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I need to solve:

$$\int \sqrt{x^2-3} \, dx.$$

So I use the substitution:

$$x=\frac{\sqrt 3}{\cos(t)}$$

$$dx= \frac{\sqrt 3 \sin(t) \, dt}{\cos^2(t)} $$

and I get

$$3\int \frac{\sqrt{\frac1 {\cos^2(t)}-1}\cdot\sin(t) \, dt}{\cos^2(t)}. $$

So I get

$$3\int \frac{\sin(t)\tan(x)\,dx}{\cos^2(x)},$$ that is $(\sin(t)\sin(t)/\cos(t))/\cos^2(t)$ and finally I have

$$3\int \frac{\sin^2(t)\,dt}{\cos^3(t)}$$

Substitute $\sin(t)=s$, so $\cos(t) \, dt=ds$. The integral becomes

$$3 \int \frac{s^2\,ds}{(1-s^2)^2};$$ $$dt= \frac{ds}{\cos t}$$

So problem is I don't know how do I get from this $\sqrt{(1/\cos^2(t))-1} = \tan(t);$

It was stupid question idk how I didn't saw that nvm, after that use partial integration $u=s$, $du=ds$, $v=1/(1-s^2)$ and i get $s/2(1-s^2)-1/2$ integral of $ds/(1-s^2)$ and the solution is $3/2(s/1-s^2-1/2\ln(1+s/1-s)$

Martin Sleziak
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user5927353
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    This is an integral, not an integer. – Arthur Jul 28 '16 at 17:25
  • It is a bit hard to read and understand the question! – Majid Jul 28 '16 at 17:25
  • $$\text{Integer}=0,1,-1,2,-2,3,-3,\ldots\ \text{Integral}=\int$$ –  Jul 28 '16 at 17:25
  • Please reedit. This is unreadable. What the heck is tgt? – fleablood Jul 28 '16 at 17:26
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    I did not read anything but the last line, so this might not be helpful. $$\sqrt{\frac1{\cos^2 x}-1}=\sqrt{\frac{1-\cos^2x}{\cos^2x}}$$ and here's where you remember what $1-\cos^2x$ is. –  Jul 28 '16 at 17:29
  • Wait... Do you wanna ask why $\sqrt{(1/\cos^2(t))-1} = \tan(t)$ or do you want to know how to continue with $3 \int \frac{s^2,ds}{(1-s^2)^2}$? –  Jul 28 '16 at 17:48
  • I finished integral,i was stupid idk how i didn't saw that 1+cos^2(t)=sin^2(t) and when i put that back i get sin^2(t)/cos^2(t) and thats tg^2(t) bcs of sqrt its tg(t).Sry it was stupid question from me – user5927353 Jul 28 '16 at 17:56
  • Please, avoid SMS language! It makes everything harder to read. – rubik Jul 28 '16 at 17:56
  • Okey,sorry my english is not that great so i will try to write better. – user5927353 Jul 28 '16 at 17:58
  • @fleablood : I would take "tgt" to mean $\operatorname{tg}t$, i.e. what in recent decades is usually denoted (at least when writing in English) as $\tan t$. A hundred years or more ago one sometimes saw $\operatorname{tg}t$ or $\operatorname{tang}t$ or other abbreviations in things written in English; one still sees them today in things written in some other languages. $\qquad$ – Michael Hardy Jul 28 '16 at 18:50
  • For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Jul 28 '16 at 20:45
  • Closely related post: http://math.stackexchange.com/questions/1445147/indefinite-integral-of-int-sqrtx2-1dx – Martin Sleziak Jul 28 '16 at 20:48

4 Answers4

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Note that $$\cos^2 \theta + \sin^2 \theta = 1$$ so dividing both sides by $\cos^2 \theta$ gives $$1 + \tan^2 \theta = \frac{1}{\cos^2 \theta} \iff \frac{1}{\cos^2 \theta} - 1 = \tan^2 \theta$$

And so, taking the square root of both sides gives you what you want.

Zain Patel
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    Since the main question appeared to be expressed as "So problem is I don't know how do I get from this $\sqrt{(1/\cos^2(t))-1} = \tan(t);$", this is the only really good answer so far. $\qquad$ – Michael Hardy Jul 28 '16 at 22:06
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Hint. As you have suggested, the change of variable $$ x=\frac{\sqrt{3}}{\cos t},\quad dx=\frac{\sqrt{3}\sin t}{\cos^2 t}\:dt, $$ gives $$ \begin{align} \int \sqrt{x^2-3}\:dx&=\sqrt{3}\int\sqrt{\frac3{\cos^2 t}-3}\:\cdot \frac{\sin t}{\cos^2 t}\:dt \\\\&=3\int\sqrt{\frac{1-\cos^2 t}{\cos^2 t}}\:\cdot \frac{\sin t}{\cos^2 t}\:dt \\\\&=3\int \frac{\sin^2 t}{\cos^3 t}\:dt \\\\&=3\int \frac{\sin^2 t\: \cos t}{(1-\sin^2 t)^2}\:dt \\\\&=3\int \frac{s^2}{(1-s^2)^2}\:ds \end{align} $$ can you take it from here?

Olivier Oloa
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$3\int \frac {\sin^2(t)}{\cos^3(t)} dt$

Rather than a substitution I suggest you do this.

$3\int \tan^2 t \sec t dt$

integration by parts:

$u = \tan t, dv = \sec t \tan t dt\\ du = sec^2 t dt, v = sec t$

$3\sec t \tan t - 3\int \sec^3t dt\\ 3\sec t \tan t - 3\int (\tan^2 t - 1)\sec t dt\\ 3\sec t \tan t + 3 \int \sec t dt - 3\int \tan^2 t \sec t dt\\ 3\sec t \tan t + 3 \ln |\sec t + \tan t| - 3\int \tan^2 t \sec t dt$

And we have come full circle:

$3\int \tan^2 t \sec t dt = 3\sec t \tan t + 3 \ln |\sec t + \tan t| - 3\int \tan^2 t \sec t dt\\ 3\int \tan^2 t \sec t dt = I\\ I = 3\sec t \tan t + 3 \ln |\sec t + \tan t| - I\\ 2 I = 3\sec t \tan t + 3 \ln |\sec t + \tan t|\\ I = \frac 32 \sec t \tan t + \frac 32 \ln |\sec t + \tan t|+C$

Alternatively, If you want to stick with your initial approach.

$\int \frac {s^2}{(1-s^2)^2} ds$

Find a partial fraction decomposition:

$\frac {s^2}{(1-s^2)^2} = \frac A{1-s} +\frac B{1+s} + \frac C{(1-s)^2} + \frac D{(1-s)^2}$

And that will integrate to

$-A\ln (1-s) + B\ln(1+s) + C(1-s)^{-1} - D(1+s)^{-1} + c$ (sorry do double use the same letter)

Hopefully, the two answers are equivalent.

Doug M
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First rescale

$$\int \sqrt{x^2-3} \, dx=3\int \sqrt{t^2-1} \, dt.$$

Then by parts,

$$\int \sqrt{t^2-1} \, dt=t\sqrt{t^2-1}-\int\frac{t^2}{\sqrt{t^2-1}}dt=t\sqrt{t^2-1}-t-\int\frac{dt}{\sqrt{t^2-1}}.$$

In the last integral, you should recognize the derivative of $\text{arcosh }t$.

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    Well, if the question is "So problem is I don't know how do I get from this $\sqrt{(1/\cos^2(t))-1} = \tan(t);$", then I don't think this addresses it. $\qquad$ – Michael Hardy Jul 28 '16 at 22:05