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Show that the area bounded by the curves $y=\frac{1}{x}$, $y=mx$ and $y=2mx$ is the same for every $m\geq 0$.

The book suggests to use polar coordinates. Therefore using $x=r \cos\theta$ $y=r \sin\theta$ we transform the curve $y=\frac{1}{x}$ to $r^2=\frac{1}{\sin\theta \cos\theta}$.

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So basically we have to integrate the curve $r^2=\frac{1}{\sin\theta \cos\theta}$ between the angles $\tan^{-1}m$ and $\tan^{-1}2m$. Therefore the shaded area is $$\int\limits_{\tan^{-1}m}^{\tan^{-1}2m}\frac{1}{2\sin\theta \cos\theta} d\theta$$

Am I right so far? and most importantly how do I evaluate the above integral?

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    $\begingroup$ The area in polar coordinates is (see here) $$\frac{1}{2}\int r(\theta)^2 d\theta$$ $\endgroup$ – Wouter Aug 12 '16 at 8:05
  • $\begingroup$ @Wouter Yes that is exactly the formula I used to get the integral $\endgroup$ – Miz Aug 12 '16 at 8:07
  • $\begingroup$ Nothing contradictory : it's what the OP has done. $\endgroup$ – Jean Marie Aug 12 '16 at 8:07
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$$\int\limits_{\tan^{-1}m}^{\tan^{-1}2m}\frac{d \theta}{2\sin\theta \cos\theta}=\int\limits_{\tan^{-1}m}^{\tan^{-1}2m}\frac{d \theta}{\sin 2\theta}=\dfrac{1}{2}[\ln{\tan{\theta}}]_{\tan^{-1}m}^{\tan^{-1}2m}$$

$$=\dfrac{1}{2}(\ln{2m}-\ln{m})=\dfrac{1}{2}\ln{2}$$

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If this is true, considering $$I(m)=\int\limits_{\tan^{-1}(m)}^{\tan^{-1}(2m)}\frac{d\theta}{2\sin(\theta) \cos(\theta)}$$ show that $$\frac{dI(m)}{dm}=0$$ using the fundamental theorem of calculus.

If $$J(m)=\int\limits_{\tan^{-1}(m)}^{\tan^{-1}(2m)}f(\theta)\,d\theta$$ then $$\frac{dJ(m)}{dm}=\frac{2 f\left(\tan ^{-1}(2 m)\right)}{4 m^2+1}-\frac{f\left(\tan ^{-1}(m)\right)}{m^2+1}$$ remembering that $$\sin(\tan^{-1}(x))=\frac{x}{\sqrt{x^2+1}}\qquad , \qquad \cos(\tan^{-1}(x))=\frac{1}{\sqrt{x^2+1}}$$

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