Question

How do you integrate? `int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))`

`int_sqrt(2)^2dx/(x^5*sqrt(x^2-1))`

Answer 1

`int_sqrt2^2dx/(x^5sqrt(x^2-1))=int_(pi/4)^(pi/3)cos^4thetad theta`, which I leave to you to solve.

Explanation:

Try letting `x=sectheta`. Note this implies that `dx=secthetatanthetad theta`.

Also, the bounds will change. Observe that `x=sqrt2=>theta=pi/4` and `x=2=>x=pi/3`.

Then:

`int_sqrt2^2dx/(x^5sqrt(x^2-1))=int_(pi/4)^(pi/3)(secthetatanthetad theta)/(sec^5thetasqrt(sec^2theta-1))`

Recall the trigonometric identity `sec^2theta-1=tan^2theta`.

`=int_(pi/4)^(pi/3)(secthetatanthetad theta)/(sec^5thetatantheta)`

`=int_(pi/4)^(pi/3)(d theta)/sec^4theta`

`=int_(pi/4)^(pi/3)cos^4thetad theta`

From here, I would use the identity `cos^2alpha=1/2(1+cos2alpha)`, or, in this case, `cos^4alpha=1/4(1+cos2alpha)^2` to further simplify the integral. It will definitely be a little messy.