# Integration in terms of x

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.

• $\int \sec tdt=\ln|\sec t+\tan t|+C$ – CY Aries Jun 7 '17 at 13:57
• If you just sub back in for $x$ without integrating, you should get your original integral back. – ziggurism Jun 7 '17 at 13:59
• – lab bhattacharjee Jun 7 '17 at 13:59
• 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 – Cato Jun 7 '17 at 14:05

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 $$\int \sec t ~dt = \ln |\sec t + \tan t| + C = \ln \left|\frac{\sqrt{x^2+9}}{3} + \frac{x}{3} \right| + C$$

• thanks so much ! – guy_sensei Jun 7 '17 at 14:19
• 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 $\ln (\sqrt{x^2+9} + x) + C$. – Daniel Schepler Jun 7 '17 at 16:49

$$\int \sec t \ dt = \int\sec t\cdot\frac{\sec t + \tan t}{\sec t + \tan t}\ dt$$

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

Hint:

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

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