The following solutions are for the problems on the arc length integral formula here. I encourage you to attempt them by yourself first before looking through the solutions.

Problem 1

The function $f(x)=\cosh x$ is called a hyperbolic cosine function, and is defined by the equation

$$f(x)=\cosh x=\frac{e^x+e^{-x}}{2}$$

Given this information, calculate the length of the function $f(x)=\cosh x$ from $x=0$ to $x=1$.

The derivative of this function $f(x)$ is $\frac{df}{dx}=\frac{e^x-e^{-x}}{2}$, and so using the arc length integral formula, we have

\[D=\int_0^1dx\sqrt{1+\frac{e^{2x}+e^{-2x}}{4}-\frac{1}{2}}=\frac{1}{2}\int_0^1dx\sqrt{e^{2x}+2+e^{-2x}}\]

If we factor out a $e^{-x}$ factor from the radical, we find

\[D=\frac{1}{2}\int_0^1dxe^{-x}\sqrt{e^{4x}+2e^{2x}+1}=\frac{1}{2}\int_0^1dxe^{-x}\sqrt{(e^{2x}+1)^2}\]

Simplifying gives

\[D=\frac{1}{2}\int_0^1dxe^{-x}(e^{2x}+1)=\int_0^1dx\frac{e^x+e^{-x}}{2}\]

Evaluating the integral expression on the right hand side gives

$$D=\left.\frac{e^x-e^{-x}}{2}\right\vert_0^1=\frac{e}{2}-\frac{1}{2e}\approx 1.407$$

Problem 2

Calculate the length of the function $f(x)=\ln(\cos x)$ from $x=0$ to $x=\frac{\pi}{4}$.

Denote the arc length we’re trying to calculate as $D$. Using the arc length integral formula and the face that $\frac{df}{dx}=-\frac{\sin x}{\cos x}=-\tan x$, we have

\[D=\int_0^{\pi/4}dx\sqrt{1+\tan^2x}=\int_0^{\pi/4}dx\sec x\]

Using the result for the integral of $\sec x$ derived earlier here, we have

\[D=\ln\vert\sec\theta+\tan\theta\vert_0^{\pi/4}\]

Evaluating this expression gives

$$D=\ln(\sqrt{2}+1)-\ln(1+0)=\ln(1+\sqrt{2})\approx0.881$$

Problem 3

Calculate the length of the function $f(x)=x^2-\frac{1}{8}\ln x$ from $x=1$ to $x=2$.

Let’s denote the arc length as $D$. Using the arc length integral formula and the fact that \(\frac{df}{dx}=2x-\frac{1}{8x}\), we have

\[D=\int_1^2dx\sqrt{1+4x^2+\frac{1}{64x^2}-\frac{1}{2}}=\int_1^2dx\sqrt{4x^2-\frac{1}{2}+\frac{1}{64x^2}}\]

Factoring the integrand gives

\[D=\int_1^2\frac{dx}{8x}\sqrt{256x^4-32x^2+1}=\int_1^2\frac{dx}{8x}\sqrt{(16x^2-1)^2}\]

Simplifying gives

\[D=\int_1^2dx\frac{16x^2-1}{8x}=2\int_1^2dx\text{ }x-\frac{1}{8}\int_1^2\frac{1}{x}\]

Evaluating the integral expressions on the right hand side gives

\[D=x^2\vert_1^2-\frac{1}{8}\ln \vert x\vert_1^2=\left(4-1\right)-\frac{1}{8}\left(\ln 2-0\right)\]

Simplifying gives

$$D=3-\frac{1}{8}\ln2\approx 2.913$$

Problem 4

This is a challenge problem.

Calculate the length of the function $f(x)=e^x$ from $x=0$ to $x=1$.

