrm{dx}$ which is a finite positive number.

Otherwise. $f(\lambda)=\int^{\pi}_{-\pi}\frac{\sin x}{\pi -x+\lambda}\ \mathrm{dx}=$ $\int^{\pi}_0 \left(\frac{\sin x}{\pi+\lambda -x}-\frac{\sin x}{\pi+\lambda +x}\right)\ \mathrm{dx}=$ $\int^{\pi}_0 \frac{2x\sin x}{(\pi +\lambda )^2-x^2}\ \mathrm{dx}$ .

$\left(\forall\right)x\in [0,\pi ],\frac{2x\sin x}{(\pi +\lambda )^2}\le \frac{2x\sin x}{(\pi +\lambda )^2-x^2}\le \frac{2x\sin x}{(\pi +\lambda )^2-\pi ^2}$ and $\int^{\pi}_0 x\sin x\ \mathrm{dx}=\pi\Longrightarrow$

$\left(\forall\right) \lambda >0,\frac{2\pi}{(\pi +\lambda )^2}\le f(\lambda )\le \frac{2\pi}{(\pi +\lambda )^2-\pi ^2}\Longrightarrow$ $f(\lambda )>0,\lim_{\lambda\to \infty }f(\lambda ) =0,\lim_{\lambda\to \infty} \lambda ^2 f(\lambda )=2\pi$ .

$4.\blacktriangleright$ Consider the function $f\ : \ [0,1]\rightarrow [0,\infty ),\ f\in C^{(1)};\ a_n=\int^1_0 \frac{f(x)}{1+x^n}\ \mathrm{dx},\ n\in N^*$ . Prove that

the sequence $a_n,\ n\in N^*$ is convergently, i.e. $a_n\rightarrow l\in R$ and $\lim_{n\to \infty} n(l-a_n)=f(1)\ln 2$ .

Remark See here and at the publishing house my book Probleme de analiza matematica (clasa a XII - a), Editura ALL, Bucuresti, 2002.

Proof. $\left(\forall\right)n\in N^*,0\le \int^1_0 f(x)\ \mathrm{dx}-\int^1_0\frac{f(x)}{x^n+1}\ \mathrm{dx}=\int^1_0\frac{x^n}{x^n+1}f(x)\ \mathrm{dx}$ $\le A\int^1_0 x^n\ \mathrm{dx}=\frac{A}{n+1}$ , where $\left(A=\sup_{0\le x\le 1} f(x)\right)\Longrightarrow$

$\lim_{n\to \infty} \int^1_0 \frac{f(x)}{x^n+1}\ \mathrm{dx}=\int^1_0 f(x)\ \mathrm{dx}\equiv l$ . Therefore, $n\left( l-a_n\right) =\int^1_0 \frac{nx^{n-1}}{x^n+1}\cdot xf(x)\ \mathrm{dx}=f(1)\ln 2-b_n$ , where

$b_n=\int^1_0 [f(x)+xf'(x)]\ln \left(x^n+1\right)\ \mathrm{dx}\le \int^1_0 [f(x)+xf'(x)]x^n\ \mathrm{dx}\le$ $A\int^1_0 x^n dx+B\int^1_0 x^{n+1}\ \mathrm{dx}=\frac{A}{n+1}+\frac{B}{n+2}\rightarrow 0$ ,

where $B=\sup_{0\le x\le 1} f'(x)$ . Therefore, $b_n\rightarrow 0$ and $\lim_{n\to \infty} n\left(l-a_n\right)=f(1)\ln 2$ .

$5.\blacktriangleright$ Some usual formulas.

$\blacksquare\ 1^{\circ}.\ \int^{b}_{a}f(x) dx =\int^{b}_{a}f(a+b-x) dx$ .

$\blacksquare\ 2^{\circ}.\ \int^{b}_{a}f(x) dx =\int^{\frac{b-a}{2}}_{-\frac{b-a}{2}}f\left(\frac{a+b}{2}-x\right) dx$ .

$\blacksquare\ 3^{\circ}.\ \int^{b}_{a}f(x)dx =\int^{\frac{b-a}{2}}_{0}\left[ f\left(\frac{a+b}{2}-x\right)+f\left(\frac{a+b}{2}+x\right)\right] dx$ .

$\blacksquare\ 4^{\circ}.\ a > 0,\ f:\left[\frac{1}{a},a\right]\rightarrow R\Longrightarrow\int^{a}_{\frac{1}{a}}f(x)dx =\int^{a}_{\frac{1}{a}}\frac{1}{x^{2}}f\left (\frac{1}{x}\right) dx$ .

$\blacksquare\ 5^{\circ}\ .$ $T > 0\ ,\ (\forall )\ x\in R$ , $f(x+T)=f(x)$ $\iff$ $\int^{a+nT}_{a}f(x) \mathrm{dx} =$ $n\int^{T}_{0}f(x+a) dx=n\int^{a+T}_{a}f(x) dx$ .

$\blacksquare\ 6^{\circ}.\ F,f: R\rightarrow R,\ F' = f\Longrightarrow\lim_{x\to\infty}\frac{1}{x}F(x) =\frac{1}{T}\int^{T}_{0}f(x)dx$ .

$\blacksquare\ 7^{\circ}$ If the function $f:[0,1]\rightarrow \mathbb R$ is integrablely, then $\lim_{n\to\infty}\ \frac 1n\cdot\sum_{k=1}^nf\left(\frac kn\right)=\int_0^1f(x)\ \mathrm {dx}$ .
Generally, if the function $f:[a,b]\rightarrow \mathbb R$ is integrablely, then for any $\lambda\in [0,1]$ , $\lim_{n\to\infty}\ \frac 1n\cdot\sum_{k=1}^nf\left[a+\frac {(k-\lambda )(b-a)}{n}\right]=\int_a^bf(x)\ \mathrm{dx}$ .

