304. Integrals ("brothers") III.

by Virgil Nicula, Jul 27, 2011, 9:43 PM

PP1. Ascertain the integrals $L\equiv\int\frac{1}{t^4-t^2+1}\ \mathrm{dt}\ \ ,\ \ J_k=\int\frac {x^k}{1+x^4}\ \mathrm{dx}$ and $I_k=\int\frac {x^k}{(1+x)(1+x^4)}\ \mathrm{dx}$ , where $k\in\overline{0,4}$ .

Proof. $L=\frac 12\cdot\left(L_1-L_2\right)$ , where $L_1\equiv\int\frac{t^2+1}{t^4-t^2+1}\ \mathrm{dt}=$ $\int\frac{1+\frac {1}{t^2}}{t^2+\frac {1}{t^2}-1}\ \mathrm{dt}=$ $\int\frac{\left(t-\frac 1t\right)'}{\left(t-\frac 1t\right)^2+1}\ \mathrm{dt}\implies$ $\boxed{L_1=\arctan\frac {t^2-1}{t}+\mathbb C}$ and $L_2\equiv\int\frac{t^2-1}{t^4-t^2+1}\ \mathrm{dt}=$

$\int\frac{1-\frac {1}{t^2}}{t^2+\frac {1}{t^2}-1}\ \mathrm{dt}=$ $\int\frac{\left(t+\frac 1t\right)'}{\left(t+\frac 1t\right)^2-3}\ \mathrm{dt}\implies$ $\boxed{L_2=\frac {1}{2\sqrt 3}\cdot\ln\left|\frac {t^2-t\sqrt 3+1}{t^2+t\sqrt 3+1}\right|+\mathbb C}$ . Observe that $\left\{\begin{array}{ccc} J_1 & = & \int\frac {x}{1+x^4}\ \mathrm{dx}=\frac 12\cdot\int\frac {\left(x^2\right)'}{1+\left(x^2\right)^2}\ \mathrm{dx}=\frac 12\cdot\arctan x^2+\mathbb C\\\\ J_3 & = & \int\frac {x^3}{1+x^4}\ \mathrm{dx}=\frac 14\cdot\int\frac {\left(1+x^4\right)'}{1+x^4}\ \mathrm{dx}=\frac 14\cdot\ln\left(1+x^4\right)+\mathbb C\end{array}\right\|$ ,

$\left\{\begin{array}{ccc} J_2+J_0 & = & \int\frac {x^2+1}{x^4+1}\ \mathrm{dx}=\int\frac {1+\frac {1}{x^2}}{x^2+\frac {1}{x^2}}\ \mathrm{dx}=\int\frac {\left(x-\frac 1x\right)'}{\left(x-\frac 1x\right)^2+2}\ \mathrm{dx}=\frac {\sqrt 2}{2}\cdot\arctan\frac {x^2-1}{x\sqrt 2}+\mathbb C\\\\ J_2-J_0 & = & \int\frac {x^2-1}{x^4+1}\ \mathrm{dx}=\int\frac {1-\frac {1}{x^2}}{x^2+\frac {1}{x^2}}\ \mathrm{dx}=\int\frac {\left(x+\frac 1x\right)'}{\left(x+\frac 1x\right)^2-2}\ \mathrm{dx}=\frac {\sqrt 2}{4}\cdot\ln\left|\frac {x^2-x\sqrt 2+1}{x^2+x\sqrt 2+1}\right|+\mathbb C\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} J_0 & = & \frac {\sqrt 2}{4}\cdot\arctan\frac {x^2-1}{x\sqrt 2}-\frac {\sqrt 2}{8}\cdot\ln\left|\frac {x^2-x\sqrt 2+1}{x^2+x\sqrt 2+1}\right|+\mathbb C\\\\ J_2 & = & \frac {\sqrt 2}{4}\cdot\arctan\frac {x^2-1}{x\sqrt 2}+\frac {\sqrt 2}{8}\cdot\ln\left|\frac {x^2-x\sqrt 2+1}{x^2+x\sqrt 2+1}\right|+\mathbb C\end{array}\right\|$ .

