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Evaluating the integral $$\int\frac{1}{\sqrt{x^{2} +9}}dx$$ so I use tan substitution
$$ x=3\tan t ~\mbox{and}~ dx = 3\sec^{2} t ~dt $$ after substituting everything in and smplifying im left with $$\int \sec t ~dt$$ but I need to have this answer in terms of x, I know $$ \sec t = \sqrt{1 + \tan^{2}t}$$ and $$\tan t = \frac{x}{3} $$ so do I just plug in $$\sqrt{1 + (\frac{x}{3})^{2}}$$ just not sure of the final steps I need to get the integral.

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    $\begingroup$ $\int \sec tdt=\ln|\sec t+\tan t|+C$ $\endgroup$ – CY Aries Jun 7 '17 at 13:57
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    $\begingroup$ If you just sub back in for $x$ without integrating, you should get your original integral back. $\endgroup$ – ziggurism Jun 7 '17 at 13:59
  • $\begingroup$ See sosmath.com/tables/integral/integ11/integ11.html $\endgroup$ – lab bhattacharjee Jun 7 '17 at 13:59
  • $\begingroup$ you can integrate sec in different ways, it is often considered a standard integral - where you suggest the tan = x/3 substitution , you seem to be heading away from the answer - learn one of the integral sec derivations, and take it from there $\endgroup$ – Cato Jun 7 '17 at 14:05
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When you reach the integral$$\int \sec t ~dt$$ you have to integrate $$\int \sec t ~dt = \ln |\sec t + \tan t| + C$$ Now, since $\tan t = \frac{x}{3}$, then $\cos t = \frac{3}{\sqrt{x^2+9}}$ and so $\sec t = \frac{\sqrt{x^2+9}}{3}$. Then $$\int \sec t ~dt = \ln |\sec t + \tan t| + C = \ln \left|\frac{\sqrt{x^2+9}}{3} + \frac{x}{3} \right| + C$$

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  • $\begingroup$ thanks so much ! $\endgroup$ – guy_sensei Jun 7 '17 at 14:19
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    $\begingroup$ After this you could "absorb" $-\ln 3$ into the constant of integration, and also observe the argument of $\ln$ is always positive anyway (even if $x$ is negative), to get a slightly prettier result $\ln (\sqrt{x^2+9} + x) + C$. $\endgroup$ – Daniel Schepler Jun 7 '17 at 16:49
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$$\int \sec t \ dt = \int\sec t\cdot\frac{\sec t + \tan t}{\sec t + \tan t}\ dt$$

And then substitute $u = \sec t + \tan t$.

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Hint:

$$\int \frac{\mathrm dt}{\sqrt{t^2+1}}=\operatorname{arg\,sh}t=\ln(t+\sqrt{t^2+1}).$$

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    $\begingroup$ What is $arg sh t $ its new to me . $\endgroup$ – Archis Welankar Jun 7 '17 at 14:03
  • $\begingroup$ The inverse of $\sinh$ (a.k.a. $\operatorname{sh}$). Do you know the hyperbolic functions? $\endgroup$ – Bernard Jun 7 '17 at 14:05
  • $\begingroup$ Yes somewhat... $\endgroup$ – Archis Welankar Jun 7 '17 at 14:06
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    $\begingroup$ Well, the derivatives of the inverse hyperbolic functions (which are logarithms anyway) are standard. $\endgroup$ – Bernard Jun 7 '17 at 14:07
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    $\begingroup$ It is arg (which stands for ‘ argument’). $\endgroup$ – Bernard Jun 7 '17 at 14:08

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