2. In this problem, let f(r) = In| sec r|.%3D(a) Graph y = f(x). Write down a definite integral that computes the arclength of the graph of f(r) from r =how you took the derivative of f(x).0 to r = T/4. Be sure to carefully explain(b) Compute the definite integral in part (a). Be sure to mention any trig identities you use and to discuss the sign of the integrand. You will need toknow that f sec rdr = In | sec z+ tan r|+ C.%3D

Name: 2. In this problem, let f(r) = In| sec r|.%3D(a) Graph y = f(x). Writ
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Description: 2. In this problem, let f(r) = In| sec r|.%3D(a) Graph y = f(x). Write down a definite integral that computes the arclength of the graph of f(r) from r =how you took the derivative of f(x).0 to r = T/4. Be sure to carefully explain(b) Compute the de

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2. In this problem, let f(r) = In| sec r|.
%3D
(a) Graph y = f(x). Write down a definite integral that computes the arclength of the graph of f(r) from r =
how you took the derivative of f(x).
0 to r = T/4. Be sure to carefully explain
(b) Compute the definite integral in part (a). Be sure to mention any trig identities you use and to discuss the sign of the integrand. You will need to
know that f sec rdr = In | sec z+ tan r|+ C.
%3D

Expert Solution

Step 1

Part(a)

The given function is,

$\begin{array}{rcl} f (x) & = & \ln |s e c x| \\ f (x) & = & - \ln |\cos x| \end{array}$

Now, differentiate the given function with respect to x up to first order.

$\begin{array}{rcl} f' (x) & = & \frac{- 1}{\cos x} \times \sin x \\ = & - \tan x \end{array}$

Step 2

So, the arclength of the given function from 0 to $\frac{π}{4}$ is given by the formula,

$\int_{a}^{b} \sqrt{1 + {(f (x))}^{2} d x}$

Substitute $- \tan x$ for $f (x)$ into the formula $\int_{a}^{b} \sqrt{1 + {(f (x))}^{2} d x}$ .

$\begin{array}{rcl} \int_{a}^{b} \sqrt{1 + {(f (x))}^{2}} & = & \int_{0}^{\frac{π}{4}} \sqrt{1 + {(- \tan x)}^{2}} d x \\ = & \int_{0}^{\frac{π}{4}} \sqrt{s e c^{2} x} d x \\ = & \int_{0}^{\frac{π}{4}} s e c x d x \\ [Since, \sec^{2} x = 1 + \tan^{2} x] \end{array}$

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