Improper Integral - # 5

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Jason76

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The answer to this is \(\displaystyle \ln 3\)

\(\displaystyle \int_{-\infty}^{-2} \dfrac{2 dx}{x^{2} - 1} \)

\(\displaystyle x = \sec\theta d\theta \)

\(\displaystyle dx = \sec\theta\tan\theta d\theta \)

\(\displaystyle sec^{2} - 1 = \tan^{2}\theta d\theta \)

\(\displaystyle \int \dfrac{2\sec\theta\tan\theta}{\tan^{2}\theta} \)

\(\displaystyle \int \dfrac{2\sec\theta}{\tan\theta} \)

\(\displaystyle \int 2\sec\theta(\cot\theta) \)

\(\displaystyle = (\ln|\sec\theta + \tan\theta|)(\ln|\sin\theta|) + C\)

Now put in terms of x values.

\(\displaystyle \sec\theta = \dfrac{x}{1}\)

\(\displaystyle 2 \ln |\dfrac{x}{1} + \dfrac{x^{2} - 1}{1}| * \ln|\dfrac{x^{2} - 1}{x}|\) evaluated at upper bound \(\displaystyle -2\) and lower bound \(\displaystyle -\infty\) :confused:
 
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What, exactly, is your difficulty? I assume you can evaluate that at x= -2. Of course, "evaluating" at \(\displaystyle -\infty\) is really taking the limit as x goes to \(\displaystyle -\infty\).

I think I would have done this slightly differently: yes, \(\displaystyle x= sec(\theta)\) (NOT "\(\displaystyle sec(\theta)d\theta\). I assume that was a typo.) then \(\displaystyle dx= sec(\theta)tan(\theta)d\theta\) and the integral becomes \(\displaystyle 2\int \frac{sec(\theta)}{tan(\theta)} d\theta\) (you forgot the "\(\displaystyle d\theta\)").

But here I would first note that when x= -2, \(\displaystyle sec(\theta)= -2\) so \(\displaystyle cos(\theta)= -\frac{1}{2}\) and \(\displaystyle \theta= -\frac{2\pi}{3}\). And as x= sec(\theta) goes to \(\displaystyle -\infty\), \(\displaystyle cos(\theta)\) must be going to 0 from below- \(\displaystyle \theta\) must be approaching \(\displaystyle -\frac{\pi}{2}\) from below: \(\displaystyle 2\int_{-\pi/2}^{-2\pi/3} \frac{sec(\theta)}{tan(\theta)} d\theta\).

I would also change to sine and cosine: \(\displaystyle sec(\theta)= \frac{1}{cos(\theta)}\) and \(\displaystyle tan(\theta)= \frac{sin(\theta)}{cos(\theta)}\) so \(\displaystyle \frac{cos(\theta)}{tan(\theta)}= \frac{1}{cos(\theta)}\frac{cos(\theta)}{sin(\theta}= \frac{1}{sin(\theta)}\). The integral becomes \(\displaystyle 2\int_{-\pi/3}^{-2\pi/3} \frac{d\theta}{sin(\theta)}\)

Multiply both numerator and denominator by \(\displaystyle sin(\theta)\) to get \(\displaystyle 2\int_{-2\pi/3}^{-\pi/2} \frac{sin(\theta)d\theta}{1- cos^2(\theta)}\) and now let v= cos(\theta). \(\displaystyle sin(\theta)d\theta= dv\), when \(\displaystyle \theta= -2\pi/3\), \(\displaystyle v= cos(-2\pi/3)= \frac{1}{2}\) and when \(\displaystyle \theta= -\pi/2\), \(\displaystyle v= cos(-\pi/2)= 0\) so the integral is \(\displaystyle \int_0^{1/2} \frac{du}{1- u^2}\).

Do that integral by "partial fractions".
 
\(\displaystyle 2 \ln |\dfrac{-2}{1} + \dfrac{-2^{2} - 1}{1}| * \ln|\dfrac{-2^{2} - 1}{-2}| |\) - \(\displaystyle 2 \ln |\dfrac{\infty}{1} + \dfrac{\infty^{2} - 1}{1}| * \ln|\dfrac{\infty^{2} - 1}{\infty} = 0 - \infty |\) indeterminate

\(\displaystyle \dfrac{d}{dx}[2 \ln |\dfrac{x}{1} + \dfrac{x^{2} - 1}{1}| * \ln|\dfrac{x^{2} - 1}{x}|]\)
 
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Jason76 said:
The answer to this is \(\displaystyle \ln 3\)

\(\displaystyle \int_{-\infty}^{-2} \dfrac{2 dx}{x^{2} - 1} \)

\(\displaystyle x = \sec\theta d\theta \)

\(\displaystyle dx = \sec\theta\tan\theta d\theta \)

\(\displaystyle sec^{2} - 1 = \tan^{2}\theta d\theta \)

\(\displaystyle \int \dfrac{2\sec\theta\tan\theta}{\tan^{2}\theta} \)

\(\displaystyle \int \dfrac{2\sec\theta}{\tan\theta} \)

\(\displaystyle \int 2\sec\theta(\cot\theta) \)

\(\displaystyle = (\ln|\sec\theta + \tan\theta|)(\ln|\sin\theta|) + C\) ............. Incorrect .... how did you get that? See HoI's response above.

Now put in terms of x values.

