$x = a\sec\theta, dx = \sec\theta \tan\theta$

$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ = $ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ = $ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ = $\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$ = $\int \tan^{\frac{-5}{2}}\theta \sec\theta$

Here is where I get stuck...I tried converting $\tan\theta$ and $\sec\theta$ in terms of $\cos\theta$ and $\sin\theta$, but that didn't seem to get me anywhere...What is my next move from here? Did I even start this problem correctly? I can't tell :(

Update with more work after initial answers:

$\int \frac{\cos\theta}{\sin^2\theta}$ $u = \sin\theta, du = \cos\theta d\theta$

I found $\sin^{-1}\theta = \frac{\sqrt{x^2-1}}{x}$

$= \int \frac{du}{u^2} = \frac{1}{ \frac{1}{3}u^3} = \frac{1}{3\sin^3\theta} = 3 \bigg( \frac{x}{\sqrt{x^2-1}} \bigg)^3$

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    $\begingroup$ In the denominator you added the exponents instead of multiplying them. $\endgroup$ – John Wayland Bales Mar 8 '19 at 0:08
  • $\begingroup$ Why would I multiply them? I guess it is because it is an exponent of an exponent instead of an exponent * an exponent $\endgroup$ – Evan Kim Mar 8 '19 at 0:28
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    $\begingroup$ You should never keep the differentials by changing the variable. You should do it always together. The problem starts from very beginning, where you use both $\theta$ and $dx$. $\endgroup$ – user Mar 8 '19 at 1:00

$$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}} = \int \frac{\tan\theta\sec\theta d\theta}{\underbrace{(\tan^2\theta)^{\frac{3}{2}}}_{\tan^3\theta}} = \int \frac{\sec\theta d\theta}{\tan^2\theta}= -\frac1{\sin \theta}=-\frac{x}{\sqrt{x^2-1}}.$$

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  • $\begingroup$ How do you get -$\frac{1}{\sin\theta}$? Please see my update $\endgroup$ – Evan Kim Mar 8 '19 at 0:54
  • $\begingroup$ $$\int \frac{\sec\theta d\theta}{\tan^2\theta}=\int \frac{\cos\theta d\theta}{\sin^2\theta}=\int \frac{d\sin\theta}{\sin^2\theta}=-\frac1{\sin^2\theta}$$ $\endgroup$ – user Mar 8 '19 at 0:57
  • $\begingroup$ I don't understand, what is the $d\sin\theta$ notation mean? It doesn't look like regular u-substitution? $\endgroup$ – Evan Kim Mar 8 '19 at 0:58
  • $\begingroup$ For simplicity imagine that I introduced new variable $u=\sin\theta$. Thus $$\int\frac{d\sin\theta}{\sin^2\theta}=\int\frac{du}{u^2}=-\frac1u=-\frac1{\sin\theta}.$$ $\endgroup$ – user Mar 8 '19 at 1:04

Note that $$(\tan^2 x)^{3/2}=\tan^3 x$$

Thus, yous should instead obtain $$\int\frac1{(x^2-1)^{3/2}}dx =\int\frac{\sec t\tan t}{\tan^3 t}dt $$

which simplifies to $$\int\frac{\cos t}{\sin^2 t}=\int\frac1{\sin ^2 t}d(\sin t)=-\frac1{\sin t}+C=-\csc t+C$$

Now reverse the substitution by the identity $\csc^2 t=\frac1{1-\cos^2 t}=\frac1{1-x^{-2}}$.

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If you do $x=\sec\theta\,\mathrm d\theta$ and $\mathrm dx=\sec(\theta)\tan(\theta)\,\mathrm d\theta$, what you get is$$\int\frac{\sec(\theta)\tan(\theta)}{\tan^3\theta}\,\mathrm d\theta=\int\frac{\cos^2\theta}{\sin^3\theta}\,\mathrm d\theta=\int\frac{\cos^2(\theta)\sin(\theta)}{\bigl(1-\cos^2(\theta)\bigr)^2}\,\mathrm d\theta.$$Now, do $\cos\theta=u$ and $\sin\theta\,\mathrm d\theta=\mathrm du$.

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