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$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$

So, make the substitution

$ x = \sqrt{a \sec \theta}$, which simplifies to $a \tan \theta$.

$2x = \sqrt{1} \sec \theta$,

$ d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$

$\int \dfrac{\sqrt{1}\tan\theta}{(\sqrt{1}\sec\theta)^3} d\theta$

Am I making the correct substitutions here? Substituting $d\theta$ a quantity of $(\sqrt{1}\sec\theta)$ will cancel from the denominator. Somewhere along the line I need to use the identity $\sin(2\theta)=2\sin(\theta)\cos(\theta).$

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    $\begingroup$ Check again, the result is $4\int\sin^2\theta\ d\theta$. $\endgroup$
    – Nosrati
    Commented Oct 3, 2018 at 2:54
  • $\begingroup$ Care to explain how $x = \sqrt{a}\sec\theta $ gets me to that point? $\endgroup$
    – DJ2
    Commented Oct 3, 2018 at 3:24
  • $\begingroup$ It is $x=\sqrt{a}\sec\theta$. $\endgroup$
    – Nosrati
    Commented Oct 3, 2018 at 3:26
  • $\begingroup$ Ah, yes @Nosrati $\endgroup$
    – DJ2
    Commented Oct 3, 2018 at 3:28

4 Answers 4

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With the substitution $x=\frac {\sec \theta }{2}$ you get $dx = \frac {\sec \theta \tan \theta }{2} d\theta $ and the integral changes to $$ \int \frac {4\tan^2 \theta \sec \theta }{ \sec ^3 \theta } d\theta =4 \int \frac {\tan^2 \theta }{ \sec ^2 \theta } d\theta =4 \int \sin ^2 \theta d\theta$$

Now you can use the double angle equality which you mentioned.

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  • $\begingroup$ Should be: $dx = \frac{1}{2} \sec \theta \tan \theta d \theta$. $\endgroup$
    – JavaMan
    Commented Oct 3, 2018 at 3:19
  • $\begingroup$ Can you explain what to do after substituting the double angle identity? $\endgroup$
    – DJ2
    Commented Oct 3, 2018 at 3:34
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    $\begingroup$ $\sin ^2 \theta =(1-\cos( 2\theta ))/2 $ and integrate the result. $\endgroup$ Commented Oct 3, 2018 at 10:39
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Take $2x = \cos(\theta)$, then you will get

$$\int\frac{\sqrt{\cos^2(\theta)-1}}{\cos^3(\theta)} d\theta= \int i\frac{\sin(\theta)}{\cos^3(\theta)} d\theta$$

then take another substitution. $t=\cos(\theta)$ to finally get : $$=\frac{i}{8x^2} + c$$

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    $\begingroup$ $\cos^2-1\neq\sin^2$. $\endgroup$
    – Nosrati
    Commented Oct 3, 2018 at 3:07
  • $\begingroup$ @Nosrati. Yes you are right. Typo on my side. Thanks for notifying. Edit made $\endgroup$
    – SJa
    Commented Oct 3, 2018 at 3:18
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    $\begingroup$ Presumably this is question is for a calculus course, so we are (probably) tacitly assuming all quantities should be real. $\endgroup$
    – JavaMan
    Commented Oct 3, 2018 at 3:20
  • $\begingroup$ @JavaMan Correct. $\endgroup$
    – DJ2
    Commented Oct 3, 2018 at 3:21
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Hint:

If trigonometric substitution is not mandatory, for (where $a$ is an arbitrary constant )

$$\dfrac{\sqrt{4x^2+a}}{x^{2n-1}}=4^{n-1}\dfrac{\sqrt{4x^2+a}}{(4x^2)^n}\cdot4x$$

set $\sqrt{4x^2+a}=y,4x^2+a=y^2,y\ dy=4x\ dx$

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Setting $x=\frac{\sqrt{\theta^{2}+1}}{2}$

$dx=\frac{\theta}{2\sqrt{\theta^{2}+1}}$,

and the funcion to integrate is:

$\frac{4\theta^{2}}{(\theta^{2}+1)^{2}}$,

whose integral is:

$2. atan(\theta)-\frac{2\theta}{\theta^{2}+1}$.

Substituting the value of

$\theta=\sqrt{4x^{2}-1}$,

we get:

$I=2. atan(\sqrt{4x^{2}-1})-\frac{\sqrt{4x^{2}-1}}{2x^{2}}$.

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