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August 30, 2014

August 30, 2014

Posted by **sammyg** on Tuesday, January 24, 2012 at 9:46pm.

y = ln (sin(x))

,[π/4,3π/4]

- calculus -
**Steve**, Wednesday, January 25, 2012 at 1:15pmy = ln sinx

y' = 1/sinx * cosx = tanx

s = Int(sqrt(1+(y')^2)dx)[pi/4,3pi/4]

= Int(sqrt(1+tan^2(x))dx)[pi/4,3pi/4=

= Int(secx dx)[pi/4,3pi/4]

= ln|secx + tanx|[pi/4,3pi/4]

= ln|-1/√2 + 1| - ln|1/√2 + 1|

= ln|(1-√2/(1+√2)|

= ln|2√2-3|

= ln(3-2√2)

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