# Integration by Trigonometric Substitution vs Table of Integral Solution

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:

https://i.stack.imgur.com/6tkR7.png

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?

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

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