$\blacksquare\ 8^{\circ}$ If $f:[0,1]\rightarrow \mathbb R$ is derivably and $f''$ is integrably , then $\lim_{n\to\infty}\ n\left[\frac 1n\cdot\sum_{k=1}^nf\left(\frac kn\right)-\int_0^1f(x)\ \mathrm{dx}\right]=\frac 12\cdot [f(1)-f(0)]$ .

Generally, if $f:[a,b]\rightarrow \mathbb R$ is derivably and $f'$ is integrably, then for any $\lambda\in [0,1]$ ,
$\lim_{n\to\infty}\ n\cdot \left\{\frac {b-a}{n}\cdot \sum_{k=1}^nf\left[a+\frac {(b-a)(k-\lambda )}{n}\right]-\int_a^bf(x)\ \mathrm{dx}\right\}=$ $\left(\frac 12-\lambda \right)(b-a)[f(b)-f(a)]$ .

$\blacksquare\ 9^{\circ}$ If $f:[0,1]\rightarrow \mathbb R$ is $2$ -derivably and $f''$ is integrably , then $\lim_{n\to\infty}\ n^2\left[\int_0^1 f(x)\ \mathrm{dx}-\frac 1n\cdot\sum_{k=1}^n f\left(\frac {2k-1}{2n}\right)\right]=$ $\frac {1}{24}\cdot [f'(1)-f'(0)]$ .

Generally, if $f:[a,b]\rightarrow \mathbb R$ is $2$ -derivablely and $f''$ is integrably, then $\lim_{n\to\infty}\ n^2\cdot \left\{ \int_a^b f(x)\ \mathrm {dx}- \frac {b-a}{n}\cdot\sum_{k=1}^n f\left[a+\frac {b-a}{n}\cdot\left(k-\frac 12\right)\right]\right\}=\frac {(b-a)^2}{24}\cdot \left[f'(b)-f'(a)\right]$ .

$\blacksquare\ 10^{\circ}$ If the function $f:[a,b]\rightarrow [c,d]$ is continue and bijective, then $\int_a^bf(x)\ \mathrm{dx}+\int_c^df^{-1}(x)\ \mathrm{dx}=bd-ac$ .

$\blacksquare\ 11^{\circ}$ (Young's inequality) If the function $f:[0,\infty )\rightarrow\mathbb R$ is continue , strict increasing and $f(0)=0$ , then for any $a>0$ , $c>0$ , $f(c)=b$ exists the relation $\int_0^af(x)\ \mathrm{dx}+\int_0^bf^{-1}(x)\ \mathrm{dx}\ge ab$ with equality if and only if $f(a)=b$ , i.e. $c=a$ .
Remark. An equivalent enunciation : $(\forall ) a>0\ ,\ c>0$ exists inequality $(a-c)f(c)\le \int_a^acf(x)\ \mathrm{dx}$ ..
Particular case. If $f(x)=x^{p-1}$ , $p>1$ , then obtain Holder's inequality - $\frac 1p+\frac 1q=1\implies \frac {a^p}{p}+\frac {b^q}{q}\ge ab$ .

$\blacksquare\ 12^{\circ}$ If the function $f:\mathbb R\rightarrow\mathbb R$ is continue and the functions $\alpha\ ,\ \beta$ belong to $\mathrm {C^1(R)}$ , then the function

$F(x)=\int_{\alpha (x)}^{\beta (x)}f(t)\ \mathrm{dt}$ is integrably and $F'(x)=f(\beta (x))\cdot\beta '(x)-f(\alpha (x))\cdot\alpha '(x)$ .

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$6.\blacktriangleright$ Applications.

$\blacksquare\ 1^{\circ}.\ \boxed{I=\int^1_0 \frac{\ln (x+1)}{x^2+1}\ \mathrm{dx}=\frac {\pi\ln2}{8}}$ . Indeed, $I\equiv\int^1_0 \frac{\ln (x+1)}{x^2+1} \ \mathrm{dx}=$ $\int^1_0 \ln(1+x)(\arctan x)'\ \mathrm{dx}=$

$\int^{\frac{\pi}{4}}_0 \ln (1+\tan x)\ dx=$ $\int^{\frac{\pi}{4}}_0 \ln \left[1+\tan \left(\frac{\pi}{4} -x\right)\right] \ \mathrm{dx}=$ $\int^{\frac{\pi}{4}}_0\ln \frac{2}{1+\tan x}\ \mathrm{dx}=$ $\frac{\pi}{4}\ln 2 -I$ $\Longrightarrow I=\frac{\pi\ln 2}{8}$ .

$\blacksquare\ 2^{\circ}.\ \boxed{\lim_{n\to \infty}\ \int^1_0 \frac{1}{x^n-x+1}\ \mathrm{dx} =\infty}$ . Indeed, $\int^1_0\frac{1}{x^n-x+1}\ dx\ge \int^{\frac{1}{\sqrt [n] n}}_0 \frac{1}{x^n-x+1}\ dx\ge \int^{\frac{1}{\sqrt [n] n}}_0\frac{1}{\frac 1n -x+1}\ dx=$

$\ln \frac{1+\frac 1n}{1+\frac 1n -\frac{1}{\sqrt [n] n}}\rightarrow \infty$ . The open question : What is the set of the numbers $t\in [0,1)$ for which the sequence $a_n=\int^1_t\frac{1}{x^n-x+1}\ \mathrm{dx}$ is convergently ?

