# arc length question calculus [closed]

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....

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.

• yea this is where i get stuck . So basically this wont be solved by the usual integration ? what other methods are there? Dec 12, 2019 at 14:56
• I would use a numerical method Dec 12, 2019 at 14:59

$$\int_{0}^{\frac{\pi}{4}} \sqrt{\cos2x} dx = \frac{1}{2}\int_{0}^{\frac{\pi}{4}} \sqrt{\cos{u}} du = \frac{1}{2}\cdot \:2\text{E}\left(\frac{u}{2}|\:2\right) = \frac{1}{2}\cdot \:2\text{E}\left(\frac{2x}{2}|\:2\right) = \text{E}\left(x|2\right)+C$$, where $$\text{E}\left(x|m\right)$$ is the elliptic integral of the second kind. See: https://math.stackexchange.com/a/19786/733593

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