I'm not sure how to phrase this question, but I find myself confused over the correctness of a particular solution in the table of integral, specifically: $$\int\frac{dx}{\sqrt{x^2-a^2}}=\ln|\sqrt{x^2-a^2}+x|+C$$

I find that if you try to solve it with trigonometric substitution, you get a different answer:

So, for the example, since the form $\sqrt{x^2-a^2}$ is present, we use the substitution $x=asec \theta$, and $dx=asec\theta tan \theta d\theta$. Doing so:

$$\int\frac{asec\theta tan \theta d\theta}{\sqrt{(asec \theta)^2-a^2}}$$ $$\int\frac{asec\theta tan \theta d\theta}{\sqrt{a^2sec^2 \theta-a^2}}$$ $$\int\frac{asec\theta tan \theta d\theta}{a\sqrt{sec^2 \theta-1}}$$ Using the trigonometric identity $tan^2 \theta=sec^2 \theta -1$ $$\int\frac{asec\theta tan \theta d\theta}{atan \theta}$$

Cancelling like terms, we get $\int sec \theta$, which finally integrates to $ln|sec \theta + tan \theta|+C$

Undoing the substitution, using this triangle:


With $sec \theta = \frac{x}{a}$ and $tan \theta = \frac{\sqrt{x^2-a^2}}{a}$, the final answer then is:

$$ln\biggl|\frac{x}{a} + \frac{\sqrt{x^2-a^2}}{a}\biggl|+C$$

Which is evidently different from the one from the table of integrals. Is my solution wrong or something?

  • $\begingroup$ looks the same to me, since $-\ln |a|$ can be absorbed into $C$? $\endgroup$ – Calvin Khor Feb 7 at 13:48
  • $\begingroup$ Oh right, that can be done. Thanks for pointing that out $\endgroup$ – Xyzar Feb 7 at 13:52

You're right. Indefinite integrals represent a family of curves(functions) which vary by some constant.

$\ln \left|\dfrac{x+\sqrt{x^2-a^2}}{a}\right| +c= \ln |x+\sqrt{x^2-a^2}| - \ln|a| +c = \ln(x+\sqrt{x^2-a^2}) +k$

$k = c-\ln |a| = $ constant


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