$\blacksquare\ 3^{\circ}.\ \boxed{\int^{\pi}_0 x\sin (\cos^2 x)\cos (\sin ^2x)\ \mathrm{dx}=\frac{{\pi}^2}{4}\sin 1}$ . Indeed, $I\equiv \int^{\pi}_0 x\sin (\cos^2x)\cos(\sin^2 x)\ \mathrm{dx}=$

$\int^{\pi}_0 (\pi -x)\sin (\cos^2x)\cos(\sin^2x)\ \mathrm{dx}=\pi J-I$ , where $J\equiv\int^{\pi}_0 \sin (\cos^2x)\cos (\sin^2 x)\ \mathrm{dx}$ . Thus, $\boxed{I=\frac{\pi}{2}\cdot J}$ and

$J=\frac 12\int^{\pi}_0 [\sin (\cos^2x+\sin^2x)+$ $\sin (\cos^2x-\sin^2x)]\ \mathrm{dx}=$ $\frac 12\int^{\pi}_0[\sin 1+\sin(\cos 2x)]\ \mathrm{dx}$ , i.e. $\boxed{J=\frac{\pi}{2}\sin 1+\frac 12\cdot K}$ , where

$K\equiv\int^{\pi}_0 \sin(\cos 2x)\ \mathrm{dx}=$ $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sin\left[ \cos 2\left(\frac{\pi}{2}-x\right)\right]\ \mathrm{dx}=$ $-\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sin (\cos 2x)\ \mathrm{dx}=-2L$ , i.e. $\boxed{K=-2L}$ , where $L\equiv\int^{\frac{\pi}{2}}_0 \sin (\cos 2x)\ \mathrm{dx}=$

$\int^{\frac{\pi}{2}}_0 \sin \left[ \cos 2\left( \frac{\pi}{2}-x\right ) \right]\ \mathrm{dx}=$ $\int^{\frac{\pi}{2}}_0\sin (-\cos 2x)\ \mathrm{dx}=-L$ , i.e. $\overline {\underline {\left| \ L=-L\ \right| }}$ . Therefore, $L=0\Longrightarrow$ $K=0\Longrightarrow$ $J=\frac{\pi}{2}\sin 1\Longrightarrow$ $I=\frac{\pi^2}{4}\sin 1$ .

$\blacksquare\ 4^{\circ}.\ \boxed{\int^{\pi}_0 x\left[ \sin^2(\sin x)+\cos^2 (\cos x)\right]\ \mathrm{dx}=\frac{{\pi}^2}{2}}$ . Indeed, $I\equiv \int^{\pi}_0 x[\sin^2(\sin x)+\cos^2(\cos x)]\ \mathrm{dx}dx=$

$\int^{\pi}_0 (\pi -x)\{\sin^2[\sin(\pi -x)]+\cos^2[\cos (\pi -x)]\}\ \mathrm{dx}=$ $\int^{\pi}_0 (\pi -x)[\sin^2(\sin x)+\cos^2(\cos x)]\ \mathrm{dx}=\pi J-I$ . Thus $\overline {\underline {\left|\ I=\frac{\pi}{2} J\ \right| }}$ , where

$J\equiv \int^{\pi}_0 [\sin^2(\sin x)+\cos^2(\cos x)]\ \mathrm{dx}=$ $\frac 12\int^{\pi}_0 [1-\cos(2\sin x)+1+\cos (2\cos x)]\ \mathrm{dx}=$ $\pi +\frac 12 K$ . Thus $\overline {\underline {\left| \ J=\pi+\frac 12K\ \right| }}$ , where

$K\equiv\int^{\pi}_0[\cos (2\cos x)-\cos (2\sin x)]\ \mathrm{dx}=$ $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\left\{\cos \left[ 2\cos\left(\frac{\pi}{2}-x\right)\right] -\cos \left[ 2\sin \left(\frac{\pi}{2}-x\right)\right]\right\}\ \mathrm{dx}=$ $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} [\cos(2\sin x)-\cos (2\cos x)]\ \mathrm{dx}=$

$2\int^{\frac{\pi}{2}}_0[\cos (2\sin x)-\cos (2\cos x)]\ \mathrm{dx}=$ $2\left\{ \int^{\frac{\pi}{2}}_0 \cos \left[ 2\sin \left( \frac{\pi}{2} -x\right)\right]\ \mathrm{dx} -\int^{\frac{\pi}{2}}_0 \cos (2\cos x)\ \mathrm{dx}\right\}=0$ .

Therefore, $K=0\Longrightarrow J=\pi\Longrightarrow I=\frac{\pi^2}{2}$ .

$\blacksquare\ 5^{\circ}.\ \boxed{\lim_{n\to \infty} \left( \frac{1^1\cdot 2^2\cdot 3^3\cdot \ldots \cdot n^n}{n^{1+2+3+\ldots +n}}\right)^{\frac{1}{n^2}}=e^{-\frac 14}}$ . For $a_n=\left[ \prod\limits_{k=1}^n\left(\frac kn\right)^{\frac kn} \right]^\frac 1n$ have $\ln\left(\lim_{n\to\infty} a_n\right)=\lim_{n\to\infty}\ln a_n=\lim_{n\to\infty}\frac 1n\sum\limits_{k=1}^n \frac kn\ln\frac kn=$

$\int^1_0x\ln x\ \mathrm{dx}=$ $\lim_{t\searrow 0} \int^1_t x\ln x\ \mathrm{dx}=$ $\lim_{t\searrow 0} \left|\left( \frac{x^2\ln x}{2}-\frac{x^2}{4}\right)\right|^1_t=$ $\lim_{t\searrow 0} \left(\frac{t^2}{4} -\frac 14 -\frac{t^2\ln t}{2}\right)=$ $-\frac 14$ $\Longrightarrow$ $\lim_{n\to \infty} a_n=$ $e^{-\frac 14}=\frac{1}{\sqrt [4] e}$ .

