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 July 22nd, 2014, 01:44 PM #1 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Integration Challenge $\displaystyle \int\frac{dx}{\sec x+1}$ July 22nd, 2014, 04:25 PM   #2
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Quote:
 Originally Posted by greg1313 $\displaystyle \int\frac{dx}{\sec x+1}$
$\displaystyle \int \frac{1}{sec(x) + 1}~dx$

$\displaystyle = \int \frac{1}{sec(x)} \cdot \frac{sec(x) - 1}{sec(x) - 1}~dx$

$\displaystyle = \int \frac{sec(x) - 1}{sec^2(x) - 1}~dx$

$\displaystyle = \int \frac{sec(x)}{tan^2(x)}~dx - \int cot^2(x)~dx$

$\displaystyle = \int \frac{cos(x)}{sin^2(x)}~dx - \int cot^2(x)~dx$

Both of these integrations are reasonably trivial so I will skip it to the penultimate expression:
$\displaystyle = (-csc(x)) - (-cot(x) - x)$

$\displaystyle = -csc(x) + cot(x) + x + C$

-Dan

Do we have spoiler tags here?

Last edited by topsquark; July 22nd, 2014 at 04:29 PM. July 22nd, 2014, 04:48 PM   #3
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Hello, greg1313!

Quote:
 $\displaystyle \int\frac{dx}{\sec x+1}$

Multiply by $\frac{\sec x - 1}{\sec x - 1}\!:$

$\dfrac{1}{\sec x + 1}\cdot \dfrac{\sec x - 1}{\sec x - 1} \:=\:\dfrac{\sec x-1}{\sec^2\!x-1} \;=\;\dfrac{\sec x -1}{\tan^2\!x} \;=\;\dfrac{\sec x}{\tan^2\!x} - \dfrac{1}{\tan^2\!x}$

$\qquad =\;\csc x\cot x - \cot^2\!x \;=\; \csc x\cot x - (\csc^2\!x-1)$

We have: $\displaystyle \:\int(\csc x\cot x - \csc^2\!x + 1)\,dx \;=\;-\csc x + \cot x + x + C$ July 22nd, 2014, 05:45 PM #4 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,301 Thanks: 960 Math Focus: Wibbly wobbly timey-wimey stuff. is there an echo in here? -Dan July 22nd, 2014, 08:22 PM #5 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Hello, Dan! Sorry for the echo. I was so proud of my solution that I didn't see your post. July 22nd, 2014, 09:44 PM   #6
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Quote:
 Originally Posted by topsquark $\displaystyle = \int \frac{1}{sec(x)} \cdot \frac{sec(x) - 1}{sec(x) - 1}~dx$

There is a typo in the first denominator.

It should be "sec(x) + 1." July 23rd, 2014, 11:04 AM   #7
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Quote:
 Originally Posted by soroban Hello, Dan! Sorry for the echo. I was so proud of my solution that I didn't see your post.
Don't worry about it. I had a good chuckle over it. Now let's see if anyone gets the $\displaystyle x - tan \left ( \frac{x}{2} \right )$ version.

-Dan Tags challenge, integration Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Etyucan Math Events 8 May 10th, 2013 07:30 AM greg1313 Calculus 0 May 10th, 2013 03:34 AM MarkFL Calculus 8 October 25th, 2012 09:43 PM sameapple Calculus 5 January 31st, 2009 07:45 PM Etyucan Calculus 1 December 31st, 1969 04:00 PM

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