Find length of curve. Integral of sec?

phantomcow2
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Homework Statement


Find the length of the curve y=ln(cosx) from 0 to pi/3

Homework Equations


integral of (1+(y')^2)^1/2

The Attempt at a Solution


First, find y'. y' is equal to -sin/cos, or simply -tan(x).

\int\sqrt{1+(-tan(x)^{2}}
= \int\sqrt{sec(x}

Homework Statement



Can this possibly be right so far? This is a horrendous integral. I've expended so much energy on this problem, but if someone can at least validate that I am on the right track, I'll post my work from here out.
 
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phantomcow2 said:
\int\sqrt{1+(-tan(x)^{2}}
= \int\sqrt{sec(x}

1+tan^2x=sec^2x \Rightarrow \sqrt{1+tan^2x}= \sqrt{sec^2x}=secx
 
:redface::cry:

Thank you so much.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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