How do you integrate $int 1/(sqrt(x^2-4))dx$ using trigonometric substitution?

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Answer 1 · 2016-03-08T00:59:22

$int1/sqrt(x^2-4)dx = ln|x+sqrt(x^2-4)|+C$

$int1/sqrt(x^2-4)dx = int1/sqrt(4(x^2/4-1))dx$

$=1/2int1/sqrt((x/2)^2-1)dx$

Let $x/2 = sec(theta)$
Then $dx = 2sec(theta)tan(theta)$

$=>int1/sqrt(x^2-4)dx = 1/2int1/sqrt(sec^2(theta)-1)*2sec(theta)tan(theta)d theta$

$=int (sec(theta)tan(theta))/sqrt(tan^2(theta))d theta$

$=int sec(theta)d theta$

$=ln|sec(theta) + tan(theta)| + C$

$=ln|x/2 + sqrt(x^2/4-1)|+C$

(For the final equality, try drawing a right triangle with angle $theta$ such that $sec(theta) = x/2$ and solve for $tan(theta)$ )

We could stop here, but we can also make this look a little nicer by getting rid of the $1/2$

$ln|x/2+sqrt(x^2/4-1)|+C =ln|1/2(x+2sqrt(x^2/4-1))|+C$

$=ln|x+sqrt(x^2-4)|+ln(1/2)+C$

But as $C$ is an arbitrary constant, we can include the $ln(1/2)$ in that to give us the final result

$ln|x+sqrt(x^2-4)|+C$

How do you integrate int 1/(sqrt(x^2-4))dxint 1/(sqrt(x^2-4))dx using trigonometric substitution?