MHB What is the length of a curve defined by a logarithmic function?

[karush]

$\begin{align*}\displaystyle
f(x)&=\ln(\sin{x})\\
\frac{\pi}{6}& \le x \le \frac{\pi}{2}\\
\end{align*}$
$\begin{align*}\displaystyle
f^\prime(x)&=\cot{x}
\end{align*}$
so
$\begin{align*}\displaystyle
L&=\int_{\pi/6}^{\pi/2} \sqrt{1-
(\cot{x})^2} \,dx \\
\therefore L&=\Biggr|-\ln(|\csc(x)+\cot(x)|)\Biggr|_{\pi/6}^{\pi/2}\\
&=\ln{\sqrt(3)+2}
\end{align*}$

ok I used W|A to get the indefinit Integral but didn't know the steps

 

[skeeter]

$\begin{align*}\displaystyle
f(x)&=\ln(\sin{x})\\
\frac{\pi}{6}& \le x \le \frac{\pi}{2}\\
\end{align*}$
$\begin{align*}\displaystyle
f^\prime(x)&=\cot{x}
\end{align*}$
so
$\begin{align*}\displaystyle
L&=\int_{\pi/6}^{\pi/2} \sqrt{1-
(\cot{x})^2} \,dx \\
\therefore L&=\Biggr|-\ln(|\csc(x)+\cot(x)|)\Biggr|_{\pi/6}^{\pi/2}\\
&=\ln{\sqrt(3)+2}
\end{align*}$

ok I used W|A to get the indefinit Integral but didn't know the steps

$\displaystyle L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \sqrt{1 + \cot^2{x}} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \sqrt{\csc^2{x}} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \csc{x} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \csc{x} \cdot \dfrac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \dfrac{\csc^2{x}+\csc{x}\cot{x}}{\cot{x}+\csc{x}} \, dx$

$\displaystyle L = -\int_{\pi/6}^{\pi/2} \dfrac{-\csc^2{x}-\csc{x}\cot{x}}{\cot{x}+\csc{x}} \, dx$

note the integrand has the form $\dfrac{u'}{u}$ ...

 

[Prove It]

$\displaystyle L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \sqrt{1 + \cot^2{x}} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \sqrt{\csc^2{x}} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \csc{x} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \csc{x} \cdot \dfrac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}} \, dx$

$\displaystyle L = \int_{\pi/6}^{\pi/2} \dfrac{\csc^2{x}+\csc{x}\cot{x}}{\cot{x}+\csc{x}} \, dx$

$\displaystyle L = -\int_{\pi/6}^{\pi/2} \dfrac{-\csc^2{x}-\csc{x}\cot{x}}{\cot{x}+\csc{x}} \, dx$

note the integrand has the form $\dfrac{u'}{u}$ ...

A slightly easier integral evaluation:

$\displaystyle \begin{align*} L &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\csc{(x)}\,\mathrm{d}x} \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{ \frac{1}{\sin{(x)}}\,\mathrm{d}x } \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{ \frac{\sin{(x)}}{\sin^2{(x)}}\,\mathrm{d}x } \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{ \frac{-\sin{(x)}}{\cos^2{(x)} - 1} \,\mathrm{d}x } \\ &= \int_{\frac{\sqrt{3}}{2}}^{0}{ \frac{1}{u^2 - 1}\,\mathrm{d}u } \textrm{ where } u = \cos{(x)} \implies \mathrm{d}u = -\sin{(x)}\,\mathrm{d}x \end{align*}$

which can now be solved with Partial Fractions.

 

[karush]

always learn cool tricks here at MHB