# Math Help - MHF Integral Bee!

1. So who gets to post the next integral?

2. Originally Posted by Random Variable
So who gets to post the next integral?

3. $\int^{2 \pi}_{0} \frac{\cos 2 \theta}{5-4 \cos \theta} \ d \theta$

4. Originally Posted by Random Variable
$\int^{2 \pi}_{0} \frac{\cos 2 \theta}{5-4 \cos \theta} \ d \theta$
$\frac{\cos (2 \theta)}{5 - 4\cos \theta}=\frac{2 \cos^2 \theta - 1}{5 - 4 \cos \theta} = \frac{-1}{2} \cos \theta - \frac{5}{8} + \frac{17}{8} \cdot \frac{1}{5 - 4 \cos \theta}.$ we also have: $J=\int_0^{2 \pi} \frac{d \theta}{5 - 4 \cos \theta} = 2 \int_0^{\pi} \frac{d \theta}{5 - 4 \cos \theta}.$ put $\tan(\theta/2) = x$ to get: $J=4\int_0^{\infty} \frac{dx}{1 + 9x^2} = \frac{2\pi}{3}.$

thus: $\int_0^{2 \pi} \frac{\cos (2 \theta)}{5 - 4\cos \theta} \ d \theta=\frac{- 5 \pi}{4} + \frac{17 \pi}{12}=\frac{\pi}{6}.$

Problem: Prove that $\int_0^{\frac{\pi}{2}} \frac{x}{1+\sin x} \ dx = \ln 2.$

5. Originally Posted by NonCommAlg
Problem: Prove that $\int_0^{\frac{\pi}{2}} \frac{x}{1+\sin x} \ dx = \ln 2.$

$\int \frac{x}{1+\sin x}dx$

$\int \frac{x(1-sin x )}{1-sin^2 x}dx$

$\int \frac{x}{cos^2 x}dx - \int \frac{x(sin x)}{cos^2 x }dx$

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 $\int \frac{sin x}{cos^2 x} dx$ first , by sub u=cos x we can see that the integrate equal $\frac{1}{cos x}$

$x(tan x) - \int \frac{sin x}{cos x} dx -\left( x\left(\frac{1}{cos x}\right) - \int sec x dx\right)$

$x(tan x) + ln (cos x) - \left ( \frac{x}{cos x} - ln (sec x + tan x ) \right)$

$x(tan x) -\frac{x}{cos x} + ln (cos x ) + ln (sec x + tan x )$

$\frac{ x (sin x ) - x }{cos x} + ln ( 1+sin x)$

$\frac{ x (sin x ) - x }{cos x} + ln ( 1+sin x) \mid ^{\frac{\pi}{2}}_{0}$

$lim_{x\rightarrow \frac{\pi}{2}} \frac{x \sin x - x }{\cos x} + ln (1 + sin \left(\frac{pi}{2}\right)- \frac{ 0 (0) - 0 }{1} - ln (1 + 0 )$

$lim_{x\rightarrow \frac{\pi}{2}} \frac{x \sin x - x }{\cos x} + ln \left(1 + 1\right)- 0 - 0$

use lopital for the limit

$lim_{x\rightarrow \frac{\pi}{2}} \frac{x\cos x+(\sin x -1 ) }{-\sin x} + ln (2)$

the limit equal zero after the we sub pi/2 so the integral equal ln(2)

here is my integral

$\int \ln (\sqrt{x^2 +1 } +1 ) dx$

6. I am deleting the post .I read the question wrongly

$-\ln|\tan(\arctan(\frac{x}{2}))|+\frac{x}{\tan(\arc tan(\frac{x}{2}))}+x\ln(\sqrt{x^2+1}+1)-\ln|x|+C$
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..)

8. Originally Posted by ynj
$-\ln|\tan(\arctan(\frac{x}{2}))|+\frac{x}{\tan(\arc tan(\frac{x}{2}))}+x\ln(\sqrt{x^2+1}+1)-\ln|x|+C$
the procedure is quite complex....
it is too late to night, you may allow me showing it tomorrow..
your solution is not correct ( or show me your work )

another thing there is no boundaries for my integral

9. $\int\ln(\sqrt{x^2+1}+1)dx=\int x'\ln(\sqrt{x^2+1}+1)dx=$

$=x\ln(\sqrt{x^2+1}+1)-\int\frac{x^2}{\sqrt{x^2+1}(\sqrt{x^2+1}+1)}dx=$

$=x\ln(\sqrt{x^2+1}+1)-\int\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}}dx=$

$=x\ln(\sqrt{x^2+1}+1)-\int\left(1-\frac{1}{\sqrt{x^2+1}}\right)dx=$

$=x\ln(\sqrt{x^2+1}+1)-x+\ln(x+\sqrt{x^2+1})+C$

10. $\ln(x+\sqrt{x^{2}+1}) = \sinh^{-1}x$

11. Originally Posted by red_dog
$\int\ln(\sqrt{x^2+1}+1)dx=\int x'\ln(\sqrt{x^2+1}+1)dx=$

$=x\ln(\sqrt{x^2+1}+1)-\int\frac{x^2}{\sqrt{x^2+1}(\sqrt{x^2+1}+1)}dx=$

$=x\ln(\sqrt{x^2+1}+1)-\int\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}}dx=$

$=x\ln(\sqrt{x^2+1}+1)-\int\left(1-\frac{1}{\sqrt{x^2+1}}\right)dx=$

$=x\ln(\sqrt{x^2+1}+1)-x+\ln(x+\sqrt{x^2+1})+C$
post an integral

12. 1) Let $f,g:[-a,a]\to\mathbb{R}$ two continuous functions, f odd, g even. Then

$\int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_0^a g(x)dx$

2) Calculate $\int_{-1}^1\frac{x^{2008}}{1+x^{2007}+\sqrt{1+x^{4014}}}d x$

Edited.

13. Originally Posted by red_dog
1) Let $f,g:[-a,a]\to\mathbb{R}$ two continuous functions, f odd, g even. Then

$\int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_{-a}^af(x)dx$
for those who are working on this problem: there's a typo in the problem. the RHS should be $\int_0^a g(x) \ dx$ instead of $\int_{-a}^a f(x) \ dx.$

14. Originally Posted by NonCommAlg
for those who are working on this problem: there's a typo in the problem. the RHS should be $\int_{-a}^a g(x) \ dx$ instead of $\int_{-a}^a f(x) \ dx.$
It should be $
\int_{ - a}^a {\tfrac{{g\left( x \right)}}
{{1 + f\left( x \right) + \sqrt {1 + f^2 \left( x \right)} }}dx} = \int_0^a {g\left( x \right) \cdot dx}
$

15. Originally Posted by PaulRS
It should be $
\int_{ - a}^a {\tfrac{{g\left( x \right)}}
{{1 + f\left( x \right) + \sqrt {1 + f^2 \left( x \right)} }}dx} = \int_0^a {g\left( x \right) \cdot dx}
$
haha ... you're right. that's what i meant! i fixed it.

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