Let’s denote the requested length as $D$. Using the arc length integral formula and the fact that $\frac{df}{dx}=e^x$, we have

\[D=\int_0^1dx\text{ }\sqrt{1+e^{2x}}\]

Notice that we can rewrite this integral as

\[D=\int_0^1dx\text{ }e^{x}\sqrt{1+e^{-2x}}\]

This is achieved through factoring out a factor of $e^x$

Let’s define the variable $u=1+e^{-2x}$, such that $du=-2e^{-2x}dx$, or equivalently, $dx=\frac{-1}{2}e^{2x}du$. However, notice from the definition of $u$ that $\frac{1}{u-1}=e^{2x}$, and so

\[dx=\frac{-1}{2}\frac{1}{u-1}du\]

Furthermore, the lower bound becomes $u_{\downarrow}=1+e^{-2(0)}=2$, and the upper bound becomes $u_{\uparrow}=1+e^{-2(1)}=1+e^{-2}$. Note also that since $\frac{1}{u-1}=e^{2x}$, this means that $\frac{1}{\sqrt{u-1}}=e^x$. Therefore, combining all of these results and making the $u$-substitution gives

\[D=-\frac{1}{2}\int_2^{1+e^{-2}}\frac{du\sqrt{u}}{(u-1)^{3/2}}\]

Now, we can define the variable $w=\sqrt{u}$ such that $dw=\frac{1}{2\sqrt{u}}du$, and so $du=2\sqrt{u}dw$. However, notice that $\sqrt{u}=w$ by the definition of $w$, and so $du=2wdw$. This means that we can rewrite the integrand expression as

\[\frac{du\sqrt{u}}{(u-1)^{3/2}}=\frac{2w^2dw}{(w^2-1)^{3/2}}\]

Similarly, the bounds of integration become $w_{\downarrow}=\sqrt{2}$ and $w_{\uparrow}=\sqrt{1+e^{-2}}$. Therefore, the integral expression for $D$ following this $w$-substitution becomes

\[D=-\int_{\sqrt{2}}^{\sqrt{1+e^{-2}}}\frac{w^2 dw}{(w^2-1)^{3/2}}\]

We see the expression $w^2-1$ in the denominator of the integrand, and so this hints to us that we might benefit from using an appropriate trigonometric substitution, choosing $w=\sec\theta$ such that $w^2-1=\tan^2\theta$ and $w^2=\sec^2\theta$. Furthermore, $dw=\sec\theta\tan\theta d\theta$, and the bounds of integration become $\theta_{\downarrow}=\sec^{-1}(\sqrt{2})$ and $\theta_{\uparrow}=\sec^{-1}(\sqrt{1+e^{-2}})$. Making this third substitution gives

\[D=-\int_{\sec^{-1}(\sqrt{2})}^{\sec^{-1}(\sqrt{1+e^{-2}})}\frac{\sec^3\theta\tan\theta d\theta}{\tan^3\theta}=-\int_{\sec^{-1}(\sqrt{2})}^{\sec^{-1}(\sqrt{1+e^{-2}})}\frac{d\theta}{\sin^2\theta\cos\theta}\]

From this point, the problem essentially boils down to using the correct trigonometric identities. Note that

\[\frac{1}{\sin^2\theta\cos\theta}=\csc^2\theta\sec\theta=(\cot^2\theta+1)\sec\theta\]

Expanding out the right hand side gives

\[\frac{1}{\sin^2\theta\cos\theta}=\sec\theta+\cot\theta\csc\theta\]

Plugging this back into our integral expression for $D$ gives

\[D=-\int_{\sec^{-1}(\sqrt{2})}^{\sec^{-1}(\sqrt{1+e^{-2}})}d\theta\text{ }\sec\theta-\int_{\sec^{-1}(\sqrt{2})}^{\sec^{-1}(\sqrt{1+e^{-2}})}d\theta\cot\theta\csc\theta\]

We found the integral of $\sec\theta$ here, and since $\frac{d}{dx}[\csc\theta]=-\csc\theta\cot\theta$, we know that the integral of $\csc\theta\cot\theta$ is $-\csc \theta$. Therefore,

\[D=\left[-\ln\vert \sec\theta+\tan\theta\vert+\csc\theta\right]_{\sec^{-1}(\sqrt{2})}^{\sec^{-1}(\sqrt{1+e^{-2}})}\]

At this point, the easiest way to simplify this numerical value is probably to use a calculator at this point. You should get that the final answer is

$$D=1-\sqrt{2}+\sqrt{1+e^2}+\sinh^{-1}1-\ln\left(1+\sqrt{1+e^2}\right)\approx 2.004$$

where $\sinh(x)$ is the hyperbolic sine function.