# Given that $x=\dfrac 1y$, show that $∫\frac {dx}{x \sqrt{(x^2-1)}} = -∫\frac {dy}{\sqrt{1-y^2}}$

Given that $x=\dfrac 1y$, show that $\displaystyle \int \frac 1{x\sqrt{x^2-1}}\,dx = -\int \frac 1{\sqrt{1-y^2}}\,dy$

Have no idea how to prove it.

here is a link to wolframalpha showing how to integrate the left side.

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Substitute $x=1/y$ and $dx/dy=-1/y^2$ to get $\int y^2/(\sqrt{1-y^2}) (-1/y^2)dy$

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Let's treat this like a $u$-substitution problem.

$$\int \frac 1{x\sqrt{x^2-1}}\,dx$$

Let $u = \frac{1}{x}$, so that $du = \frac{-1}{x^2}dx$. Then

$$\int \frac {x}{x^2\sqrt{x^2-1}}dx = -\int \frac {\frac{1}{u}}{\sqrt{\frac{1}{u^2} - 1}}du = -\int \frac{1}{u\sqrt{1/u^2 - 1}}du = -\int \frac{1}{\sqrt{1 - u^2}}du$$

I just happen to use $u$ because that's how I teach it, but calling it $y$ is just the same. Most importantly, there is no need to actually integrate anything.

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Given that $x=1/y$

Put $x = \dfrac{1}{y},dx = \dfrac{-1}{y^2}.dy$:

$$\int \dfrac{1}{(x\sqrt{(x^2-1)})}dx$$ $$=\int \dfrac{1}{(\frac{1}{y} \sqrt{((\frac{1}{y})^2-1)})}\cdot\dfrac{-1}{y^2}dy$$

$$\int \dfrac{1}{\frac{1}{y} \sqrt{\frac{1-y^2}{y^2}}}\cdot \dfrac{-1}{y^2}dy$$