$\blacksquare\ 6^{\circ}.\ \boxed{\int^1_0 \frac{\arcsin \sqrt x}{x^2-x+1}\ \mathrm{dx}=\frac{\pi ^2\sqrt 3}{18}}$ . Indeed, $I\equiv\int^1_0 \frac{\arcsin \sqrt x}{x^2-x+1}\ \mathrm{dx}=$ $\int^1_0 \frac{\arcsin \sqrt{1-x}}{(1-x)^2-(1-x)+1}\ \mathrm{dx}=$

$\int^1_0 \frac{\arcsin \sqrt{1-x}}{x^2-x+1}\ \mathrm{dx}$ $\Longrightarrow$ $I=\frac 12\int^1_0 \frac{\arcsin \sqrt x+\arcsin \sqrt{1-x}}{x^2-x+1}\ \mathrm{dx}=$ $\frac{\pi}{4}\int^1_0 \frac{1}{x^2-x+1}\ \mathrm{dx}=$ $\frac{\pi}{4}\int^1_0 \frac{\left(x-\frac 12\right)'}{\left(x-\frac 12\right)^2+\frac 34}\ \mathrm{dx}=$

$\frac{\pi}{4}\int^{\frac 12}_{-\frac 12} \frac{1}{x^2+\frac 34}\ \mathrm{dx}=$ $\frac{\pi}{2}\int^{\frac 12}_0 \frac{1}{x^2+\frac 34}\ \mathrm{dx}=$ $\left(\frac{\pi}{2}\cdot \frac{2}{\sqrt 3}\cdot \arctan \frac{2x}{\sqrt 3}\right)^{\frac 12}_0=$ $\frac{\pi}{\sqrt 3}\cdot \frac{\pi}{6}=$ $\frac{\pi ^2\sqrt 3}{18}$ .

$\blacksquare\ 7^{\circ}.\ \boxed{\int^{\sqrt 3}_{\frac{1}{\sqrt 3}} \frac{\arctan\ x}{x}\ \mathrm{dx}=\frac{\pi}{4} \ln 3}$ . Indeed, $J\equiv\int^{\sqrt 3}_{\frac{1}{\sqrt 3}} \frac{\arctan x}{x}\ \mathrm{dx}=$ $\int^{\sqrt 3}_{\frac{1}{\sqrt 3}} \frac{\frac{\pi}{2}-\arctan \frac 1x}{-\frac 1x}\ \left (\frac 1x\right )^{'}\ \mathrm{dx}=$

$\int^{\sqrt 3}_{\frac{1}{\sqrt 3}} \frac{\frac{\pi}{2}-\arctan x}{x}\ \mathrm{dx}$ $\Longrightarrow$ $J=\frac{\pi}{4}\int^{\sqrt 3}_{\frac{1}{\sqrt 3}} \frac 1x\ \mathrm{dx}=$ $\left(\frac{\pi}{4}\ln x\right)^{\sqrt 3}_{\frac{1}{\sqrt 3}}=\frac{\pi\ln 3}{4}$ .

$\blacksquare\ 8^{\circ}.\ \boxed{\int^2_{\frac 12} \frac{x^n -1}{x^{n+2}+1}\ \mathrm{dx}=0}\ ,\ n\in N^*$ . Indeed, $J\equiv\int^2_{\frac 12} \frac{1-x^n}{1+x^{n+2}}\ \mathrm{dx}=$ $\int^2_{\frac 12}\frac{1-x^n}{1+x^{n+2}}(-x^2)\left(\frac 1x\right)^{'}\ \mathrm{dx}=$

$\int^2_{\frac 12} \frac{1-\frac{1}{x^n}}{1+\frac{1}{x^{n+2}}}\left(\frac 1x\right)^{'} \ \mathrm{dx}=$ $\int^2_{\frac 12} \frac{x^n-1}{x^{n+2}+1}\ \mathrm{dx}=-J\Longrightarrow J=0$ .

$\blacksquare\ 9^{\circ}.\ \left(\forall\right)k\in \overline {1,n}$ the functions $f_k\ : \ [a,b]\rightarrow (0,\infty )$ are increasing $\Longrightarrow$ $\boxed{\prod\limits_{k=1}^n \int^b_a f_k(x)\ dx\le (b-a)^{n-1}\cdot \int^b_a\prod\limits_{k=1}^nf_k(x)\ \mathrm{dx}}$ (Chebyshev's inequality).

Indeed, denote $X.s.s.Y$ $\Longleftrightarrow$ $XY>0\ \vee\ X=Y=0$ , i.e. $X,Y$ have the same sign. Let $\{ m,n\}\subset N^*$ so that $\left( \forall \right) i\in \overline {1,m}$ , $\left(\forall\right)j\in \overline {1,n}$

we have $a_{ij}>0$ and $\left(\forall\right) \{i,l\}\subset \overline {1,m}$ and $\left(\forall\right) \{j,k\}\subset \overline {1,n},\ (a_{ij}-a_{ik}).s.s.(a_{lj}-a_{lk})$ . Then $\prod\limits_{i=1}^m\sum\limits_{j=1}^na_{ij}\le \sum\limits_{j=1}^n\prod\limits_{i=1}^m a_{ij}$

(the Cebasev's generalized inequality). Apply this inequality to the Riemann's sums.