Denote $K=\int\frac {1}{1+x}\ \mathrm{dx}=\ln\left|1+x\right|+\mathbb C$ . Thus, the system $(*)\left\{\begin{array}{cc} (1) & I_0+I_1=J_0\\\\ (2) & I_0+I_4=K\\\\ (3) & I_1+I_2=J_1\\\\ (4) & I_2+I_3=J_2\\\\ (5) & I_3+I_4=J_3\end{array}\right\|$ has extended matrix $\overline A=(A|B)$ , where $\overline A=\left(\begin{array}{ccccc} 1 & 1 & 0 & 0 & 0\\\\ 1 & 0 & 0 & 0 & 1\\\\ 0 & 1 & 1 & 0 & 0\\\\ 0 & 0 & 1 & 1 & 0\\\\ 0 & 0 & 0 & 1 & 1\end{array}\right|\left|\begin{array}{c} J_0\\\\ K\\\\ J_1\\\\ J_2\\\\ J_3\end{array}\right)$

and its determinant $\Delta=\det A$ is $\Delta =\left|\begin{array}{ccccc} 1 & 1 & 0 & 0 & 0\\\\ 1 & 0 & 0 & 0 & 1\\\\ 0 & 1 & 1 & 0 & 0\\\\ 0 & 0 & 1 & 1 & 0\\\\ 0 & 0 & 0 & 1 & 1\end{array}\right|\stackrel{(C_2:=C_2-C_1)}{\ \ =\ \ }$ $\left|\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0\\\\ 1 & -1 & 0 & 0 & 1\\\\ 0 & 1 & 1 & 0 & 0\\\\ 0 & 0 & 1 & 1 & 0\\\\ 0 & 0 & 0 & 1 & 1\end{array}\right|=$ $\left|\begin{array}{cccc} -1 & 0 & 0 & 1\\\\ 1 & 1 & 0 & 0\\\\ 0 & 1 & 1 & 0\\\\ 0 & 0 & 1 & 1\end{array}\right|\stackrel{(C_4:=C_4+C_1)}{\ \ =\ \ }$ $\left|\begin{array}{cccc} -1 & 0 & 0 & 0\\\\ 1 & 1 & 0 & 1\\\\ 0 & 1 & 1 & 0\\\\ 0 & 0 & 1 & 1\end{array}\right|=$

$-\left|\begin{array}{ccc} 1 & 0 & 1\\\\ 1 & 1 & 0\\\\ 0 & 1 & 1\end{array}\right|\implies$ $\boxed{\ \Delta=-2\ne 0\ }$ . Thus, the equation $A\cdot X=B$ has the solution $X=A^{-1}\cdot B$ , i.e. $\left(\begin{array}{c} I_0\\\\ I_1\\\\ I_2\\\\ I_3\\\\ I_4\end{array}\right)=\frac 12\cdot \left(\begin{array}{ccccc} 1 & 1 & -1 & 1 & -1\\\\ 1 & -1 & 1 & -1 & 1\\\\ -1 & 1 & 1 & 1 & -1\\\\ 1 & -1 & -1 & 1 & 1\\\\ -1 & 1 & 1 & -1 & 1\end{array}\right)\cdot \left(\begin{array}{c} J_0\\\\ K\\\\ J_1\\\\ J_2\\\\ J_3\end{array}\right)$ .

Example, $I_2=\int\frac {x^2}{(1+x)(1+x^4)}\ \mathrm{dx}=$ $\frac 12\cdot \left(-J_0+K+J_1+J_2-J_3\right)=$ $\frac 12\cdot \left[K+J_1+\left(J_2-J_0\right)-J_3\right]$ $\implies$

$2\cdot I_2=2\cdot \int\frac {x^2}{(1+x)(1+x^4)}\ \mathrm{dx}=$ $\frac {\sqrt 2}{4}\cdot\ln\left|\frac {x^2-x\sqrt 2+1}{x^2+x\sqrt 2+1}\right|+\frac 14\ln\frac {(1+x)^4}{1+x^4}+\frac 12\cdot\arctan x^2+\mathbb C$ . Indeed,

$2x^2=\left(-1+x+x^2-x^3\right)(1+x)+\left(1+x^4\right)$ $\iff$ $\frac {2x^2}{(1+x)(1+x^4)}=\frac {-1+x+x^2-x^3}{1+x^4}+\frac {1}{1+x}$ .

