Question

Determine the length of $y=log\left(secx\right)$ between $0\le x\le \frac{\pi }{4}$.

• A. $log(\sqrt{2}-1)$
• B. $log(\sqrt{3}+1)$
• C. $log(\sqrt{2}+1)$
• D. $log(\sqrt{3}-1)$

Answer 1

Answer: (C)

$log(\sqrt{2}+1)$

Given $f\left(x\right)=y=log\left(secx\right)$ on $[0,\frac{\pi }{4}]$
The arc length is given by $L={\int }_a^b\sqrt{\left(f^{\prime }\right(x){)}^2+1}dx$
We have $f\left(x\right)=log\left(secx\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{secx}\left(secxtanx\right)=tanx$
$\therefore L={\int }_0^{\pi /4}\sqrt{(tanx{)}^2+1}dx$
$={\int }_0^{\pi /4}\sqrt{ta{n}^{2}x+1}dx$
$={\int }_0^{\pi /4}\sqrt{se{c}^{2}x}dx$ ($\because se{c}^{2}x-ta{n}^{2}x=1$)
$={\int }_0^{\pi /4}secxdx$
$=\left[log\right(tanx+secx){]}_0^{\pi /4}$
$=\left[log\right(tan\left(\pi /4\right)+sec\left(\pi /4\right)\left)\right]-\left[log\right(tan0+sec0\left)\right]$
$=\left[log\right(1+\sqrt{2}\left)\right]-\left[log\right(1\left)\right]$
$=log(1+\sqrt{2})$
$\therefore L=log(\sqrt{2}+1)$