Question

integral of 1/(sqrt(x^2-4))

Answer 1

\[I=\int\limits \frac{ 1 }{ \sqrt{x^2-4} }dx\]
put\[x=\frac{ 1 }{ y },dx=\frac{ -1 }{ y^2 }dy\]
\[I=\int\limits \frac{ \frac{ -1 }{ y^2 }dy }{ \sqrt{\frac{ 1 }{ y^2 }-4} }\]
\[I=\int\limits \frac{- y~dy }{ y^2\sqrt{1-4y^2} }\]
put\[\sqrt{1-4y^2}=t,1-4y^2=t^2,1-t^2=4y^2,y^2=\frac{ 1 }{ 4 }\left( 1-t^2 \right)\]
\[2y~dy=\frac{ 1 }{ 4 }\left( -2t~dt \right)\]
\[y~dy=\frac{ -1 }{ 4 }t~dt\]
\[I=\frac{ 1 }{ 4 }\int\limits \frac{ t~dt }{ \frac{ 1 }{ 4 } \left( 1-t^2 \right)t}=\int\limits \frac{ dt }{ \left( 1+t \right) \left(1-t \right)}\]
complete it.

Answer 2

With all due respect, I'd not introduce another variable, especially not ' y ', prior to attempts to integrate \[\int\limits\limits \frac{ 1 }{ \sqrt{x^2-4} }dx.\] x and y represent variables by convention; in contrast, a and b represent parameters or constants.

This integral has the form \[I=\int\limits\limits \frac{ 1 }{ \sqrt{x^2-a^2} }dx,\]and there is a standard formula for the integral listed in "A Brief Table of Integrals" in the Calculus textbook I use for reference.

To evaluate \[I=\int\limits\limits \frac{ 1 }{ \sqrt{x^2-4} }dx,\]I'd use trig substitution, letting x=2 sec theta (or \[x=2*\sec \theta \]

Answer 3

because then the expression under the radical sign becomes (2 sec theta)^2 - 2^2, or
\[4\sec^2 \theta - 4, ~or~4(\sec^2-1)=??\]