Remark. Can solve system $(*)$ . $I_2\stackrel{(3)}{=}J_1-I_1\stackrel{(1)}{=}J_1-\left(J_0-I_0\right)\stackrel{(2)}{=}$ $J_1-J_0+\left(K-I_4\right)\stackrel{(5)}{=}$ $J_1-J_0+K-\left(J_3-I_3\right)\stackrel{(4)}{=}$ .0

$J_1-J_0+K-J_3+\left(J_2-I_2\right)\implies$ $\boxed{I_2=\frac 12\cdot \left[K+J_1+\left(J_2-J_0\right)-J_3\right]}$ .

PP2. Ascertain the integrals $\int\frac {x^4}{\left(x^4+ax^2+1\right)\left(x^4+bx^2+1\right)}\ \mathrm{dx}$ , where $a\ne b$ and $\{a,b\}\subset (2,\infty )$ .

Proof. Denote $A=\int\frac {x^4}{\left(x^4+ax^2+1\right)\left(x^4+bx^2+1\right)}\ \mathrm{dx}$ and $B=\int\frac {x^2}{\left(x^4+ax^2+1\right)\left(x^4+bx^2+1\right)}\ \mathrm{dx}$ . Observe that :

$\blacktriangleright\ A+B=\int\frac {\left(x-\frac 1x\right)'}{\left[\left(x-\frac 1x\right)^2+a+2\right]\left[\left(x-\frac 1x\right)^2+b+2\right]}\ \mathrm{dx}=$ $E\left(x-\frac 1x\right)$ , where $E\stackrel{(y=x-\frac 1x)}{\ =\ }\int\frac {1}{\left[y^2+(a+2)\right]\left[y^2+(b+2)\right]}\ \mathrm{dy}=$

$\frac {1}{a-b}\cdot\int \left[\frac {1}{y^2+(b+2)}-\frac {1}{y^2+(a+2)}\right]\ \mathrm {dy}=$ $\frac {1}{a-b}\cdot \left(\frac {1}{\sqrt {b+2}}\cdot\arctan\frac {y}{\sqrt {b+2}}-\frac {1}{\sqrt {a+2}}\cdot\arctan\frac {y}{\sqrt {a+2}}\right)+\mathbb C$ .

Therefore, $A+B=\frac {1}{a-b}\cdot$ $\left(\frac {1}{\sqrt {b+2}}\cdot\arctan\frac {x^2-1}{x\sqrt {b+2}}-\frac {1}{\sqrt {a+2}}\cdot\arctan\frac {x^2-1}{x\sqrt {a+2}}\right)$ $+\mathbb C$ .

$\blacktriangleright\ A-B=\int\frac {\left(x+\frac 1x\right)'}{\left[\left(x+\frac 1x\right)^2+a-2\right]\left[\left(x+\frac 1x\right)^2+b-2\right]}\ \mathrm{dx}=$ $F\left(x+\frac 1x\right)$ , where $F\stackrel{(y=x+\frac 1x)}{\ =\ }\int\frac {1}{\left[y^2+(a-2)\right]\left[y^2+(b-2)\right]}\ \mathrm{dy}=$

$\frac {1}{a-b}\cdot\int \left[\frac {1}{y^2+(b-2)}-\frac {1}{y^2+(a-2)}\right]\ \mathrm {dy}=$ $\frac {1}{a-b}\cdot \left(\frac {1}{\sqrt {b-2}}\cdot\arctan\frac {y}{\sqrt {b-2}}-\frac {1}{\sqrt {a-2}}\cdot\arctan\frac {y}{\sqrt {a-2}}\right)+\mathbb C$ .

Therefore, $A-B=\frac {1}{a-b}\cdot$ $\left(\frac {1}{\sqrt {b-2}}\cdot\arctan\frac {x^2+1}{x\sqrt {b-2}}-\frac {1}{\sqrt {a-2}}\cdot\arctan\frac {x^2+1}{x\sqrt {a-2}}\right)$ $+\mathbb C$ . In conclusion,

$\blacktriangleright\ 2A=$ $\frac {1}{a-b}$ $\left(\frac {1}{\sqrt {b+2}}\arctan\frac {x^2-1}{x\sqrt {b+2}}-\frac {1}{\sqrt {a+2}}\arctan\frac {x^2-1}{x\sqrt {a+2}}\right)+\frac {1}{a-b}$ $\left(\frac {1}{\sqrt {b-2}}\arctan\frac {x^2+1}{x\sqrt {b-2}}-\frac {1}{\sqrt {a-2}}\arctan\frac {x^2+1}{x\sqrt {a-2}}\right)$ $+\mathbb C$ .

