How do I start this integral?

integral from [-2 to 1]

sqrt((x^2)-1) / x dx

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- Sep 16th 2011, 10:05 PMSneakyintegral question
How do I start this integral?

integral from [-2 to 1]

sqrt((x^2)-1) / x dx - Sep 16th 2011, 10:33 PMProve ItRe: integral question
$\displaystyle \displaystyle \int_{-2}^1{\frac{\sqrt{x^2 - 1}}{x}\,dx} = \frac{1}{2}\int_{-2}^1{\frac{2x\sqrt{x^2 - 1}}{x^2}\,dx}$

Now make the substitution $\displaystyle \displaystyle u = x^2 - 1 \implies du = 2x\,dx$, and making note that $\displaystyle \displaystyle u(-2) = 3, u(1) = 0$, the integral becomes

$\displaystyle \displaystyle \begin{align*} \frac{1}{2}\int_{-2}^1{\frac{2x\,\sqrt{x^2 - 1}}{x^2}\,dx} &= \frac{1}{2}\int_3^0{\frac{\sqrt{u}}{u + 1}\,du} \\ &= -\frac{1}{2}\int_0^3{\frac{\sqrt{u}}{u+1}\,du} \\ &= -\int_0^3{\frac{u}{2\sqrt{u}\,(u + 1)}\,du} \end{align*}$

Now make the substitution $\displaystyle \displaystyle v = \sqrt{u} \implies dv = \frac{1}{2\sqrt{u}}\,du$ and noting that $\displaystyle \displaystyle v(0) = 0$ and $\displaystyle \displaystyle v(3) = \sqrt{3}$, the integral becomes

$\displaystyle \displaystyle \begin{align*} -\int_0^3{\frac{u}{2\sqrt{u}\,(u + 1)}\,du} &= -\int_0^{\sqrt{3}}{\frac{v^2}{v^2 + 1}\,dv} \\ &= -\int_0^{\sqrt{3}}{1 - \frac{1}{v^2 + 1}\,dv} \\ &= -\left[v - \arctan{v}\right]_0^{\sqrt{3}} \\ &= -\left[\left(\sqrt{3} - \arctan{\sqrt{3}}\right) - \left(0 - \arctan{0}\right)\right] \\ &= -\left(\sqrt{3} - \frac{\pi}{3} - 0 + 0\right) \\ &= \frac{\pi}{3} - \sqrt{3} \end{align*}$ - Sep 17th 2011, 05:39 AMskeeterRe: integral question
- Sep 17th 2011, 10:02 AMSneakyRe: integral question
oh i mis-read it, its -2 to -1. But I think I can fix it myself.

Also I have a question about how to change the integral bounds when you do a change of variable. What if you change u=cos(x), and the integral bound was -inf to inf, then what would be the new integral bounds? - Sep 17th 2011, 10:09 AMProve ItRe: integral question
- Sep 17th 2011, 10:23 AMSneakyRe: integral question
For the integral ln(x) / x dx, i let u=ln(x), then got

[e to infinity]

integral u du

[1 to infinity]

then i got (after integrating and subbing in the bounds)

ln^2(infinity)/2 - 0

Is this right so far? - Sep 17th 2011, 10:29 AMProve ItRe: integral question
If you change the bounds, you don't convert the function back to a function of x, leave it in terms of u and substitute the u values.

But you are correct that the integral does not converge. - Sep 17th 2011, 10:32 AMSneakyRe: integral question
If i do that I end up with (u^2) / 2 |[1 to inf]

then it ends up being infinity - 1/2

It doesn't look correct. - Sep 17th 2011, 04:21 PMSneakyRe: integral question
Does this mean the answer is infinity?

- Sep 17th 2011, 10:02 PMProve ItRe: integral question