Care with the secant substitution Recall that the

Chapter 4, Problem 75AE

(choose chapter or problem)

QUESTION:

Care with the secant substitution Recall that the substitution $x=a \sec \theta$ implies that $x \geq a$ (in which case $0 \leq \theta<\pi / 2$ and $\tan \theta \geq 0$ ) or $x \leq-a$ (in which case $\pi / 2<\theta \leq \pi$ and $\tan \theta \leq 0$ ).

Show that

$\int \frac{d x}{x \sqrt{x^{2}-1}}= \begin{cases}\sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1\end{cases}$

Questions & Answers

QUESTION:

Show that

$\int \frac{d x}{x \sqrt{x^{2}-1}}= \begin{cases}\sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1\end{cases}$

ANSWER:

Problem 75AE

Answer

We have to show

$\int\limits_{\ }^{\ }\frac{dx}{x\sqrt{{x}^{2}\ -1}}\ dx\ \ ={\{}_{-sec{\ }^{-1}\ x\ +C=-tan{\ }^{-1}\ \sqrt{{x}^{2}\ -1}\ +C\ \ if\ x<-1}^{\ se{c}^{-1\ }x+C\ =tan{\ }^{-1}\ \sqrt{{x}^{2}\ -1}\ +C\ \ if\ x>1}$

Substitute $x=sec\ \theta{}\Rightarrow{}dx\ =\ sec\ \theta{}\ tan\ \theta{}\ d\theta{}$ $\$

Therefore $\int\limits_{\ }^{\ }\frac{dx}{x\sqrt{{x}^{2}\ -1}}\ dx=\int\limits_{\ }^{\ }\frac{sec\ \theta{}\ tan\ \theta{}}{sec\ \theta{}\sqrt{{sec\ }^{2}\theta{}\ -1}}\ d\theta{}$