Question

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

Answer 1

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!