Just wondering if you could help me integrate the following:
[(cosh (x^0.5)) / (x^0.5)] dx
with respect to x
cheers
Hello, daaavo!
$\displaystyle \int \frac{\cosh\left(x^{\frac{1}{2}}\right)}{x^{\frac{ 1}{2}}}\,dx$
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: . $\displaystyle 2\sinh\left(x^{\frac{1}{2}}\right) + C$
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:
$\displaystyle \int{e^{u}}du+\int{e^{-u}}du=2sinh(u)$
$\displaystyle =\boxed{2sinh(\sqrt{x})}$
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
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 $\displaystyle \frac{1}{24} (x^2 + 1)^{24} + C$.