$\blacksquare\ 10^{\circ}.\ \boxed{\lim_{x\to \infty}\ \frac{1}{x^3}\int^x_{-x} \left| t^2-t-2\right|\ \mathrm{dt}=\frac 23}$ . Indeed, denote $F(x)=\int^{x}_{-x} |t^2-t-2|\ \mathrm{dt}\Longrightarrow$

$\left(\forall\right) x\in R,\ F(-x)=-F(x)$ and $F(x)\equiv\left \{ \begin{array}{cc} -\frac 23 x^3+4x & 0\le x\le 1\ .\\\\ x^2+\frac 73 & 1<x<2\ .\\\\ \frac 23 x^3-4x+9 & 2\le x\ . \end{array}\right.\ \Longrightarrow \lim_{x\to \infty} \frac{F(x)}{x^3}=\frac 23$ .

$\blacksquare\ 11^{\circ}.\ \boxed{\min_{x\in R}\ \int^1_0\left| t^2+x\right|\ \mathrm{dt}=\frac 14}$ . Indeed, observe that $F(x)\equiv\int^1_0 \left|t^2+x\right|\ dt=$ $\left \{ \begin{array}{cc} -x-\frac 13 & x\le -1\ .\\\\ \frac 13\left(1+3x-4x\sqrt{-x}\right) & -1<x<0\ .\\\\ x+\frac 13 & 0\le x\ .\end{array}\right\|$

$\min_{x\le-1} F(x)=F(-1)=\frac 23\ ;\ \min_{-1<x<0}$ $F(x)=F\left(-\frac 14\right)=\frac 14\ ;\ \min_{0\le x} F(x)=F(0)=\frac 13$ . Therefore, $\min_{x\in R} F(x)=\frac 14=F\left(-\frac 14\right)$ .

$\blacksquare\ 12^{\circ}.$ PP1. Find the values of $k\in (0,1)$ such that the areas of the three parts bounded by the graph of $y=-x^4+2x^2$ and the line $y=k$ are all equal.

Proof. Let: $A(a,k)\ ;\ B(b,k)\ ;\ C(-b,k)\ ;\ D(-a,k)$ . We need $\int_{a}^{b}(-x^4+2x^2-k)\ dx =$ $\int_{b}^{-b}(x^4-2x^2+k)\ dx =$ $\int_{-b}^{-a}(-x^4+2x^2-k)\ dx$ .

Let: $x=-t \Rightarrow$ $\int_{a}^{b}(-x^4+2x^2-k)\ dx =$ $\int_{-b}^{-a}(-x^4+2x^2-k)\ dx\ (1)$ . We find $k$ satisfied $\int_{a}^{b}(-x^4+2x^2-k)\ dx =$ $\int_{b}^{-b}(x^4-2x^2+k)\ dx$

$\iff$ $\int_{a}^{b}(-x^4+2x^2-k)\ dx -\int_{b}^{-b}(x^4-2x^2+k)\ dx =0$ $\iff$ $\int_{a}^{-b}\left(x^4-2x^2+k\right)\ dx=0$ $\iff$ $\left(\frac{x^5}{5}-2\frac{x^3}{3}+kx\right)_{a}^{-b}=0$ $\implies k$ .

$\blacksquare\ 13^{\circ}.$ PP2. Let $f(x)=\left\{ \begin{array}{ccc} 1 & \mathrm{if} & x=0\\\\ \frac 1x\cdot \min\limits_{n\in N} |n-x| & \mathrm{if} & x>0\end{array}\right\|$ and $a_n=\int^n_0 f(x)\ \mathrm{dx}$ , where $n\in N^*$ . Prove that $\lim_{n\to \infty} a_n=\infty ,\ \lim_{n\to \infty} \frac{a_n}{\ln n}=\frac 14$ .

$\blacksquare\ 14^{\circ}.$ PP3. Calculate very short and simply $\int_{0}^{\pi}\left|\sin x+\cos x\right|\ \mathrm{dx}$ .

Proof. Denote $I=\int_{0}^{\pi}\left|\sin x-\cos x\right|\ \mathrm{dx}$ and observe that $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left|\sin x-\cos x\right|\ \mathrm{dx}=$ $\int_{0}^{\frac{\pi}{2}}\left| \sin x-\cos x\right|\ \mathrm{dx}+$

$\int_{0}^{\frac{\pi}{2}}\left|\sin x+\cos x\right|\ \mathrm{dx}=$ $2+\int_{0}^{\frac{\pi}{2}}\left|\sin x-\cos x\right|\ \mathrm{dx}=$ $2+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left|\frac{\sqrt 2}{2}(\cos x-\sin x)-\frac{\sqrt 2}{2}(\cos x+\sin x)\right|\ \mathrm{dx}=$

$2+\sqrt 2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left|\sin x\right|\ \mathrm{dx}=$ $2+2\sqrt 2\int_{0}^{\frac{\pi}4}\sin x\ \mathrm{dx}=$ $2+2\sqrt 2\left|\cos x\right|_{\frac {\pi}{4}}^0=$ $2+2\sqrt 2\cdot\left(1-\frac{\sqrt 2}{2}\right)=2\sqrt 2$ .

