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Integrate; $$\int \frac{1}{\sqrt{x^2 - a^2}} \, dx $$

I can get the answer by substituting $\sec t = \dfrac{x}{a}$ but I am left with $\ln\left|\dfrac{x + \sqrt{x^2 + a^2}}{a}\right|$. Granted, this is right but how would I get there without having the 1/a in the log?

Also is there a way to find $\int \sec t \space dt$ without having to multiply by $\dfrac{\sec t + \tan t}{\sec t + \tan t}$?

and finally, more of a general question; are there and integrals in which only a hyperbolic substitution can be used to integrate?

Michael Hardy
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Robert S
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2 Answers2

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Regarding $\int\sec x\,dx$, you can do a tangent half-angle substitution, that is: $t=\tan \frac{x}{2}$ and thus $\cos x=\frac{1-t^2}{1+t^2}$.

Alexandre Cartaxo
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Recall that the log of a quotient is equal to the log of the numerator minus the log of the denominator: $\ln\frac{f(x)}{g(x)} = \ln f(x) - \ln g(x)$ Therefore the $1/a$ is absorbed into the constant from the indefinite integral.

The method you have given for finding $\int \sec{t} \space dt$ is the simplest method I know of, but there is another way using sin and cos substitutions shown here: Find the integration of $\sec(x)$ and prove it

See this answer regarding hyperbolics: Are there integrals you can't solve without inverse hyperbolic substitution?

Community
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Austin Freeman
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