We have:
int_-3^-2 1/(sqrt(x^2-1))dx∫−2−31√x2−1dx
We use the fundamental theorem of calculus:
int_a^bf(x)dx=F(b)-F(a)∫baf(x)dx=F(b)−F(a) when F'(x)=f(x)
What is int1/(sqrt(x^2-1))dx?
We use the trigonometric substitution.
Since the variable is getting subtracted by one, this is the secant case.
We draw a right triangle:
We see that:
sec(theta)=x
=>theta=arcsec(x)
=>sec(theta)tan(theta)d theta=dx
tan(theta)=sqrt(x^2-1)/1
=>tan(theta)=sqrt(x^2-1)
Substitute.
=>int1/(tan(theta))sec(theta)tan(theta)d theta Simplify
=>intsec(theta)d theta
This is one of the "basic" integrals you should memorize.
=>lnabs(sec(theta)+tan(theta)) substitute
=>lnabs(sec(arcsec(x))+tan(arcsec(x)))
=>lnabs(x+tan(arcsec(x)))
Therefore:
int_-3^-2 1/(sqrt(x^2-1))dx=[lnabs(x+tan(arcsec(x)))]_-3^-2
=>lnabs(-3+tan(arcsec(-3)))-lnabs(-2+tan(arcsec(-2)))
=>1.76274717404-1.31695789692
=>0.44578927712
That is the answer!
