How would you evaluate the integral #int_(-2)^(-7) sqrt(x^2 - 1)/x^3 dx#?

1 Answer
Noah G
Mar 12, 2017

The integral #int_(-7)^(-2)# is equivalent to #-0.34#.

Explanation:

Use the trig substitution #x = sectheta#. Then #dx = sec thetatantheta d theta#.

#int_(sec(-7))^(sec(-2)) sqrt(((sec theta)^2 - 1))/sec^3theta * secthetatantheta d theta#

#int_(sec(-7))^(sec(-2)) sqrt(tan^2theta)/sec^2theta * tan theta d theta#

#int_(sec(-7))^(sec(-2)) tantheta/sec^2theta * tan theta d theta#

#int_(sec(-7))^(sec(-2)) tan^2theta/sec^2theta d theta#

#int_(sec(-7))^(sec(-2)) (sin^2theta/cos^2theta)/(1/cos^2theta)#

#int_(sec(-7))^(sec(-2)) sin^2theta#

Now apply the power reduction formula #sin^2x = (1 - cos2x)/2#.

#int_(sec(-7))^(sec(-2)) (1 - cos(2theta))/2#

#1/2int_(sec(-7))^(sec(-2)) 1 - cos2theta d theta#

To integrate #cos2theta#, we perform a u-substitution of #u = 2theta -> du = 2d theta# to get.

#int1/2cosudu#

#1/2sin(2theta)#

We now put this back together.

#1/2[theta - 1/2sin(2theta)]_(sec(-7))^(sec(-2))#

We must now reverse the substitution.

#1/2[sec^-1(x) - sqrt(x^2 - 1)/x]_(-7)^(-2)#

An approximation of this gives

#-0.34#.

Hopefully this helps!

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