$\blacktriangleright\ 2B=$ $\frac {1}{a-b}$ $\left(\frac {1}{\sqrt {b+2}}\arctan\frac {x^2-1}{x\sqrt {b+2}}-\frac {1}{\sqrt {a+2}}\arctan\frac {x^2-1}{x\sqrt {a+2}}\right)-\frac {1}{a-b}$ $\left(\frac {1}{\sqrt {b-2}}\arctan\frac {x^2+1}{x\sqrt {b-2}}-\frac {1}{\sqrt {a-2}}\arctan\frac {x^2+1}{x\sqrt {a-2}}\right)$ $+\mathbb C$ .

Remark. The following decomposition will immediately lead to the answer:

$\frac{x^4}{(x^4 + ax^2 + 1)(x^4 + bx^2 + 1)} =$ $\frac{1}{2(b-a)}\cdot \left[\frac{\left(x-\frac 1x\right)'}{\left( x - \frac 1x \right)^{2} + a + 2} + \frac{\left(x+\frac 1x\right)'}{\left( x+\frac 1x \right)^{2} + a - 2}- \frac{\left(x-\frac 1x\right)'}{\left( x - \frac 1x \right)^{2} + b + 2}-\frac{\left(x+\frac 1x\right)'}{\left( x+\frac 1x \right)^{2} + b - 2}\right]$ . Indeed,

$A=\frac{x^{4}}{(x^{4}+ax^{2}+1)(x^{4}+bx^{2}+1)}=$ $\frac{x^2}{b-a}\cdot \left(\frac{1}{x^4+ax^2+1} - \frac{1}{x^4+bx^2+1} \right)=$ $\frac{1}{2 \left( b-a \right)} \cdot\left[\frac{ \left( x^2+1 \right) + \left( x^2-1 \right) }{x^4+ax^2+1} - \frac{ \left( x^2+1 \right) + \left( x^2-1 \right) }{x^4+bx^2+1} \right]=$

$\frac{1}{2 \left( b-a \right)}\cdot \left(\frac{x^2+1}{x^4+ax^2+1} + \frac{x^2-1}{x^4+ax^2+1} - \frac{x^2+1}{x^4+bx^2+1} - \frac{x^2-1}{x^4+bx^2+1} \right)=$

$\frac{1}{2 \left( b-a \right)}\cdot \left[\frac{x^2+1}{(x^2-1)^2+(a+2)x^2} + \frac{x^2-1}{(x^2+1)^2+(a-2)x^2} - \frac{x^2+1}{(x^2-1)^2+(b+2)x^2} - \frac{x^2-1}{(x^2+1)^2+(b-2)x^2}\right]=$

$\frac{1}{2 \left( b-a \right)} \cdot\left[\frac{1+\frac{1}{x^2}}{\left( x-\frac{1}{x}\right)^{2}+a+2} + \frac{1-\frac{1}{x^2}}{\left( x+\frac{1}{x}\right)^{2}+a-2} - \frac{1+\frac{1}{x^2}}{\left( x-\frac{1}{x}\right)^{2}+b+2} - \frac{1-\frac{1}{x^2}}{\left( x+\frac{1}{x}\right)^{2}+b-2} \right]$ . Then, $\displaystyle{ \int A \ \mathrm{dx} }$ is reduced to a sum of $\displaystyle{ \int \frac{1}{u^2+1} \ \mathrm{du} = \arctan u +\mathbb C}$ .

PP3. Ascertain the integrals $L_k=\int\frac {x^k}{1+x^6}\ \mathrm{dx}$ , where $k\in\overline{0,5}$ .

Proof.

