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$$\int \frac{dz}{z^3 \sqrt{z^2 - 4}}$$

$z = 4\sec\theta$

$dz = 4\sec\theta \tan d\theta$

$$\int \frac{\sec\theta \tan\theta}{4^3 \sec^3 \theta \tan \theta}$$

$$ \frac{1}{4^3} \int \frac{d \theta}{\sec^2 \theta}$$

I am again stuck, I have no idea how to proceed.

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Use trig identities. – Ataraxia Jul 3 '13 at 1:51
Euler's subtitutions. – Mlazhinka Shung Gronzalez LeWy Jul 3 '13 at 1:51
$\frac{1}{\sec\theta} = \cos(\theta)$.. – Cameron Williams Jul 3 '13 at 1:51

First step. Erase that trigonometric substitution. Then you can make $\sqrt{z^2-4}=z+t$, which is the first Euler's substitution.

the second Euler substitution also work. This would mean to make $\sqrt{z^2-4}=(z-2)t$.

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If I do this how do I find the value of cos theta – Dantheman Jul 3 '13 at 2:00
@Dantheman see my edit. – Ataraxia Jul 3 '13 at 2:30
@ZettaSuro, we need $z^2-4=4\tan^2\theta\implies z^2=4(1+\tan^2\theta)=4\sec^2\theta\implies z=\pm2\sec\theta$ – lab bhattacharjee Jul 3 '13 at 3:37

Putting $z=2\sec\theta, dz=2\sec\theta\tan\theta d\theta$

$$\int \frac{dz}{z^3\sqrt{z^2-4}}=\int \frac{2\sec\theta\tan\theta d\theta}{(2\sec\theta)^32\tan\theta}=\frac18\int\cos^2\theta d \theta=\frac1{16}\int(1+\cos2\theta)d\theta$$


Now, $\cos\theta =\frac2z\implies \theta =\arccos \frac2z $

$\implies \sin\theta=\sqrt{1-\left(\frac2z\right)^2}=\frac{\sqrt{z^2-4}}z$

$\implies \sin2\theta=2\sin\theta\cos\theta=2\cdot \frac{\sqrt{z^2-4}}z\cdot \frac2z$

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Putting $z^2-4=u^2, zdz=udu$

$$\int \frac{dz}{z^3\sqrt{z^2-4}}=\int \frac{udu}{(u^2+4)^2u}=\frac{du}{(u^2+2^2)^2}$$

This can be addressed using Trigonometric identities $u=2\tan\theta$ like my other answer

Alternatively, $$\text{Let }I_n=\int \frac {du}{(u^2+a^2)^n}$$

$$=\int du\cdot\frac1{(u^2+a^2)^n}-\int\left(\int du \frac{d\frac1{(u^2+a^2)^n}}{du}\right)du $$

$$=\frac u{(u^2+a^2)^n}+2n\int \frac{u^2}{(u^2+a^2)^{n+1}}du$$

$$=\frac u{(u^2+a^2)^n}+2n\left(\int \frac1{(u^2+a^2)^n}-a^2\int \frac1{(u^2+a^2)^{n+1}} du\right)$$

$$\implies I_n=\frac u{(u^2+a^2)^n}+2n\left(I_n-a^2I_{n+1}\right)$$

$$\implies 2a^2n\cdot I_{n+1}=\frac u{(u^2+a^2)^n}+(2n-1)I_n $$

$$n=1\implies 2a^2 \int \frac {du}{(u^2+a^2)^2}=\frac u{(u^2+a^2)}+\int \frac {du}{(u^2+a^2)}=\frac u{(u^2+a^2)}+\frac1a\cdot \arctan \frac ua+K$$

$$a=2\implies 8 \int \frac {du}{(u^2+4)^2}=\frac u{(u^2+4)}+\int \frac {du}{(u^2+4)}=\frac u{(u^2+4)}+\frac12\cdot\arctan \frac u2+K$$

Now, $u=\sqrt{z^2-4}, u^2+4=z^2$

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