So who gets to post the next integral?
$\displaystyle \frac{\cos (2 \theta)}{5 - 4\cos \theta}=\frac{2 \cos^2 \theta - 1}{5 - 4 \cos \theta} = \frac{-1}{2} \cos \theta - \frac{5}{8} + \frac{17}{8} \cdot \frac{1}{5 - 4 \cos \theta}.$ we also have: $\displaystyle J=\int_0^{2 \pi} \frac{d \theta}{5 - 4 \cos \theta} = 2 \int_0^{\pi} \frac{d \theta}{5 - 4 \cos \theta}.$ put $\displaystyle \tan(\theta/2) = x$ to get: $\displaystyle J=4\int_0^{\infty} \frac{dx}{1 + 9x^2} = \frac{2\pi}{3}.$
thus: $\displaystyle \int_0^{2 \pi} \frac{\cos (2 \theta)}{5 - 4\cos \theta} \ d \theta=\frac{- 5 \pi}{4} + \frac{17 \pi}{12}=\frac{\pi}{6}.$
Problem: Prove that $\displaystyle \int_0^{\frac{\pi}{2}} \frac{x}{1+\sin x} \ dx = \ln 2.$
$\displaystyle \int \frac{x}{1+\sin x}dx$
$\displaystyle \int \frac{x(1-sin x )}{1-sin^2 x}dx$
$\displaystyle \int \frac{x}{cos^2 x}dx - \int \frac{x(sin x)}{cos^2 x }dx$
let x=u and dv=1/cos^2 x use integrate by parts for the left integral and for the right integral u=x and v=x(sin x)/cos^2 x
but for the right integral you want to evaluate $\displaystyle \int \frac{sin x}{cos^2 x} dx $ first , by sub u=cos x we can see that the integrate equal $\displaystyle \frac{1}{cos x}$
$\displaystyle x(tan x) - \int \frac{sin x}{cos x} dx -\left( x\left(\frac{1}{cos x}\right) - \int sec x dx\right) $
$\displaystyle x(tan x) + ln (cos x) - \left ( \frac{x}{cos x} - ln (sec x + tan x ) \right)$
$\displaystyle x(tan x) -\frac{x}{cos x} + ln (cos x ) + ln (sec x + tan x ) $
$\displaystyle \frac{ x (sin x ) - x }{cos x} + ln ( 1+sin x) $
$\displaystyle \frac{ x (sin x ) - x }{cos x} + ln ( 1+sin x) \mid ^{\frac{\pi}{2}}_{0}$
$\displaystyle lim_{x\rightarrow \frac{\pi}{2}} \frac{x \sin x - x }{\cos x} + ln (1 + sin \left(\frac{pi}{2}\right)- \frac{ 0 (0) - 0 }{1} - ln (1 + 0 ) $
$\displaystyle lim_{x\rightarrow \frac{\pi}{2}} \frac{x \sin x - x }{\cos x} + ln \left(1 + 1\right)- 0 - 0 $
use lopital for the limit
$\displaystyle lim_{x\rightarrow \frac{\pi}{2}} \frac{x\cos x+(\sin x -1 ) }{-\sin x} + ln (2) $
the limit equal zero after the we sub pi/2 so the integral equal ln(2)
here is my integral
$\displaystyle \int \ln (\sqrt{x^2 +1 } +1 ) dx $
my answer is
$\displaystyle -\ln|\tan(\arctan(\frac{x}{2}))|+\frac{x}{\tan(\arc tan(\frac{x}{2}))}+x\ln(\sqrt{x^2+1}+1)-\ln|x|+C$
the procedure is quite complex....
it is too late to night, you may allow me showing it tomorrow..
oh...it is not correct........sorry,there is a obvious mistake.(edited later..)
$\displaystyle \int\ln(\sqrt{x^2+1}+1)dx=\int x'\ln(\sqrt{x^2+1}+1)dx=$
$\displaystyle =x\ln(\sqrt{x^2+1}+1)-\int\frac{x^2}{\sqrt{x^2+1}(\sqrt{x^2+1}+1)}dx=$
$\displaystyle =x\ln(\sqrt{x^2+1}+1)-\int\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}}dx=$
$\displaystyle =x\ln(\sqrt{x^2+1}+1)-\int\left(1-\frac{1}{\sqrt{x^2+1}}\right)dx=$
$\displaystyle =x\ln(\sqrt{x^2+1}+1)-x+\ln(x+\sqrt{x^2+1})+C$
1) Let $\displaystyle f,g:[-a,a]\to\mathbb{R}$ two continuous functions, f odd, g even. Then
$\displaystyle \int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_0^a g(x)dx$
2) Calculate $\displaystyle \int_{-1}^1\frac{x^{2008}}{1+x^{2007}+\sqrt{1+x^{4014}}}d x$
Edited.