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July 22nd, 2014, 01:44 PM   #1
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Integration Challenge

$\displaystyle \int\frac{dx}{\sec x+1}$
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July 22nd, 2014, 04:25 PM   #2
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Quote:
Originally Posted by greg1313 View Post
$\displaystyle \int\frac{dx}{\sec x+1}$
$\displaystyle \int \frac{1}{sec(x) + 1}~dx$

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

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

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

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

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.
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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: $\displaystyle \:\int(\csc x\cot x - \csc^2\!x + 1)\,dx \;=\;-\csc x + \cot x + x + C$

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July 22nd, 2014, 05:45 PM   #4
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is there an echo in here?

-Dan
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July 22nd, 2014, 08:22 PM   #5
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Hello, Dan!

Sorry for the echo.

I was so proud of my solution
that I didn't see your post.
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July 22nd, 2014, 09:44 PM   #6
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Quote:
Originally Posted by topsquark View Post

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

There is a typo in the first denominator.

It should be "sec(x) + 1."
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July 23rd, 2014, 11:04 AM   #7
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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
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