Integral of $\frac{1}{\sqrt{x^2-1}}$

Question

I know that I should use substitution, but I am not really sure how.

$$\int \frac{1}{\sqrt{x^2-1}}dx$$

Answer 1

Hint

Try to use the fact that $$\cosh^2(x)-\sinh^2(x)=1.$$

Answer 2

There are really $12$ possibilities: $$\begin{align}1-\sin^2\theta&=\cos^2\theta\\ 1-\cos^2\theta&=\sin^2\theta\\ 1+\tan^2\theta&=\sec^2\theta\\ \sec^2\theta-1&=\tan^2\theta\\ 1+\cot^2&=\csc^2\theta\\ \csc^2\theta-1&=\cot^2\theta\\ \sinh^2\theta+1&=\cosh^2\theta\\ \cosh^2\theta-1&=\sinh^2\theta\\ 1-\tanh^2\theta&=\text{sech}^2\theta\\ 1-\text{sech}^2\theta&=\tanh^2\theta\\ \coth^2\theta-1&=\text{csch}^2\theta\\ 1+\text{csch}^2\theta&=\coth^2\theta\end{align}$$ So of these, $4$ have the form $x^2-1$, so you could try them all. $x=\sec\theta$, $dx=\sec\theta\tan\theta\,d\theta$ leads to $$\int\frac1{\sqrt{x^2-1}}dx=\int\sec\theta\,d\theta=\ln|\tan\theta+\sec\theta|+C=\ln|\sqrt{x^2-1}+x|+C$$ Then if $x=\csc\theta$, $dx=-\csc\theta\cot\theta\,d\theta$, $$\int\frac1{\sqrt{x^2-1}}dx=-\int\csc\theta\,d\theta=-\ln|\csc\theta+\cot\theta|+C=-\ln|x+\sqrt{x^2-1}|+C$$ Or if $x=\cosh\theta$, then $dx=\sinh\theta\,d\theta$, $$\int\frac1{\sqrt{x^2-1}}dx=\int d\theta=\theta+C=\cosh^{-1}\theta+C=\ln|x+\sqrt{x^2-1}|+C$$ And finally if $x=\coth\theta$m we have $dx=-\text{csch}^2\theta\,d\theta$, $$\int\frac1{\sqrt{x^2-1}}dx=-\int\text{csch}\theta\,d\theta=-\ln|\text{csch}\theta+\coth\theta|+C=-\ln|\sqrt{x^1-1}+x|+C$$ The decreasing function substitutions have an extra minus sign, but consideration that the limits would be reversed by such a substitution will hopefully resolve that discrepancy.

Answer 3

$\DeclareMathOperator{\arcosh}{arcosh}$

Assuming that the integrand is defined on the interval $(1, \infty)$ (rather than $\mathbb{R} \setminus \{-1, 1\}$), we can substitute for $x$ the function $$ g(u) = \left\{ \begin{array}{l} (0, \infty) \to (1, \infty) \\ u \mapsto \cosh(u) \end{array} \right. $$ for which we have $$ g'(u) = \sinh(u) \text{,} $$ $$ g^{-1}(u) = \arcosh(u) \text{.} $$ At this point you should convince yourself that we are indeed allowed to apply this substitution here! Now using the hint given by Surb this yields $$ \int \frac{1}{\sqrt{x^2 - 1}} dx = \bigg[ \int \frac{\sinh(u)}{\sqrt{\cosh^2(u) - 1}} du \bigg]_{u = \arcosh(x)} = \bigg[ \int \frac{\sinh(u)}{\sqrt{\sinh^2(u)}} du \bigg]_{u = \arcosh(x)} = $$ $$ = \bigg[ \int 1 du \bigg]_{u = \arcosh(x)} = \bigg[ u + C \bigg]_{u = \arcosh(x)} = \arcosh(x) + C \text{.} $$