$\blacktriangleleft\blacktriangleright\ L_2=\int\frac {x^2}{1+x^6}\ \mathrm{dx}=$ $\frac 13\cdot\int\frac {\left(x^3\right)'}{1+x^6}\ \mathrm{dx}\implies$ $\boxed{L_2=\frac 13\cdot\arctan x^3+\mathbb C}\ (1)\ .$

$\blacktriangleleft$ $\blacktriangleright\ L_0+2L_2+L_4=$ $\int\frac {1+2x^2+x^4}{1+x^6}\ \mathrm{dx}=$ $\int\frac {\left(1+x^2\right)^2}{1+x^6}\ \mathrm{dx}=$ $\int\frac {1+x^2}{1-x^2+x^4}\ \mathrm{dx}=$

$\int\frac {1+\frac {1}{x^2}}{\frac {1}{x^2}-1+x^2}\ \mathrm{dx}=$ $\int\frac {\left(x-\frac {1}{x}\right)'}{\left(x-\frac 1x\right)^2+1}\ \mathrm{dx}\implies$ $\boxed{L_0+2L_2+L_4=\arctan\frac {x^2-1}{x}+\mathbb C}\ (2)\ .$

$\blacktriangleleft$ $\blacktriangleright\ L_4-L_0=$ $\int\frac {x^4-1}{1+x^6}\ \mathrm{dx}=$ $\int\frac {x^2-1}{1-x^2+x^4}\ \mathrm{dx}=$ $\int\frac {1-\frac {1}{x^2}}{\frac {1}{x^2}-1+x^2}\ \mathrm{dx}=$ $\int\frac {\left(x+\frac {1}{x}\right)'}{\left(x+\frac 1x\right)^2-3}\ \mathrm{dx}\implies$

$\boxed{L_4-L_0=\frac {1}{2\sqrt 3}\cdot\ln\left|\frac {x^2-x\sqrt 3+1}{x^2+x\sqrt 3+1}\right|+\mathbb C}\ (3)$ . From $(1)\ ,\ (2)\ ,\ (3)$ obtain that $L_0\ ,\ L_2\ ,\ L_4$ , i.e. $L_2\odot\begin{array}{c} \nearrow\\\\ \searrow\end{array}$ $\begin{array}{cc}L_4+L_0 & \searrow\\\\ L_4-L_0 & \nearrow\end{array}\odot\implies$ $L_0\ \wedge\ L_4\ .$

$\blacktriangleleft$ $\blacktriangleright\ L_3+L_1=\int\frac {x\left(x^2+1\right)}{1+x^6}\ \mathrm{dx}=$ $\int\frac {x}{1-x^2+x^4}\ \mathrm{dx}=$ $\frac 12\cdot\int\frac {\left(x^2\right)'}{1-\left(x^2\right)+\left(x^2\right)^2}\ \mathrm{dx}=$ $G\left(x^2\right)+\mathbb C$ , where

$G=\int\frac {1}{t^2-t+1}\ \mathrm{dt}=$ $\int\frac {\left(t-\frac 12\right)'}{\left(t-\frac 12\right)^2+\frac 34}\ \mathrm{dt}=$ $\frac {2}{\sqrt 3}\arctan\frac {2t-1}{\sqrt 3}+\mathbb C$ . In conclusion, $\boxed{L_3+L_1=\frac {2}{\sqrt 3}\arctan\frac {2x^2-1}{\sqrt 3}+\mathbb C}\ (4)\ .$

$\blacktriangleleft$ $\blacktriangleright\ L_3-L_1=$ $\int\frac {x^3-x}{\left(1+x^2\right)\left(1-x^2+x^4\right)}\ \mathrm{dx}=$ $\int\frac {1-\frac {1}{x^2}}{\left(x+\frac 1x\right)\left(\frac {1}{x^2}-1+x^2\right)}\ \mathrm{dx}=$ $\int\frac {\left(x+\frac 1x\right)'}{\left(x+\frac 1x\right)\left[\left(x+\frac {1}{x}\right)^2-3\right]}\ \mathrm{dx}=$ $F\left(x+\frac 1x\right)+\mathbb C$ ,

where $F=\int\frac {1}{t\left(t^2-3\right)}\ \mathrm{dt}=$ $\frac 12\cdot \int\frac {\left(t^2\right)'}{t^2\left(t^2-3\right)}\ \mathrm{dt}=$ $H\left(t^2\right)+\mathbb C$ , where $H(y)=\int\frac {1}{y(y-3)}\ \mathrm{dy}=$ $\frac 13\cdot \int\frac {y-(y-3)}{y(y-3)}\ \mathrm{dy}=$ $\frac 13\cdot \int\left(\frac {1}{y-3}-\frac 1y\right)\ \mathrm{dy}\implies$

