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we have to calculate the arc length of the following function $y=\sqrt{(\cos2x)} dx$ in the interval $[0 ,\pi/4]$. I know the arc length formula but following it becomes an integral thats really complex...need help....

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I got the following integral $$\int_0^{\frac{\pi }{4}} \sqrt{\sin (2 x) \tan (2 x)+1} \, dx$$ no hope that this has an algebraic solution.

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  • $\begingroup$ yea this is where i get stuck . So basically this wont be solved by the usual integration ? what other methods are there? $\endgroup$ Dec 12, 2019 at 14:56
  • $\begingroup$ I would use a numerical method $\endgroup$ Dec 12, 2019 at 14:59
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$\int_{0}^{\frac{\pi}{4}} \sqrt{\cos2x} dx = \frac{1}{2}\int_{0}^{\frac{\pi}{4}} \sqrt{\cos{u}} du = \frac{1}{2}\cdot \:2\text{E}\left(\frac{u}{2}|\:2\right) = \frac{1}{2}\cdot \:2\text{E}\left(\frac{2x}{2}|\:2\right) = \text{E}\left(x|2\right)+C$, where $\text{E}\left(x|m\right)$ is the elliptic integral of the second kind. See: https://math.stackexchange.com/a/19786/733593

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    $\begingroup$ This is not the arc length. The arc length of $f$ in $[a,b]$ is given by the integral $\int_a^b \sqrt{1+(f')^2}\,dx$. $\endgroup$ Dec 12, 2019 at 16:11

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