Evaluate involving arcsech and square roots

**Karl Harder** · August 27th 2009, 04:26 AM

How do i evaluate

$<br /> \int_{1}^{x}dz\sqrt{\frac{z}{z^3-1}}<br />$

It is part of a previous question where i prooved that the derivative of

$\frac{dy}{dx}sech^-1(x) = -\frac{1}{x\sqrt{x^2-1}}$

I'm sure that i have to do some rearranging of the problem to get it into the derivative of [tex] sech^-1(x) [\math].

I have tried to do the problem and so far the solution i have is

$\int_{1}^{x}dz\sqrt{\frac{z}{z^3-1}} = \int_{1}^{x}dz\sqrt{\frac{z}{z^3}} + \int_{1}^{x}dz\sqrt{\frac{z}{-1}} = \int_{1}^{x}dz\sqrt{\frac{1}{z^2}} - \int_{1}^{x}dz\sqrt{z}$

Either i have made a mistake somewhere or i am going about the problem the wrong way but i have hit a brick wall. Can someone please give me some feedback on what i need to do or what i have done wrong

**Defunkt** · August 27th 2009, 04:51 AM

$\sqrt \frac{z}{z^3-1} \neq \sqrt \frac{z}{z^3} + \sqrt \frac{z}{-1}$ !!!!

**Karl Harder** · August 27th 2009, 05:53 AM

Originally Posted by Defunkt

$\sqrt \frac{z}{z^3-1} \neq \sqrt \frac{z}{z^3} + \sqrt \frac{z}{-1}$ !!!!

So if I can't do that then how can i rearrange the question to integrate it?

**DeMath** · August 27th 2009, 09:08 AM

Maybe $\int_1^x {\sqrt {\frac{{{z^3}}}{{z - 1}}} dz}$ ??

**halbard** · August 27th 2009, 11:21 AM

If you put $u=z^{3/2}$ then you should obtain

$\int_1^x\sqrt{\frac z{z^3-1}}\mathrm dz=\int_1^{x^{3/2}}\frac2{3\sqrt{u^2-1}}\mathrm du$ . Would this help?

**DeMath** · August 27th 2009, 12:13 PM

Originally Posted by halbard

If you put $u=z^{3/2}$ then you should obtain

$\int_1^x\sqrt{\frac z{z^3-1}}\mathrm dz=\int_1^{x^{3/2}}\frac2{3\sqrt{u^2-1}}\mathrm du$ . Would this help?

If $u=z^{3/2}$ then

$\int_1^x\sqrt{\frac z{z^3-1}}\, dz=\int_1^{x^{3/2}}\frac{2u^{8/3}}{3\sqrt{u^2-1}}\, du$

**halbard** · August 27th 2009, 01:29 PM

Originally Posted by DeMath

This is wrong.
If you put $u=z^{3/2}$ then

$\int_1^x\sqrt{\frac z{z^3-1}}\, dz=\color{red}{\int_1^{x^{3/2}}\frac{2u^{8/3}}{3\sqrt{u^2-1}}\, du}$

This is wrong.

OK, let's do this slowly...

Put $u=z^{3/2}$ or $z=u^{2/3}$ , so that $\mathrm dz={\textstyle\frac23}u^{-1/3}\mathrm du$ .

Then $\int_1^x\sqrt{\frac z{z^3-1}}\mathrm dz=\int_1^x\frac{z^{1/2}}{\sqrt{z^3-1}}\mathrm dz=\int_1^{x^{3/2}}\frac{(u^{2/3})^{1/2}}{\sqrt{(u^{2/3})^3-1}}{\textstyle\frac23}u^{-1/3}\mathrm du$ $={\textstyle\frac23}\int_1^{x^{3/2}}\frac{u^{1/3}}{\sqrt{u^2-1}}u^{-1/3}\mathrm du={\textstyle\frac23}\int_1^{x^{3/2}}\frac{u^{1/3}u^{-1/3}}{\sqrt{u^2-1}}\mathrm du$

and the rest is obvious. Notice it is $u^{1/3}u^{-1/3}$ and not $u^3u^{-1/3}$ in the numerator.

The answer is ${\textstyle\frac23}\cosh^{-1}(x^{3/2})$ .

Differentiate it: $\mathrm d\bigl({\textstyle\frac23}\cosh^{-1}(x^{3/2})\bigr)=\frac2{3\sqrt{(x^{3/2})^2-1}}\mathrm d(x^{3/2})=\frac2{3\sqrt{x^3-1}}\frac32\sqrt x\,\mathrm dx=\sqrt{\frac x{x^3-1}}\mathrm dx$ $=\mathrm d\left(\int_1^x\sqrt{\frac z{z^3-1}}\mathrm dz\right)$ .

Who's the daddy?

Incidentally, putting $z=u^{-2/3}$ gives

$\int_1^{x^{-3/2}}\frac{u^{-1/3}}{\sqrt{u^{-2}-1}}({\textstyle -\frac23}u^{-5/3})\mathrm du={\textstyle\frac23}\int_1^{x^{-3/2}}-\frac1{u\sqrt{1-u^2}}\mathrm du={\textstyle\frac23}\mathop{\textrm{sech}}\nolim its^{-1}(x^{-3/2})$ which is quite probably what the question setter was after.

**DeMath** · August 27th 2009, 02:41 PM

Yes, you are right

Sorry, I was very inattentive.

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