How do you use substitution to integrate $\frac{2}{x \sqrt{4 ln x - {(ln x)}^{2}}}$ $2/(xsqrt(4lnx- (lnx)^2 ))$ ?

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How do you use substitution to integrate $\frac{2}{x \sqrt{4 ln x - {(ln x)}^{2}}}$ $2/(xsqrt(4lnx- (lnx)^2 ))$ ?

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Answer 1 · 2015-07-23T23:58:42

The answer turned out to be $4 arcsin (\frac{\sqrt{ln x}}{2}) + C$ $4arcsin(sqrt(lnx)/2) + C$ . Just a note; Wolfram Alpha gives a much more complicated answer, and it's not particularly nice of a result when, for example, you are typing it in for an online homework problem.

Wolfram Alpha gives its alternative answer as:

$= \frac{4 \sqrt{ln x - 4} \sqrt{ln x} ln (\sqrt{ln x - 4} + \sqrt{ln x})}{\sqrt{- (ln x - 4) ln x}}$ $= (4sqrt(lnx - 4)sqrt(lnx)ln(sqrt(lnx - 4) + sqrt(lnx)))/sqrt(-(lnx - 4)lnx)$

$= \frac{4 \sqrt{ln x - 4} \sqrt{ln x} ln (\sqrt{ln x - 4} + \sqrt{ln x})}{\sqrt{4 ln x - {ln}^{2} x}}$ $= (4sqrt(lnx - 4)sqrt(lnx)ln(sqrt(lnx - 4) + sqrt(lnx)))/sqrt(4lnx - ln^2x)$

which... let's face it, looks terrible! :)

I feel like trig substitution should be done somewhere here... Let's say for now...

Let:
$u = ln x$ $u = lnx$
$d u = \frac{1}{x} d x$ $du = 1/xdx$

$\Rightarrow \int \frac{2}{x \sqrt{4 u - u^{2}}} d x$ $=> int 2/(xsqrt(4u - u^2))dx$

$= 2 \int \frac{1}{\sqrt{4 u - u^{2}}} d u$ $= 2int 1/(sqrt(4u - u^2))du$

Okay, so this looks a little better. What if we factored something out?

$= 2 \int \frac{1}{\sqrt{u} \sqrt{4 - u}} d u$ $= 2int 1/(sqrtusqrt(4 - u))du$

$= 2 \int \frac{1}{\sqrt{u} \sqrt{2^{2} - {(\sqrt{u})}^{2}}} d u$ $= 2int 1/(sqrtusqrt(2^2 - (sqrtu)^2))du$

Now, what if we said...
$\sqrt{u} = 2 sin θ$ $sqrtu = 2sintheta$
$u = 4 {sin}^{2} θ$ $u = 4sin^2theta$
$d u = 8 sin θ cos θ d θ$ $du = 8sinthetacosthetad theta$
$\sqrt{4 - u} = \sqrt{2^{2} - 2^{2} {sin}^{2} θ} = 2 cos θ$ $sqrt(4 - u) = sqrt(2^2 - 2^2sin^2theta) = 2costheta$

More manageable now.

$= 2int 1/(cancel(2)cancel(sintheta)cancel(2)cancel(costheta))*cancel(8)^2cancel(sinthetacostheta)d theta$

$= 4int d theta$

Huh? How... easy?

$= 4theta$

Now, we had $2sintheta = sqrtu$ , so:

$theta = arcsin((sqrtu)/2)$

Thus:

$theta = arcsin(sqrt(lnx)/2)$

And so we have:

$= color(blue)(4arcsin(sqrt(lnx)/2) + C)$

That is our answer. It seems rather intuitive considering the resemblance of the original with $1/(sqrt(1-u^2))$ .

Let's try to differentiate this and see if we get back to the original.

$(dy)/(dx)[arcsinu] = 1/sqrt(1-u^2)((du)/(dx))$

Thus, with $u = sqrt(lnx)/2$ , we have, after some Chain Rule action...

Cancel out terms:
$4(d)/(dx)[arcsin(sqrt(lnx)/2)] = cancel(4)*1/(sqrt(1-((sqrtlnx)/2)^2))*1/cancel(2)*1/(cancel(2)sqrt(lnx))*1/x$

Shift values around:
$= 1/(xsqrt(lnx)sqrt(1-((sqrtlnx)/2)^2))$

Multiply out the square:
$= 1/(xsqrt(lnx)sqrt(1-(lnx)/4))$

Distribute into the square root:
$= 1/(xsqrt(lnx-(lnx)^2/4))$

Factor $sqrt(1/4)$ out:
$= 1/(xsqrt(1/4)sqrt(4lnx-(lnx)^2))$

Shift that around:
$= sqrt4/(xsqrt(4lnx-(lnx)^2))$

Done!
$= color(green)(2/(xsqrt(4lnx-(lnx)^2)))$

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How do you use substitution to integrate $\frac{2}{x \sqrt{4 ln x - {(ln x)}^{2}}}$ $2/(xsqrt(4lnx- (lnx)^2 ))$ ?

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