\(\displaystyle \sec\theta = \dfrac{x}{1}\)

\(\displaystyle 2 \ln |\dfrac{x}{1} + \dfrac{x^{2} - 1}{1}| * \ln|\dfrac{x^{2} - 1}{x}|\) evaluated at upper bound \(\displaystyle -2\) and lower bound \(\displaystyle -\infty\) :confused:
Click to expand...
.
 
True [except for typo's] but I would note
\(\displaystyle \frac{2}{x^2-1} = \frac{1}{x-1} - \frac{1}{x+1}\)
and hope I didn't make any typo's
 
HallsofIvy said:
What, exactly, is your difficulty? I assume you can evaluate that at x= -2. Of course, "evaluating" at \(\displaystyle -\infty\) is really taking the limit as x goes to \(\displaystyle -\infty\).

I think I would have done this slightly differently: yes, \(\displaystyle x= sec(\theta)\) (NOT "\(\displaystyle sec(\theta)d\theta\). I assume that was a typo.) then \(\displaystyle dx= sec(\theta)tan(\theta)d\theta\) and the integral becomes \(\displaystyle 2\int \frac{sec(\theta)}{tan(\theta)} d\theta\) (you forgot the "\(\displaystyle d\theta\)").

But here I would first note that when x= -2, \(\displaystyle sec(\theta)= -2\) so \(\displaystyle cos(\theta)= -\frac{1}{2}\) and \(\displaystyle \theta= -\frac{2\pi}{3}\). And as x= sec(\theta) goes to \(\displaystyle -\infty\), \(\displaystyle cos(\theta)\) must be going to 0 from below- \(\displaystyle \theta\) must be approaching \(\displaystyle -\frac{\pi}{2}\) from below: \(\displaystyle 2\int_{-\pi/2}^{-2\pi/3} \frac{sec(\theta)}{tan(\theta)} d\theta\).

I would also change to sine and cosine: \(\displaystyle sec(\theta)= \frac{1}{cos(\theta)}\) and \(\displaystyle tan(\theta)= \frac{sin(\theta)}{cos(\theta)}\) so \(\displaystyle \frac{cos(\theta)}{tan(\theta)}= \frac{1}{cos(\theta)}\frac{cos(\theta)}{sin(\theta}= \frac{1}{sin(\theta)}\). The integral becomes \(\displaystyle 2\int_{-\pi/3}^{-2\pi/3} \frac{d\theta}{sin(\theta)}\)

Multiply both numerator and denominator by \(\displaystyle sin(\theta)\) to get \(\displaystyle 2\int_{-2\pi/3}^{-\pi/2} \frac{sin(\theta)d\theta}{1- cos^2(\theta)}\) and now let v= cos(\theta). \(\displaystyle sin(\theta)d\theta= dv\), when \(\displaystyle \theta= -2\pi/3\), \(\displaystyle v= cos(-2\pi/3)= \frac{1}{2}\) and when \(\displaystyle \theta= -\pi/2\), \(\displaystyle v= cos(-\pi/2)= 0\) so the integral is \(\displaystyle \int_0^{1/2} \frac{du}{1- u^2}\).

Do that integral by "partial fractions".
Click to expand...

or

\(\displaystyle \displaystyle{\int\dfrac{d\theta}{sin(\theta)}} = -ln[cosec(\theta) + cot(\theta)] + C\)
 
Jason76 said:
The answer to this is \(\displaystyle \ln 3\)

\(\displaystyle \int_{-\infty}^{-2} \dfrac{2 dx}{x^{2} - 1} \)

\(\displaystyle x = \sec\theta d\theta \)

\(\displaystyle dx = \sec\theta\tan\theta d\theta \)

\(\displaystyle sec^{2} - 1 = \tan^{2}\theta d\theta \)

\(\displaystyle \int \dfrac{2\sec\theta\tan\theta}{\tan^{2}\theta} \)

\(\displaystyle \int \dfrac{2\sec\theta}{\tan\theta} \)

\(\displaystyle \int 2\sec\theta(\cot\theta) \)

\(\displaystyle = (\ln|\sec\theta + \tan\theta|)(\ln|\sin\theta|) + C\)

Now put in terms of x values.

\(\displaystyle \sec\theta = \dfrac{x}{1}\)

\(\displaystyle 2 \ln |\dfrac{x}{1} + \dfrac{x^{2} - 1}{1}| * \ln|\dfrac{x^{2} - 1}{x}|\) evaluated at upper bound \(\displaystyle -2\) and lower bound \(\displaystyle -\infty\) :confused:
Click to expand...

Way more straight forward:

\(\displaystyle \int_{-\infty}^{-2} \dfrac{2 dx}{x^{2} - 1} \)

= \(\displaystyle \int_{-\infty}^{-2} \left [\dfrac{dx}{x - 1} - \dfrac {dx}{x + 1} \right ] \) .... and continue....
 
Jason76 said:
\(\displaystyle \int_{-\infty}^{-2} \dfrac{2 dx}{x^{2} - 1} \)

\(\displaystyle x = \sec\theta d\theta \)

\(\displaystyle dx = \sec\theta\tan\theta d\theta \)

\(\displaystyle sec^{2} - 1 = \tan^{2}\theta d\theta \)

\(\displaystyle \int \dfrac{2\sec\theta\tan\theta}{\tan^{2}\theta} \)

\(\displaystyle \int \dfrac{2\sec\theta}{\tan\theta} \)

\(\displaystyle \int 2\sec\theta(\cot\theta) \)
Click to expand...

Jason76, you must include the \(\displaystyle \ "d\theta" \ \) as part of your integrals. Here, it's in the
fifth, sixth, and seventh lines of this edited quote box.
 
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