My Math Forum Integration Challenge

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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.

Now let's see if anyone gets the $\displaystyle x - tan \left ( \frac{x}{2} \right )$ version.

-Dan

 Tags challenge, integration

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