$H=\frac 13\cdot\ln\left|\frac {y-3}{y}\right|+\mathbb C$ . In conclusion, $F(t)=\frac 13\cdot\ln\left|\frac{t^2-3}{t^2}\right|+\mathbb C$ and $\boxed {L_3-L_1=\frac 13\cdot\ln\frac {x^4-x^2+1}{\left(x^2+1\right)^2}+\mathbb C}\ (5)$ . From $(4)\ ,\ (5)$ obtain that $L_1\ ,\ L_3\ .$

$\blacktriangleleft\blacktriangleright\ L_5=$ $\int\frac {x^5}{1+x^6}\ \mathrm{dx}=$ $\frac 16\cdot \int\frac {\left(1+x^6\right)'}{1+x^6}\ \mathrm{dx}\implies$ $\boxed{L_5=\frac 16\cdot\ln\left(1+x^6\right)+\mathbb C}$ .

PP4. Ascertain the integrals $\int\frac{1-2e^x\sin x}{\left(e^x+\cos x\right)\left(e^x-\sin x\right)}\ \mathrm{dx}$ and $\int\frac{x^{2}}{(x \sin x + \cos x)(x\cos x - \sin x)}\ \mathrm {dx}$ .

Proof.

$\blacktriangleright\ \int\frac{1-2e^x\sin x}{\left(e^x+\cos x\right)\left(e^x-\sin x\right)}\ \mathrm{dx}=$ $\int\frac {\left(e^x-\sin x\right)^2-\left(e^x-\cos x\right)\left(e^x+\cos x\right)}{\left(e^x+\cos x\right)\left(e^x-\sin x\right)}\ \text{d}x=$

$\int\left(\frac {\text{e}^x-\sin x}{\text{e}^x+\cos x}-\frac {\text{e}^x-\cos x}{\text{e}^x-\sin x}\right)\ \text{d}x=$ $\int\left[\frac {\left(\text{e}^x+\cos x\right)^{\prime}}{\text{e}^x+\cos x}-\frac {\left(\text{e}^x-\sin x\right)^{\prime}}{\text{e}^x-\sin x}\right]\ \text{d}x=\ln\left|\frac {\text{e}^x+\cos x}{\text{e}^x-\sin x}\right|+\mathbb C$ .

$\blacktriangleright\ \int\frac{x^{2}}{(x \sin x + \cos x)(x\cos x - \sin x)}\ \mathrm{dx}=$ $\int\frac{x^{2}\cos ^{2} x - x\sin x\cos x+ x^{2}\sin^{2} x + x\sin x\cos x}{(x\sin x + \cos x)(x\cos x-\sin x)}\ \mathrm{dx}=$

$\int\left(\frac{x\cos x}{x\sin x + \cos x} + \frac{x \sin x}{x\cos x - \sin x}\right)\ \mathrm{dx}=$ $\int\left(\frac{x\cos x}{x\sin x+ \cos x} - \frac{-x\sin x}{x\cos x-\sin x}\right)\ \mathrm{dx}=$ $\ln\left|\frac {x\sin x +\cos x}{\cos x-\sin x}\right| + \mathbb C$

PP5. Evaluate the definite integral $\int_0^{\frac {\pi}{2}}\frac {\sin^2x}{a^2\sin^2x+b^2\cos^2x}\ \mathrm{dx}$ .

Proof. Let $I=\int_0^{\frac {\pi}{2}}\frac {\sin^2x}{a^2\sin^2x+b^2\cos^2x}\ \mathrm{dx}$ and $J=\int_0^{\frac {\pi}{2}}\frac {\cos^2x}{a^2\sin^2x+b^2\cos^2x}\ \mathrm{dx}$ , where $a>0$ , $b>0$ and $a\ne b$ . Thus, $I+J=\int_0^{\frac {\pi}{2}}\frac {1}{a^2\sin^2x+b^2\cos^2x}\ \mathrm{dx}=$

$\int_0^{\frac {\pi}{2}}\frac {1}{a^2\tan^2x+b^2}\cdot\left(\tan x\right)'\ \mathrm{dx}=$ $\int_0^{\infty}\frac {1}{a^2t^2+b^2}\ \mathrm{dt}=$ $\frac {1}{a^2}\cdot \int_0^{\infty}\frac {1}{t^2+\left(\frac ba\right)^2}\ \mathrm{dt}=$ $\frac {1}{ab}\cdot \left|\arctan\frac {at}{b}\right|^{\infty}_0\implies$ $\left\{\begin{array}{c} I+J=\frac {\pi}{2ab}\\\\ a^2I+b^2J=\frac {\pi}{2}\end{array}\right\|\ \implies\ \boxed{\ aI=bJ=\frac {\pi}{2(a+b)}\ }$ .

