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

Question

Elijah Abram · Accepted Answer

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

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:

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!

Evelyn Agard · Answer

undefined To integrate ∫ dx / √(x^2 - 1) from -2 to -3, you first perform the antiderivative of 1/√(x^2 - 1), which yields arcsinh(x). Then, you substitute the upper limit (-3) into the antiderivative and subtract the result of substituting the lower limit (-2). This gives you the value of the definite integral.