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Arclength of the curve $y= \ln( \sec x)$ $ 0 \le x \le \pi/4$

I know that I have to find its derivative which is easy, it is $\tan x$

Then I put it into the arclength formula

$$\int \sqrt {1 - \tan^2 x}$$

From here I am not sure what to do, I put it in wolfram and it got something massive looking. I know I can't use u substitution and I am pretty certain I have to algebraicly manipulate this before I can continue but I do not know how.

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Jordan , your integral should have + –  Theorem Jun 11 '12 at 21:29
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One big clue that your arc-length formula was wrong, incidentally, is that whenever $\tan x$ (or more generally, $y'$) was greater than one it would have required you to take the square root of a negative number! It's useful to keep an eye out for sanity checks to be able to pick up when you might be going down the wrong path. –  Steven Stadnicki Jun 11 '12 at 21:31
    
Oh I did not realize that. Also I wouldn't have noticed that it would have been negative anyways, I am not really familiar enough with trig to know the values at each x. –  user138246 Jun 11 '12 at 21:34
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@Jordan You don't need to know the trig - remember that the arc length formula in general doesn't have to have trig in it. If the formula had been $\int \sqrt{1-\left(y'\right)^2}$, (instead of the same formula with a plus sign, as many have pointed out) then anywhere that $y'\gt 1$ you'll run into a negative square root. –  Steven Stadnicki Jun 11 '12 at 21:36
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4 Answers

up vote 4 down vote accepted

The arclength formula is

$$\mathrm S_a^b(f) =\int_a^b \sqrt{1+f'(x)^2}dx$$

You have

$$f(x) = \log \sec x$$

This means

$$f'(x) = \tan x$$

Then you need to find

$$\mathrm S =\int_0^{\pi/4} \sqrt{1+\tan^2 x}dx$$

Remember that

$$1+\tan^2 x=\sec ^2 x$$

Also, remember the secant is positive in the first quadrant, so

$$\mathrm S =\int_0^{\pi/4} \sqrt{\sec^2 x}dx$$

$$\mathrm S =\int_0^{\pi/4} \sec xdx$$

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As others have noted, it should be

$$\int_0^{\pi/4} \sqrt{1+\tan^2 x} \ \ dx$$ Recall $1+\tan^2 x = \sec^2 x$ $$\int_0^{\pi/4} \sqrt{\sec^2 x} \ \ dx$$ Since all trig functions are positive in the first quadrant, we can simply rewrite the integrand as $$\int_0^{\pi/4} {\sec x} \ \ dx$$

Which is a (relatively) well known integral that evaluates to $$\log (\tan(x) + \sec (x))$$

Now simply evaluate at the endpoints - you should get around $.8814$ assuming I didn't make a button-punching error.

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You made a mistake. Arc length is given by the integral:

$$ \ell = \int_a^b \sqrt{1+\left(y'\right)^2}\,dx $$

So if $y' = \tan x$, arc length is:

$$ \ell = \int_a^b \sqrt{1+\tan^2 x}\,dx = \int_a^b \sqrt{\sec^2 x}\,dx = \int_a^b |\sec x| \,dx $$

For $a = 0$, $b = \frac{\pi}{4}$:

$$ \ell = \int_0^{\pi/4} \sec x \,dx $$

The absolute value is gone as $\sec x$ is positive in $[0, \frac{\pi}{4}]$.

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The arclength formula should be $\int \sqrt{1+\tan^2 x}\ dx$.

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