PP6. Let $a_n=\int_{1}^{e} \ln^{n} x\ \mathrm{dx}$ , where $n\in\mathbb N$ . Evaluate $a_n$ and show that $a_n\searrow 0\ ,\ \frac {a_n}{n!}\rightarrow 0$ .

Proof. Let $n\in\mathbb N$ . I"ll use the partial integration $\left\{\begin{array}{ccc} u(x)=\ln^{n+1}x & \implies & u'(x)=(n+1)\cdot\frac {\ln ^nx}{x}\\\\ v'(x)=1 & \implies & v(x)=x\end{array}\right\|$ . Therefore, $a_{n+1}=\int_{1}^{e} \ln ^{n+1}x\ \mathrm{dx}=$

$\left| x\cdot\ln^{n+1}x\right|_1^e-(n+1)\cdot \int_1^e\ln^nx\ \mathrm{dx}\implies$ $\boxed{\begin{array}{c} a_0=e-1\\\\ a_{n+1}=e-(n+1)a_n\ ,\ n\in\mathbb N\end{array}}\ (*)$ . Observe that $a_{n+1}-a_n=\int_1^e\ln^nx\left(\ln x-1\right)\ \mathrm{dx}<0$

$\implies$ the sequence $a_n$ is (strict) decreasing, in particulat has limit, i.e. $a_n\searrow l\in \left[0,a_0\right)$ . Suppose $l>0$ . Then for $n\rightarrow \infty$ in the recurrence relation $(*)$

obtain that $l=-\infty$ , what is absurd. In conclusion, $\boxed{a_n\searrow 0}$ . Denote $\boxed{b_n=\frac {a_n}{n!}}\ ,\ n\in\mathbb N$ . Thus, $b_{n+1}=\frac {e}{(n+1)!}-b_n$ , where $b_0=e-1$ .

Prove easily that $\boxed{b_n=e\cdot\sum_{k=0}^n\frac {(-1)^k}{k!}-1}$ . Since $\sum_{k=0}^n\frac {(-1)^k}{k!}\rightarrow\frac 1e$ obtain that $b_n\rightarrow 0$ . Hence $\boxed{a_n=n!\cdot \left[e\cdot\sum_{k=0}^n\frac {(-1)^k}{k!}-1\right]}$ .

PP7. Prove that $K\equiv \int_{\frac 12}^2\frac {\ln x}{x^2+1}\ \mathrm{dx}=0$ and ascertain $L\equiv \int_0^{\frac {\pi}{4}}\sqrt{\tan x}\ \mathrm{dx}$ .

Proof. $K=\int_{\frac 12}^2\frac {\ln x}{x^2+1}\ \mathrm{dx}=$ $\int_{\frac 12}^1\frac {\ln x}{x^2+1}\ \mathrm{dx}+\int_1^2\frac {\ln x}{x^2+1}\ \mathrm{dx}=$ $\int_{\frac 12}^1\frac {\ln x}{x^2+1}\ \mathrm{dx}+$ $\int_1^{\frac 12}\frac {-\ln t}{\frac 1{t^2}+1}\left(-\frac 1{t^2}\right)\ \mathrm{dt}=$

