# Math Help - Integration

1. ## Integration

Hi,

$\text{Show that}\,\,\int_0^\frac{\pi}{6} \frac{1}{2\cos^2\,\theta-1}=\frac{1}{2}\ln(2 + \sqrt3)$

$\int \frac{1}{2\cos^2 \theta - 1} = \frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K$

$\frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K$

$F|_a^b\Rightarrow F |_0^\frac{\pi}{6}$

$= \frac{1}{2} \bigg(\log\bigg(\sin\bigg(\frac{\pi}{6}\bigg) + \cos\bigg(\frac{\pi}{6}\bigg)\bigg) - \log\bigg(\cos\bigg(\frac{\pi}{6}\bigg) - \sin\bigg(\frac{\pi}{6}\bigg)\bigg)\bigg)\bigg -
\frac{1}{2} \bigg(\log\bigg(\sin\bigg(0\bigg) + \cos\bigg(0\bigg)\bigg) - \log\bigg(\cos\bigg(0\bigg) - \sin\bigg(0\bigg)\bigg)\bigg)$

$= \frac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2} - \log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)\bigg)$

2. Your last line should actually look like:

$\dfrac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2}\bigg) - \dfrac{1}{2}\log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)$

$\dfrac{1}{2}\bigg(\log\bigg(\frac{\sqrt{3}+1}{2}\b igg) - \log\bigg(\frac{\sqrt{3}-1}{2}\bigg)\bigg)$

now use: $\log(A)-\log(B)= \log\dfrac{A}{B}$.

Then rationalize the denominator and simplify.

3. Originally Posted by harish21
Your last line should actually look like:

$\dfrac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2}\bigg) - \dfrac{1}{2}\log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)$

$\dfrac{1}{2}\bigg(\log\bigg(\frac{\sqrt{3}+1}{2}\b igg) - \log\bigg(\frac{\sqrt{3}-1}{2}\bigg)\bigg)$

now use: $\log(A)-\log(B)= \log\dfrac{A}{B}$.

Then rationalize the denominator and simplify.
$\dfrac{1}{2}\bigg(\log\bigg(\frac{\sqrt{3}+1}{2}\b igg) - \log\bigg(\frac{\sqrt{3}-1}{2}\bigg)\bigg)$

$\log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)$

$\log \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

4. NO..

$\dfrac{1}{2}\bigg(\log\bigg(\frac{\sqrt{3}+1}{2}\b igg) - \log\bigg(\frac{\sqrt{3}-1}{2}\bigg)\bigg)$

$= \dfrac{1}{2} \log \bigg(\dfrac{\frac{\sqrt{3}+1}{2}}{\frac{\sqrt{3}-1}{2}}\bigg)$

$=\dfrac{1}{2} \log \dfrac{\sqrt{3}+1}{\sqrt{3}-1}$

5. I wasn't done. Even if my answer was going to be incorrect. I'm not sure of the reason for the slow connection.

6. Originally Posted by Hellbent
Hi,

$\text{Show that}\,\,\int_0^\frac{\pi}{6} \frac{1}{2\cos^2\,\theta-1}=\frac{1}{2}\ln(2 + \sqrt3)$

$\int \frac{1}{2\cos^2 \theta - 1} = \frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K$

$\frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K$

$F|_a^b\Rightarrow F |_0^\frac{\pi}{6}$

$= \frac{1}{2} \bigg(\log\bigg(\sin\bigg(\frac{\pi}{6}\bigg) + \cos\bigg(\frac{\pi}{6}\bigg)\bigg) - \log\bigg(\cos\bigg(\frac{\pi}{6}\bigg) - \sin\bigg(\frac{\pi}{6}\bigg)\bigg)\bigg)\bigg -
\frac{1}{2} \bigg(\log\bigg(\sin\bigg(0\bigg) + \cos\bigg(0\bigg)\bigg) - \log\bigg(\cos\bigg(0\bigg) - \sin\bigg(0\bigg)\bigg)\bigg)$

$= \frac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2} - \log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)\bigg)$
This may help

$2\cos(\theta)-1=\cos(2\theta)$

This gives the integral

$\int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta=\frac {1}{2}\ln|\sec(2\theta)+\tan(2\theta)|=\frac{1}{2} \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi)}{3}|-\ln|\sec(0)+\tan(0)|=\ln|2+\sqrt{3}|$

7. Originally Posted by Hellbent
I wasn't done. Even if my answer was going to be incorrect. I'm not sure of the reason for the slow connection.
OKay.