$\blacksquare\ 15^{\circ}.$ PP4. Ascertain $\lim_{n\to\infty} \frac{\int_{\frac{1}{n+1}}^{\frac 1n} \arctan nx\ dx}{\int_ {\frac{1}{n+1}}^{\frac 1n} \arcsin nx\ dx}=\frac 12$ without the calculation of the two definite integrals.

Proof. $\lim_{n\to\infty} \frac{\int_{\frac{1}{n+1}}^{\frac 1n} \arctan nx\ \mathrm{dx}}{\int_ {\frac{1}{n+1}}^{\frac 1n} \arcsin nx\ \mathrm{dx}}= \lim_{n\to\infty} \frac{\arctan n\xi_n}{\arcsin n\zeta_n}= \frac{\frac{\pi}{4}}{\frac{\pi}{2}} = \frac{1}{2}$ , where $\xi_n,\zeta_n \in \left[\frac{1}{n+1},\frac{1}{n}\right]\ \implies\ n\xi_n \to 1\ \ \wedge\ \ n\zeta_n \to\ 1$ .

$\blacksquare\ 16^{\circ}.$ PP5. Prove that if $f: [0,1]\rightarrow R$ is a continue and monotonous function, then $\left(\forall\right)a\in R\ ,\ \int_0^1 |f(x)-a|\ \mathrm{dx}\ge \int_0^1\left|f(x)-f\left(\frac 12\right)\right|\ \mathrm{dx}\ .$

Proof. Let $f$ increasing and $a<f\left(\frac 12\right)$ . In other cases the prove is the same. If $f(0)\le a<f\left(\frac 12\right)$ , then $a=f(b)$ and

$\int_0^1 |f(x)-a|\ \mathrm{dx} - \int_0^1\left|f(x)-f\left(\frac 12\right)\right|\ \mathrm{dx} =$ $\int_0^b |f(x)-a|\ \mathrm{dx} + \int_b^{\frac 12} |f(x)-a|\ \mathrm{dx}$ $+\int_{\frac 12}^1 |f(x)-a|\ \mathrm{dx} -$

$\int_0^b\left|f(x)-f\left(\frac 12\right)\right|\ \mathrm{dx}$ $-\int_b^{\frac 12}\left|f(x)-f\left(\frac 12\right)\right|\ \mathrm{dx}$ $-\int_{\frac 12}^1\left|f(x)-f\left(\frac 12\right)\right|\ \mathrm{dx} =$ $-b\left[f\left(\frac 12\right) - a\right] + \frac{1}{2}\cdot \left(\frac 12-a\right)$

$-\int_b^{\frac 12} \left|f(x)-f\left(\frac 12\right)\right|\ \mathrm{dx} +$ $\int_b^{\frac 12} |f(x)-a|\ \mathrm{dx} \ge -b\left[f\left(\frac 12\right) - a\right] +$ $\frac{1}{2}\cdot \left (\frac 12-a\right) -$ $\int_b^{\frac 12} \left|f(x)-f\left(\frac 12\right)\right|\ \mathrm{dx} \ge$

$-b\left[f\left(\frac 12\right) - a\right] + \frac 12\cdot \left(\frac 12-a\right)$ $-\int_b^{\frac 12} \left|a-f\left(\frac 12\right)\right|\ \mathrm{dx} = 0$ . If $a\le f(0)$ , then

$\int_0^1 |f(x)-a|\ \mathrm{dx} \ge$ $\int_0^1 |f(x)-f(0)|\ \mathrm{dx} \ge$ $\int_0^1 \left|f(x)-f\left(\frac 12\right)\right|\ \mathrm{dx}$ since above.

$\blacksquare\ 17^{\circ}.$ PP6. Find the value of $a$ for which $\int_{0}^{\frac{\pi}{2}}|a\sin x-\cos x|\ dx\ (a>0)$ is minimized.

Proof. Define $x\ .s.s.\ y\Longleftrightarrow xy>0\ \vee \ x=y=0\ .$ (same sign). If $a\le 0$ , then $F(a)=\int_{0}^{\pi}{2}(\cos x-a\sin x)\ dx=1-a$ and

$\boxed{\ \min_{a\le 0}F(a)=F(0)=1\ }$ . If $a>0$ , then $a\sin x-\cos x\ .s.s.\ \tan x-\frac{1}{a}=\tan x-\tan\left(\arctan\frac{1}{a}\right)\ .s.s.\ x-\arctan\frac{1}{a}$ and in this

case $F(a)=\int_{0}^\frac{\pi}{2}\left|a\sin x-\cos x\right|\ dx=$ $\int_{0}^{\arctan\frac{1}{a}}(\cos x-a\sin x)\ dx+$ $\int_{\arctan\frac{1}{a}}^\frac{\pi}{2}(a\sin x-\cos x)\ dx=$ $-1-a+2\sqrt{a^{2}+1}$ .

But $F'(a)=-1+\frac{2a}{\sqrt{a^{2}+1}}\ .s.s.\ 2a-\sqrt{a^{2}+1}\ .s.s.\ 4a^{2}-(a^{2}+1)=$ $3a^{2}-1\ .s.s.\ a-\frac{\sqrt 3}{3}$ , i.e. $a_{\mathrm{min}}=\frac{\sqrt3}{3}$ . Thus,

$\boxed{\ \min_{a>0}F(a)=F\left(\frac{\sqrt 3}{3}\right)=\sqrt 3-1\ }$ . In conclusion, $\min_{a\in R}F(a)=F\left(\frac{\sqrt 3}{3}\right)=\sqrt 3-1$ .

$\blacksquare\ 18^{\circ}.$ PP7. Find the value of $a\in R$ for which $F(a)\equiv\int_{0}^{1}|xe^{x}-a|\ dx$ is minimized.

