Question

Find the arc length of the curve y = ln (cos x) for 0 < x < pi/4

Answer 1

\[\int_{0}^{\pi/4}\sqrt{1+[f'(x)]^2}dx\]

Answer 2

ln(cos(x))' = -\(\cfrac{sin(x)}{cos(x)}\)

Answer 3

or simply: -tan(x)

\[\int_{0}^{\pi/4}\sqrt{1+[-tan(x)]^2}dx\]
\[\int_{0}^{\pi/4}\sqrt{sec^2(x)}dx\]
\[\int_{0}^{\pi/4}sec(x)dx\]

Answer 4

and sec(x) has a naughty little secret ...

Answer 5

\[ \int_{0}^{\pi/4}sec(x)dx=\int_{0}^{\pi/4}sec(x)\frac{sec(x)+tan(x)}{sec(x)+tan(x)}dx\]
\[ \int_{0}^{\pi/4}\frac{sec^2(x)+sec(x)tan(x)}{sec(x)+tan(x)}dx\]
\[ \int_{0}^{\pi/4}\frac{sec^2(x)+sec(x)tan(x)}{tan(x)+sec(x)}dx\]
now the top is the derivative of the bottom and its just an ln|u| up

Answer 6

\[ln|tan(\pi/4)+sec(\pi/4)|-ln|tan(0)+sec(0)|\]
\[ln|1+\sqrt{2}|-ln|0+1|\]
\[ln(1+\sqrt{2})\]
maybe?