Another Integration problem! Watch

chickster
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#1
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Ahh... integration is doing my head in! Any ideas on how to integrate:

 - \frac{1}{cos{x}}
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darkness9999
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(Original post by chickster)
Ahh... integration is doing my head in! Any ideas on how to integrate:

 - \frac{1}{cos{x}}
 - \frac{1}{cos{x}}\equiv -\sec x

and we all know the answer to that... dont we :p:
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Unbounded
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(Original post by darkness9999)
 - \frac{1}{cos{x}}\equiv -\sec x

and we all know the answer to that... dont we :p:
No... :erm:
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From memory, It's something like \mathrm{tanh}^{-1} \left(\tan \frac{x}{2}\right)\right)
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fishpie57
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sec squared x integrated is tan x... Umm... then I'm stuck.

Don't forget the +c!
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Lou Reed
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(Original post by GHOSH-5)
No... :erm:
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From memory, It's something like \mathrm{tanh}^{-1} \left(\tan \frac{x}{2}\right)\right)
\displaystyle\int \sec x = \ln|\sec x + \tan x| + C

from the formula book
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Lou Reed
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(Original post by chickster)
.
so you know i replied
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chickster
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I am confused, how do people know the integral of -sec x? is it actually what GHOSH-5 said, as that seems complicated to work out!
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Unbounded
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(Original post by Lou Reed)
\displaystyle\int \sec x = \ln|\sec x + \tan x| + C

from the formula book
I think they're equivalent. I could be wrong.
(Original post by chickster)
I am confused, how do people know the integral of -sec x? is it actually what GHOSH-5 said, as that seems complicated to work out!
That was my point. Unless you're doing further maths, there's no way you'd be asked to do a question like this, unless told what  \displaystyle\int \sec \theta \ \mathrm{d}\theta is. Perhaps the question is incorrectly constructed.
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fishpie57
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(Original post by Lou Reed)
\displaystyle\int \sec x = \ln|\sec x + \tan x| + C

from the formula book
Ah, the formula book - the mathematics equivalent of the instruction manual. "When in doubt, look in the formula book."

I'll remember that now, thanks! :yep:
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Lou Reed
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(Original post by chickster)
I am confused, how do people know the integral of -sec x? is it actually what GHOSH-5 said, as that seems complicated to work out!
im pretty sure its what i said, ive got my C4 book open in front of me
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darkness9999
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(Original post by GHOSH-5)
No... :erm:
Spoiler:
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From memory, It's something like \mathrm{tanh}^{-1} \left(\tan \frac{x}{2}\right)\right)
Spoiler:
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\displaystyle\int \sec x \ dx = \displaystyle\int \sec x \frac{sec x + tan x}{sec x + tan x}\ dx

u = \sec x + \tan x

du = (\sec x \tan x + \sec^2 x) dx

so

\displaystyle\int \sec x \frac{\sec x + \tan x}{\sec x + \tan x}\ dx

becomes

\Rightarrow \displaystyle\int \frac{(\sec^2 x + \sec x \tan x)}{\sec x + \tan x}\ dx

\Rightarrow \displaystyle\int \frac{du}{u}

\Rightarrow \ln \left|u\right| + C

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chickster
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that makes sense, dunno if i would be able to remember to put the sec x + tanx, spose something to try and learn...
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Unbounded
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(Original post by darkness9999)
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\displaystyle\int \sec x \ dx = \displaystyle\int \sec x \frac{sec x + tan x}{sec x + tan x}\ dx

u = \sec x + \tan x

du = (\sec x \tan x + \sec^2 x) dx

so

\displaystyle\int \sec x \frac{\sec x + \tan x}{\sec x + \tan x}\ dx

becomes

\Rightarrow \displaystyle\int \frac{(\sec^2 x + \sec x \tan x)}{\sec x + \tan x}\ dx

\Rightarrow \displaystyle\int \frac{du}{u}

\Rightarrow \ln \left|u\right| + C

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Indeed  \mathrm{tanh}^{-1} \tan \frac{x}{2} \equiv \ln \left( \dfrac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} \right)

 = \ln \left( \dfrac{1+t}{1-t} \right) \equiv \ln \left( \dfrac{1+t^2+2t}{1-t^2}\right) \equiv \ln (\sec x + \tan x)
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Glutamic Acid
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\displaystyle \int \frac{1}{\cos x} \, \text{d}x = \int \frac{\cos x}{\cos^2 x} \, \text{d}x = \int \frac{\cos x}{1 - \sin^2 x} \, \text{d}x. Let u = sin x and hit it with partial fractions / a standard result / a hyperbolic sub.
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darkness9999
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#15
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(Original post by GHOSH-5)
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Indeed  \mathrm{tanh}^{-1} \tan \frac{x}{2} \equiv \ln \left( \dfrac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} \right)

 = \ln \left( \dfrac{1+t}{1-t} \right) \equiv \ln \left( \dfrac{1+t^2+2t}{1-t^2}\right) \equiv \ln (\sec x + \tan x)
looks good:yep: I dnt know anything about  \mathrm{tanh}^{-1}:o:
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Unbounded
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(Original post by darkness9999)
looks good:yep: I dnt know anything about  \mathrm{tanh}^{-1}:o:
Nor do I, I just shoved down what Wolfram decided the integral was. There's no chance of this coming up in a standard a-level maths exam, no question, really. So the OP must have an incorrect question in from of them, or something :erm:
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nuodai
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(Original post by chickster)
that makes sense, dunno if i would be able to remember to put the sec x + tanx, spose something to try and learn...
If this is for A-level/IB/Highers/etc., you'll get a formula book in your exam, and \displaystyle \int \sec x\ \mbox{d}x will almost 100% definitely be in it.
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chickster
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(Original post by nuodai)
If this is for A-level/IB/Highers/etc., you'll get a formula book in your exam, and \displaystyle \int \sec x\ \mbox{d}x will almost 100% definitely be in it.
Doing maths degree doesn't allow for the luxury of formula books, we are meant to derive everything I suppose, lol.
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nuodai
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(Original post by chickster)
Doing maths degree doesn't allow for the luxury of formula books, we are meant to derive everything I suppose, lol.
Well doesn't doing a maths degree merit the ability to derive (or at least remember) this? I like Glutamic Acid's method anyway, it's quite tidy.
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