Proof. Denote the functions $f: [0,1]\rightarrow R\ ,\ f(x)=xe^{x}$ and $g: [0,1]\rightarrow R\ ,\ g(x)=a$ . Then $F(a)=\int_{0}^{1}|f(x)-g(x)|\ dx=$ $[\Gamma_{f,g}]$ - the area

of the surface $\Gamma_{f,g}$ comprised between the graphs of the above functions. The function $f$ is strict increasing, convex and for any $x\in [0,1]$ , $0\le f(x)\le e\ .$

Observe that $P(x)=(x-1)e^{x}-ax\in \int \left(xe^{x}-a\right)\ dx$ and $P(0)=-1$ , $P(1)=-a\ .$ Distinguish the cases :

$1.\blacktriangleright\ a\le 0\Longrightarrow$ $g(x)\le f(x)\ ,\ 0\le x\le 1$ $\Longrightarrow$ $F(a)=P(1)-P(0)=1-a\ge 1\ .$

$2.\blacktriangleright\ a\ge e$ $\Longrightarrow$ $f(x)\le g(x)\ ,\ 0\le x\le 1$ $\Longrightarrow$ $F(a)=P(0)-P(1)=a-1\ge e-1\ .$

$3.\blacktriangleright\ a\in (0,e)\Longrightarrow$ Exists uniquely $c=c(a)\in (0,1)$ so that $f(c)=a$ , i.e. $ce^{c}=a\Longrightarrow$

$F(a)=P(0)-2P(c)+P(1)=-1-a-2\left[(c-1)e^{c}-ac\right]=2e^{c}+2ac-3a-1\ .$

Remark. It is well-known the relation $\boxed{\ (\forall )a\in R\ ,\ \int_{0}^{1}|h(x)-a|\ dx\ge \int_{0}^{1}\left|h(x)-h\left(\frac{1}{2}\right)\right|\ dx\ }$ , where

$h: [0,1]\rightarrow R$ is continue and monotonous. For $h(x)=xe^{x}$ obtain Kunny's problem. See here the my topic.

Proof. Suppose w.l.o.g. that $h$ is increasing. Then $(\forall ) x\in \left[0,\frac{1}{2}\right]$ , $\left|h\left(x+\frac 12\right)-h(x)\right|\le\left| h\left(x+\frac 12\right)-a\right|+|h(x)-a|$ $\Longrightarrow$

$\int_{0}^{\frac{1}{2}}\left[h\left(x+\frac{1}{2}\right)-h(x)\right]\ \mathrm{dx}\le$ $\int_{0}^{\frac{1}{2}}\left|h\left(x+\frac{1}{2}\right)-a\right|\ \mathrm{dx}+$ $\int_{0}^{\frac{1}{2}}|h(x)-a|\ \mathrm{dx}$ $\Longrightarrow$ $\int_{\frac{1}{2}}^{1}h(x)\ \mathrm{dx}-$ $\int_{0}^{\frac{1}{2}}h(x)\ \mathrm{dx}\le$

$\int_{\frac{1}{2}}^{1}|h(x)-a|\ \mathrm{dx}+$ $\int_{0}^{\frac{1}{2}}|h(x)-a|\ \mathrm{dx}$ $\Longrightarrow$ $\int_{0}^{1}|h(x)-a|\ \mathrm{dx}\ge$ $\int_{\frac{1}{2}}^{1}\left[h(x)-h\left(\frac{1}{2}\right)\right]\ \mathrm{dx}+$ $\int_{0}^{\frac{1}{2}}\left[h\left(\frac{1}{2}\right)-h(x)\right]\ \mathrm{dx}=$

$\int_{0}^{1}\left|h(x)-h\left(\frac{1}{2}\right)\right]\ \mathrm{dx}$ . In conclusion, $\int_{0}^{1}|h(x)-a|\ \mathrm{dx}\ge$ $\int_{0}^{1}\left|h(x)-h\left(\frac{1}{2}\right)\right|\ \mathrm{dx}$ . Answer : $\boxed{\ a=h\left(\frac{1}{2}\right)=\frac{\sqrt e}{2}\ }$ .

See here, here and XXII - message

This post has been edited 306 times. Last edited by Virgil Nicula, Jul 17, 2017, 7:24 AM

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August 2018

462. Diverse probleme de geometrie.

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460. Working page VI.

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459.Working page V.

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457. Relatii metrice si inegalitati geometrice.

456. Probleme de Algebra.

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456* Integrals.

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454. Mathematical note: "trig. ineg."- GMB 6/1992, page 201.

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453. Working page III.

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452. Working page II.

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451. ABC with C(O,R), C(I,r) and N-Nagel point ==> ON=R-2r.

450. Nota matematica: "Asupra unei inegalitati remarcabile".

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449. Working page I.

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448. Extensions or generalizations.

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447. Locuri geometrice remarcabile.

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446. Mathematics "instant" for the middle school.

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444. Remarkable points in a triangle.

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435. Geometry problems in space.

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433. Some properties of remarkable lines in a triangle II.

432. Geometry problems with perpendicularities II.

431. Trigonometry in geometry II.

430. Trigonometrical identities.

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429. Bretschneider's_formula

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428. Some metrical problems from the contests II.

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427. Selected nice or well-known geometry problems.

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426. Problems from and for baccalaureate.

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425. Geometrical extremity II.

424. Some nice problems of E. E. Q. Rodriguez and others.

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423. Geometry problems.

