d/dx ln(cosx) = -tanx
1 + (dy/dx)^2 = 1 + tan^2x = sec^2x
∫√sec^2x = ∫secx = secx tanx
all of those answers seem to indicate that you are doing
∫ tanx dx = -ln(cosx)
or something...
Did I miss something?
What is the length of the arc of the curve y = ln (cos(x)) for 0 is less than or equal to x and x is less than or equal to pi over 3 ? (2 points)
A) the natural log of the quantity 1 plus the square root of 3
B) the natural log of the quantity 2 plus the square root of 3
C) the natural log of the quantity 3 plus the square root of 2
D) the natural log of the quantity 3 plus the square root of 3
3 answers
I think you interpreted it wrong. The answers are...
A) In ( 1+ square root of 3 )
B) In ( 2 + square root of 3 )
C) In ( 3 + square root of 2 )
D) In (3 + square root of 3 )
A) In ( 1+ square root of 3 )
B) In ( 2 + square root of 3 )
C) In ( 3 + square root of 2 )
D) In (3 + square root of 3 )
yes, I see that. But my integral has the value 2√3
so where did those logs come from?
so where did those logs come from?