$\int_{\frac 12}^1\frac {\ln x}{x^2+1}\ \mathrm{dx}+$ $\int_1^{\frac 12}\frac {\ln t}{t^2+1}\ \mathrm{dt}=$ $\int_{\frac 12}^1\frac {\ln x}{x^2+1}\ \mathrm{dx}-\int_{\frac 12}^1\frac {\ln x}{x^2+1}\ \mathrm{dx}=0$ . $L=\int_0^{\frac {\pi}{4}}\sqrt{\tan x}\ \mathrm{dx}\ \stackrel{(x:=\sqrt{\tan x})}{=}\ 2J$ ,

where $J=\int_0^1\frac {x^2}{1+x^4}\ \mathrm{dx}=$ $\left|\left(\frac {\sqrt 2}{4}\cdot\arctan\frac {x^2-1}{x\sqrt 2}+\frac {\sqrt 2}{8}\cdot\ln\left|\frac {x^2-x\sqrt 2+1}{x^2+x\sqrt 2+1}\right|\right)\right|_0^1\implies$ $\boxed{L=\frac {\sqrt 2}{2}\cdot\left[\frac {\pi}{2}+\ln\left(\sqrt 2-1\right)\right]}$ .

PP8. Ascertain an indefinite integral of the function $f:\mathbb R\rightarrow \mathbb R$ , where $f(x)=\frac {xe^{\arctan x}}{\left(x^2+1\right)^n}$ and $n\in\mathbb N\ ,\ n\ge 2$ .

Proof. $\left\{\begin{array}{ccc} u(x)=e^{\arctan x} & \implies & u'(x)=\frac {e^{\arctan x}}{x^2+1}\\\\ v'(x)=\frac x{\left(x^2+1\right)^n} & \implies & v(x)=-\frac 1{2(n-1)\left(x^2+1\right)^{n-1}}\end{array}\right\|$ $\implies$ $I_n=\int\frac {xe^{\arctan x}}{\left(x^2+1\right)^n}\ \mathrm{dx}=$ $\frac {-e^{\arctan x}}{2(n-1)\left(x^2+1\right)^{n-1}}+$ $\frac 1{2(n-1)}\cdot J_n$ , where $J_n=\int\frac {e^{\arctan x}}{\left(x^2+1\right)^n}\ \mathrm{dx}=$

$F_{n-1}(\arctan x)+\mathcal C$ , where $F_n=\int e^t\cos ^{2n}t\ \mathrm{dt}=$ $\int e^t\cos ^{2(n-1)}t\left(1-\sin^2t\right)\ \mathrm{dt}=$ $F_{n-1}-K$ , where $K=\int e^t\cos^{2(n-1)}t\sin^2t\ \mathrm{dt}$ , Thus,

$\left\{\begin{array}{ccc} u(x)=e^t\sin t & \implies & u'(x)=e^t(\sin t+\cos t)\\\\ v'(x)=\sin t\cos^{2(n-1)}t & \implies & v(x)=-\frac 1{2n-1}\cdot\cos ^{2n-1}t\end{array}\right\|$ $\implies$ $K=-\frac 1{2n-1}\cdot e^t\sin t\cos^{2n-1}t+\frac 1{2n-1}\cdot\left(F_n+L\right)$ , where $L=\int e^t\sin t\cos^{2n-1}\ \mathrm{dt}$ . From

$\left\{\begin{array}{ccc} u(x)=e^t & \implies & u'(x)=e^t\\\\ v'(x)=\sin t\cos^{2n-1}t & \implies & v(x)=-\frac 1{2n}\cdot\cos ^{2n}t\end{array}\right\|$ obtain that $L=\frac 1{2n}\cdot \left(F_n-e^t\cos^{2n}t\right)$ . In conclusion,

$F_n=F_{n-1}-\left\{-\frac 1{2n-1}\cdot e^t\sin t\cos^{2n-1}t+\frac 1{2n-1}\cdot\left[F_n+\frac 1{2n}\cdot \left(F_n-e^t\cos^{2n}t\right)\right]\right\}$ $\implies$ $\boxed{\left(4n^2+1\right)F_n=2n(2n-1)F_{n-1}+2ne^t\sin t\cos^{2n-1}t+e^t\cos^{2n}t}$ . See here

This post has been edited 142 times. Last edited by Virgil Nicula, Nov 20, 2015, 9:26 PM

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

251. Problems of Analytical Geometry.

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245. An inequality in an acute triangle.

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

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240. Some interesting geometrical inequalities.

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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.

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232. Some geometrical inequalities (own).

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231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

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

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227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

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

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219. A very nice geometrical inequality.

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199. Some properties of symmedian. Two extensions.

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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).

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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).

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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

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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.

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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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by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

304. Integrals ("brothers") III.

by Virgil Nicula, Jul 27, 2011, 9:43 PM

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