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419. Some geometry problems for hight school I

418. Some nice problems with polynomials/equations III

417 <bis> 418 (educatie: "14.27.28.32.57"/\"127.248").

417. The incircle of ABC and other two incircles.

416. Cateva solutii ale unei probleme de tip slicing.

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407. Cyclical quadrilaterals.

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400. Some nice geometry problems for the high school I.

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399. Algebra (I).

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396. Own inegalities stronger than the Panaitopol's.

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395. Metrical relations in a quadrilateral.

394. A cevian in a triangle and two incircles.

393. Some problems with circles.

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392. Art of Problem Solving (geometry).

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391. CRUX Mathematicorum.

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390. Probleme de analiza matematica.

389. Some geometric properties in a triangle I.

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388. Some interesting "slicing" problems.

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387. Algebra (II).

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386. Barycentrical coordinates.

385. Some nice geometry problems.

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382. Peruan "jewels" (Israel Diaz, M. O. Sanchez a.s.o.)

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373. Extension of Chebyshev's inequality.

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372. The distancies : OI , OI_a , ON a.s.o.

371. Some problems from NMO (final stage) - 2013, Romania.

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369. Mixtilinear circles of ABC.

364. Some interesting inequalities I

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362. Fermat's problem and other metrical problems.

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361. Arhimedes theorem of the broken chord in a circle.

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358. Some properties of the remarkable lines in a triangle.

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357. Morley theorem.

356. Order relation between s & (2R+r) in an acute triangle.

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355. Some problems with polynomials or equations I

354. Gradul II - Univ. B.B. Cluj, aug. 1998 Prof I sau II.

353. Some problems of the analytical geometry.

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352. Some definite integrals.

351. Problems with areas.

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349. Another interesting limits.

348. Subiectele de la BAC 2012 /Mate-Info.

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347. Two isogonal lines in a triangle.

346. Some easy integrals.

345. Characteristic function of a set.

344. Some trigonometry problems.

343. Problems with collinear points and concurrent lines.

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342. Exponential or logarithmic equations/inequations.

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341. Some nice and interesting "slicing" problems.

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339. Derivatives. Rolle's function.

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336. Some problems of the analytical geometry.

335. Some difficult geometric/trigonometric inequalities.

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333. Problems with the common tangent for two circles.

332. Some metrical problems from the contests I.

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331. Some nice problems with polynomials/equations I.

November 2011

330. A triangle and an its transversal through a point.

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328. A general identities with areas in a triangle ABC.

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327. Some geometrical/algebraic extremum problems.

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325. Without the l'Hospital's rules.

324. Some determinants.

323. Easy, nice and interesting problems for high school.

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321. Harmonical quadrilateral.

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319. Some problems with induction.

318. IRAN - 2011 (geometrie).

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316. Some properties of isogonal rays in an angle of ABC.

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315. Indian Team Selection Test 2010 ST1 P1 and extension.

314. USAMTS 2009 Round 2 #5.

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312. Very nice (famous trigonometrical problems) !

311. Some problems from USAMO and other contests.

310. Some properties of the tangential quadrilateral.

309. Beautiful problem ! IRAN Selection Test for IMO, 2004.

308. Some nice and easy properties in a triangle.

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307. Very nice proofs !

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305. Position of the roots w.r.t. a given real number.

304. Integrals ("brothers") III.

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301. Some trigonometrico-geometrical inequalities.

300. Lagrange's multiplier. Examples and problems.

299. A class of the recurrent sequencies.

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296. Some nice Vittasko's problems.

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293. A difficult trigonometric equation (third point !).

292. Length of the angled bisector and some means.

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289. Nice identities in an algebra/ geometry/trigonometry.

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281. Identity with R in an acute triangle (ILL - 1974).

280. Common roots of two polynomial equations.

279. O.M. Holland , O.M. Ireland 1988 and TST USA 2000, etc.

278. In memoria profesorului Alexandru Lupas.

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276. Chinese TST 2009 and Second Round 2006.

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274. An inequality in a triangle for its interior point.

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272. Outside of triangle. Extension of Fermat's point.

271. Relations concerning Fermat points.

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270. Morrocan M.O. 2010 and Croatia TST 2009.

269. Some nice and easy properties in a trapezoid.

268. Saraghin contest, 2011.

267. Germany 2002, Russia 2003, IMO 2003, Aus.-Polish 1992.

266. Rectangles erected on sides of a triangle. Concurrence.

265. Really nice problem !

264. Identities with complex numbers.

263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

261. IMO 2009, problem 2 and JBMO 2007, problems 1 & 2.

260. Some nice, simple identities and their applications.

259. ABC equilateral triangle ==> DEF equilateral triangle.

258. Some difficult and nice problems with complex numbers.

257. Seeking roots of 2th equation with complex coef.

256. An extremum problems with complex numbers.

March 2011

255. A line which is parallelly with OH.

254. An interesting C. Mateescu's geometrical inequality.

253. An old algebraic inequality.

252. OI=f(A,R) in certain conditions.

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

247. An remarkable inequality with median.

246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

February 2011

232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

222. An extension of probl. 2 from IMO 2009.

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

January 2011

219. A very nice geometrical inequality.

218. Some problems from the Suiss IMO Selection Team 2006.

217. A nice and interesting Murray S. Klamkin's problem.

216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

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202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

November 2010

179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

October 2010

170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

120. Two equal neighbouring angles/sides in a quadrilateral.

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

July 2010

72. Colinearity in a triangle - H , P , M .

70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

64. Problems of the type "instant" (at most one minute).

63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

45. \cos B+\cos C=1 <==> nice property.

44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

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