integral, facil - Foro fmat.cl

2 Páginas:

< 1 2

integral, facil, solo para novatos (:

master_c	Aug 22 2011, 08:56 PM Publicado: #11
Invitado	CITA(danielomalmsteen @ Aug 22 2011, 07:55 PM) $TEX: $$\int\limits_{0}^{1}{\frac{\arctan xdx}{x+1}}=\left. \ln \left( x+1 \right)\arctan x \right\|_{0}^{1}-\int\limits_{0}^{1}{\frac{\ln \left( x+1 \right)dx}{1+x^{2}}}\underbrace{=}_{x\to \tan x}\frac{\pi \ln 2}{4}-\int\limits_{0}^{\frac{\pi }{4}}{\ln \left( \tan x+1 \right)dx}$$$ $TEX: $$\int\limits_{0}^{\frac{\pi }{4}}{\ln \left( \tan x+1 \right)dx}\underbrace{=}_{x=\frac{\pi }{4}-t}\int\limits_{0}^{\frac{\pi }{4}}{\ln \left( \frac{1-\tan t}{1+\tan t}+1 \right)dt}=\int\limits_{0}^{\frac{\pi }{4}}{\ln \left( \frac{2}{1+\tan t} \right)dt}=\frac{\pi \ln 2}{4}-\int\limits_{0}^{\frac{\pi }{4}}{\ln \left( 1+\tan x \right)dx}$$$ , luego $TEX: $$\int\limits_{0}^{\frac{\pi }{4}}{\ln \left( \tan x+1 \right)dx}=\frac{\pi \ln 2}{8}$$$ Asi que la integral vale $TEX: $$\frac{2\pi \ln 2}{8}-\frac{\pi \ln 2}{8}=\frac{\pi \ln 2}{8}$$$ Para la Anterior: $TEX: $$\int\limits_{0}^{1}{\frac{\arctan xdx}{x}}=\int\limits_{0}^{1}{\frac{1}{x}\sum\limits_{k=0}^{\infty }{\frac{\left( -1 \right)^{k}x^{2k+1}}{2k+1}}}dx=\sum\limits_{k=0}^{\infty }{\frac{\left( -1 \right)^{k}}{2k+1}}\int\limits_{0}^{1}{x^{2k}dx}=\sum\limits_{k=0}^{\infty }{\frac{\left( -1 \right)^{k}}{\left( 2k+1 \right)^{2}}}=C$$$ El cambio de la suma por la integral queda justificado, pues la funcion es positiva en el intervalo dado y C es la constante de catalan la mi9sma solucion que le di para el primero xd