How would you evaluate the integral #int_(-2)^(-7) sqrt(x^2 - 1)/x^3 dx#?
1 Answer
The integral
Explanation:
Use the trig substitution
#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
#int_(sec(-7))^(sec(-2)) (1 - cos(2theta))/2#
#1/2int_(sec(-7))^(sec(-2)) 1 - cos2theta d theta#
To integrate
#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!