Yes, MHF is kind of slow. Most users are experiencing the same problem.

However, I think you should be able to proceed from here:

$\dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)$

8. $\dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)$
Proceed to where? Ha! I thought this was done, even though it doesn't resemble this $\frac{1}{2}\ln(2 + \sqrt{3})$

9. Originally Posted by TheEmptySet
This may help

$2\cos(\theta)-1=\cos(2\theta)$

This gives the integral

$\int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta=\frac {1}{2}\ln|\sec(2\theta)+\tan(2\theta)|=\frac{1}{2} \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi)}{3}|-\ln|\sec(0)+\tan(0)|=\ln|2+\sqrt{3}|$
It doesn't help. How does the substitution help? My already poor understanding is being muddled. I don't see how it gives the integral...

$\int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta=\frac {1}{2}\ln|\sec(2\theta)+\tan(2\theta)|=\frac{1}{2} \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi)}{3}|-\ln|\sec(0)+\tan(0)|=\ln|2+\sqrt{3}|$

$\ln|2+\sqrt{3}|$
$\text{Is the above the same as}\, \frac{\pi\,sec(2\theta)}{6}\text{?}$

10. Originally Posted by Hellbent
It doesn't help. How does the substitution help? My already poor understanding is being muddled. I don't see how it gives the integral...

$\int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta=\frac {1}{2}\ln|\sec(2\theta)+\tan(2\theta)|=\frac{1}{2} \ln|\sec(\frac{\pi}{3})+\tan(\frac{\pi)}{3}|-\ln|\sec(0)+\tan(0)|=\ln|2+\sqrt{3}|$

$\ln|2+\sqrt{3}|$
$\text{Is the above the same as}\, \frac{\pi\,sec(2\theta)}{6}\text{?}$
From trigonomerty we have the identities that

$2\cos^2(\theta)-1=\cos(2\theta)$

We also have the identity that

$\frac{1}{\cos(2\theta)}=\sec(2\theta)$

This gives

$\frac{1}{2\cos(\theta)-1}=\frac{1}{\cos(2\theta)}=\sec(2\theta)$

Now we need to integrate

$\int_{0}^{\frac{\pi}{6}}\sec(2\theta)d\theta$

FYI most people just memorize this anti-derivative but here is the substitution to work it out. We will do it in two steps. first let
$w=2\theta \implies dw=2\theta$

$\frac{1}{2}\int_{0}^{\frac{\pi}{3}}\sec(w)dw$

Now here is the part most folks memorize.

let $\displaystyle u=\sec(w)+\tan(w) \implies du=\sec(w)\tan(w)+\sec^2(w)dw=\sec(w)(\tan(w)+\sec (w))dw=\sec(w)udw$

This gives that

$\displaystyle \frac{du}{u}=\sec(w)dw$ now the integral becomes

$\displaystyle \frac{1}{2}\int \frac{du}{u}=\ln|u|=\ln|\sec(w)+\tan(w)| \bigg|_{0}^{\frac{\pi}{3}}=\ln|2+\sqrt{3}|$

11. Originally Posted by TheEmptySet
From trigonomerty we have the identities that

$2\cos(\theta)-1=\cos(2\theta)$
Correction: It might be a typo, but

That should be $2\cos^{2}(\theta)-1=\cos(2\theta)$

12. What's the relationship between these two: $\dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)$ and $\ln|2+\sqrt{3}|$?

13. Originally Posted by Hellbent
What's the relationship between these two: $\dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)$ and $\ln|2+\sqrt{3}|$?
$\dfrac{\sqrt{3}+1}{\sqrt{3}-1} = \dfrac{\sqrt{3}+1}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$

$= \dfrac{3+2\sqrt{3}+1}{3-1} = \dfrac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$

therefore,

$\dfrac{1}{2} \log \bigg(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\bigg) = \dfrac{1}{2} log( 2+\sqrt{3})$

14. Argh! I know I'm tremendously slow, but now that I see it, it's rather trvial.
Thinking of it my rationalization was different and wrong. I used the incorrect conjugate.

Ah, is TheEmpySet's method another approach or was he just hinting at what is directly above this post?