How do you integrate #int (dx) / ( sqrt(x^(2) - 1 ) # from -2 to -3?

1 Answer
Apr 20, 2018

#int_-3^-2 1/(sqrt(x^2-1))dx~~0.44578927712#

Explanation:

We have:

#int_-3^-2 1/(sqrt(x^2-1))dx#

We use the fundamental theorem of calculus:

#int_a^bf(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:

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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!