Just wondering if you could help me integrate the following:

[(cosh (x^0.5)) / (x^0.5)] dx

with respect to x

cheers

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- Mar 6th 2008, 03:31 PMdaaavohyperbolic integration
Just wondering if you could help me integrate the following:

[(cosh (x^0.5)) / (x^0.5)] dx

with respect to x

cheers - Mar 6th 2008, 04:00 PMSoroban
Hello, daaavo!

Quote:

$\displaystyle \int \frac{\cosh\left(x^{\frac{1}{2}}\right)}{x^{\frac{ 1}{2}}}\,dx$

We have: .$\displaystyle \int\cosh\left(x^{\frac{1}{2}}\right)\cdot\frac{dx }{x^{\frac{1}{2}}} $

Let $\displaystyle u \:=\:x^{\frac{1}{2}} \quad\Rightarrow\quad du \:=\:\frac{1}{2}x^{-\frac{1}{2}}dx\quad\Rightarrow\quad\frac{dx}{x^{\f rac{1}{2}}} \:=\:2\,du$

Substitute: .$\displaystyle \int\cosh u\,(2\,du) \;\;=\;\;2\int\cosh u\,du \;\;=\;\;2\sinh u + C$

Back-substitute: . $\displaystyle 2\sinh\left(x^{\frac{1}{2}}\right) + C$

- Mar 6th 2008, 04:05 PMgalactus
You could rewrite $\displaystyle \frac{cosh(\sqrt{x})}{\sqrt{x}}=\frac{\frac{1}{2}e ^{\sqrt{x}}+\frac{1}{2}e^{-\sqrt{x}}}{\sqrt{x}}$

Now, let $\displaystyle u=\sqrt{x}, \;\ 2du=\frac{1}{\sqrt{x}}dx$

You get:

$\displaystyle \int{e^{u}}du+\int{e^{-u}}du=2sinh(u)$

$\displaystyle =\boxed{2sinh(\sqrt{x})}$ - Mar 6th 2008, 04:18 PMdaaavo
Thanks for your help - would you also possibly be able to guide me through this one too.... I'm trying to teach myself hyperbolic integration for fun, but the examples in my textbook are pretty poor...

Integrate: 2x((x^2)+1)^23) dx

Thanks very much for all of your help - Mar 6th 2008, 08:44 PMmr fantastic
No hyperbolic functions needed here.

Make the substitution $\displaystyle u = x^2 + 1$. Then $\displaystyle \frac{du}{dx} = 2x \Rightarrow dx = \frac{du}{2x}$.

Then the integral becomes:

$\displaystyle \int (2x) \, u^{23} \, \frac{du}{2x} = \int u^{23} \, du$.

So the answer will be $\displaystyle \frac{1}{24} (x^2 + 1)^{24} + C$.