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Find the length of the curve $r=\sqrt{1 + \cos (2\theta)}, 0\le\theta\le\frac{\pi\sqrt{2}}{4}$.

I generated the integral: $\int_0^{\frac{\pi\sqrt{2}}{4}} {\sqrt{2 - \cos (2\theta)}}\text{d}\theta$

Is it correct? How would I solve this integral?

The length of the curve $=\frac{\pi}{2}$.

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We know that $ x=r\cos \theta=\sqrt{1+\cos 2\theta}\cos \theta=\sqrt{2} \cos^2 \theta$
and $ y=r\sin \theta=\sqrt{1+\cos 2\theta}\sin \theta=\sqrt{2} \sin\theta \cos \theta$

So $ \dfrac{dy}{dx}= \dfrac{dy}{d\theta}\cdot\dfrac{d\theta}{dx}=-\cot 2\theta$

Therefore the required arc length is \begin{align} & \int_{{\pi\sqrt{2}}/{4}}^0 \sqrt{1+\Big(\dfrac{dy}{dx}\Big)^2}dx \\ &= \int_{{\pi\sqrt{2}}/{4}}^0 \sqrt{1+\cot^2 2\theta}\quad(-\sqrt2\sin2\theta) d\theta\\ &=-\sqrt2 \int_{{\pi\sqrt{2}}/{4}}^0 dx \\ &=\frac{\pi}{2} \end{align}

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Arc length is given by $$ L =\int \sqrt{r^2+r^{'2} }d \theta$$ Now $$ r=\sqrt 2 \cos \theta;\,r'=-\sqrt 2 \sin \theta;\,\sqrt{r^2+r^{'2} }= \sqrt 2$$ $$L= \sqrt 2 \int_0^{\frac{\pi\sqrt{2}}{4} }d \theta =\sqrt 2\cdot {\frac{\pi\sqrt{2}}{4} }= {\frac{\pi}{2} }.$$

This is an arc of circle of diameter $\sqrt 2$ subtending $ {\dfrac{\pi\sqrt{2}}{4} }$ radians of angle at the origin and double that $ =\dfrac{\pi}{\sqrt{2}}$ radians at circle center.

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Note $r=\sqrt{1 + \cos (2\theta)}=\sqrt2 \cos\theta$, which is a circle with the radius $\frac1{\sqrt2}$ and the range $0\le\theta\le\frac{\pi\sqrt{2}}{4}$ spans a circle sector of angle $\alpha=\frac{\pi\sqrt{2}}{2}$. Thus, the circumference length of the sector is

$$\alpha r = \frac{\pi\sqrt{2}}{2}\cdot \frac1{\sqrt2}=\frac\pi2 $$

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    $\begingroup$ Shouldn't it be $\sqrt{1+ \cos (2\theta)}=\sqrt{2} \cos \theta$ for $0 \leq \theta \leq \frac{\pi \sqrt{2}}{4}$? $\endgroup$ – Dunkelheit Apr 29 '20 at 20:36
  • $\begingroup$ @Dunkelheit correct thx for spotting it $\endgroup$ – Quanto Apr 29 '20 at 20:54

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