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Let $x=\sec\theta$, so $\mathrm{d}x=\tan\theta \sec\theta\, \mathrm{d}\theta$. Then $$\int {\mathrm{d}x \over x^2\sqrt{x^2 -1}}=\int {{\tan\theta \sec\theta} \over {\sec^2\theta \sqrt{\sec^2\theta -1}}}\, \mathrm{d}\theta = \int {\tan\theta \over {\sec\theta \sqrt{\sec^2\theta -1}}}\,\mathrm{d}\theta.$$

I really don't know how to continue this. I was thinking about u-sub, where $u=\tan\theta$ and $\mathrm{d}u=\sec^2\theta\,\mathrm{d}\theta$, but then I don't know what to do with the term $\sec\theta$.

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  • $\begingroup$ Also, $sec^2(\theta )-1=tan^2(\theta )$ $\endgroup$ – Cbjork May 2 '15 at 20:39
  • $\begingroup$ The solution actually depends upon how $\theta$ is dependent on x or vice versa. Is there any other information given? If no , just take it out as constant. $\endgroup$ – Mann May 2 '15 at 20:42
  • $\begingroup$ Wrote it wrong originally, it was supposed to be x^2 not sec^2theta $\endgroup$ – Jay May 2 '15 at 20:43
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Set $x=\cosh(t)$, we then have $dx = \sinh(t)dt$. Hence, the integral is $$\int \dfrac{\sinh(t)dt}{\cosh^2(t)\sinh(t)} = \int \text{sech}^2(t)dt = \tanh(t) + \text{constant}$$


If you want to use $x=\sec(t)$, we then have $dx = \tan(t)\sec(t)dt$. Hence, the integral becomes \begin{align} I & = \underbrace{\int \dfrac{\tan(t)\sec(t)dt}{\sec^2(t) \sqrt{\sec^2(t)-1}} = \int \dfrac{\tan(t)\sec(t)dt}{\sec^2(t) \tan(t)}}_{\sec^2(t) = 1+\tan^2(t)} = \int \cos(t)dt = \sin(t) + \text{constant}\\ & = \sqrt{1-1/x^2} + \text{constant} \end{align}

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use the change of variable $$u = \frac 1x, x = \frac 1u, dx = -\frac1{u^2} \, du$$ then $$\int \frac{dx}{x^2\sqrt{1-x^2}} = -\int\frac{du}{u^2x^2\sqrt{1/u^2-1}}= -\int \frac {u\,du}{\sqrt{1-u^2}}=\sqrt{1-u^2}={\sqrt{1-1/x^2}}$$

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Your substitution is right. You have $$\int\frac{\tan\left(\theta\right)}{\sec\left(\theta\right)\sqrt{\sec^{2}\left(\theta\right)-1}}d\theta=\int\cos\left(\theta\right)d\theta $$ because $\sqrt{\sec^{2}\left(\theta\right)-1}=\tan\left(\theta\right) $, then $$\int\frac{1}{x^{2}\sqrt{x^{2}-1}}dx=\sin\left(\sec^{-1}\left(x\right)\right)+C $$ and using $$\sin\left(\sec^{-1}\left(x\right)\right)=\sqrt{1-\frac{1}{x^{2}}} $$ we have $$\int\frac{1}{x^{2}\sqrt{x^{2}-1}}dx=\sqrt{1-\frac{1}{x^{2}}}+C. $$

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Hint: $\frac{1}{sec\theta }=cos\theta$

So: $$\int\frac{tan\theta \:\:\:\:d\theta \:\:}{sec\theta \:\:\:\sqrt{sec^2\theta \:\:\:-1}}=\int\:cos\theta\:d\theta=sin\theta+C=\frac{\sqrt{x^2-1}}{x}+C=\sqrt{1-\frac{1}{x^2}}+C$$

Where you don't understand ask me.

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