Question

integral 1 to infinity sqrt[x^2-1]/x^2
do i do trig substitution of sin^-1?

Answer 1

\[\int\limits_{1}^{\infty} \sqrt{x^2-1}/x^2\]

Answer 2

\[\int\limits_{1}^{\infty}\frac{ \sqrt{x^2-1} }{ x^2 }\]

Answer 3

Do a double u substituion with your first u=sqrt(x^2-1)

Answer 4

so like u=u^2-1 du=2u?

Answer 5

my mistake, you only need to do the u substitution once.
\[\int\limits_{1}^{\infty}\frac{\sqrt{x^2-1}}{x^2}\]
\[u=\sqrt(x^2-1)\]
\[u^2+1=x^2\]
\[du=\frac{1}{2\sqrt{x^2-1}}dx\]
\[2*u*du=dx\]
\[\frac{du}{2\sqrt{u-1}}dx\]
\[\int\limits_{1}^{\infty}\frac{\sqrt{x^2-1}}{x^2}=\int\limits_{1}^{\infty}\frac{2u^2}{u^2+1}du\]

From there, you can do integration by parts with \[\int\limits_{1}^{\infty}udv=uv - \int\limits_{1}^{\infty}vdu\]
where u = 2u^2
and dv = 1/(u^2+1)du
You'll get an arctan.

Answer 6

kk and i am going to give it a go. thank u. i pratice the rest

Answer 7

If you know trig substitution this could be easier. I'll show you the work for it to see if you can follow. Give me a bit.

Answer 8

From that triangle, try to create a substitution that can replace \[\frac{\sqrt{x^2-1}}{x^2}\]
From trig, you know that sin(x) = opposite over hypotenuse and all the other soh cah toa mumbo jumbo.
\[sin(\theta)=\frac{1}{x}\]
Take the derivative
\[-cos(\theta)d\theta=\frac{-1}{x^2}dx\]
Negatives cancel and you get
\[cos(\theta)d\theta=\frac{1}{x^2}dx\]
Now you have something that can substitute for the \[\frac{dx}{x^2}\] portion, but you need something for the \[\sqrt{x^2-1}\]
Going back to our triangle, you know that cot(theta) = adjacent over hypotenuse
\[cot(\theta)=\sqrt(x^2-a)\]
Now, you can substitute back into the equation.
\[\int\limits_{1}^{\infty}\frac{\sqrt{x^2-1}}{x^2}=\int\limits_{1}^{\infty}cos(\theta)cot(\theta)d\theta\]
Mess around with the trig a little to make it easier
\[\int\limits_{1}^{\infty}cos(\theta)cot(\theta)d\theta=\int\limits_{1}^{\infty}cos(\theta)\frac{cos(\theta)}{sin(\theta)}d\theta=\]
remember that cos^2(x)=1-sin^2(x)
\[\int\limits_{1}^{\infty}\frac{cos^2(\theta)}{sin(\theta)}d\theta=\int\limits_{1}^{\infty}\frac{1-sin^2(\theta)}{sin(\theta)}d\theta=\int\limits_{1}^{\infty}\frac{1}{sin(\theta)}-sin(\theta)d\theta\]
Now, that integral is easier to solve.

Do note, trig substitution is one of the most difficult ones to comprehend so if you don't get it, you don't really NEED it. It just sometimes makes it easier.