# Thread: Having Trouble with Integral?

1. ## Having Trouble with Integral?

Hi, am doing a question looking for the length of a curve, the function given is $\displaystyle y=e^x + 1/4e^-x$, from $\displaystyle [1,0]$ is my interval.

I got all the parts down i got $\displaystyle dy/dx=e^x-1/4e^-x$then from the length of curve formula,

$\displaystyle L = \int\ {\sqrt{1 + (e^x -1/4e^x)^2}}$

my problem is the integral under the square root, i tried substitution i tried parts i couldnt get anything, any clues on how to start off this integral. Thank You

$\displaystyle f(x)=e^x+\frac{1}{4}e^{-x}$
$\displaystyle f(x)=\frac{2e^x+\frac{1}{2e^x}}{2}$
$\displaystyle f(x)=\frac{e^{\ln2}e^x+\frac{1}{e^{\ln2}e^x}}{2}$
$\displaystyle f(x)=\frac{e^{(x+\ln2)}+e^{-(x+\ln2)}}{2}$
$\displaystyle f(x)=\cosh(x+\ln2)$
and is thus just a translated hyperbolic cosine, so the curve is a catenary. That might make the integral easier using hyperbolic function identities, but I can't say for sure.

--Kevin C.

3. In fact, it will make the integral easier. With $\displaystyle y=\cosh(x+\ln2)$, we find $\displaystyle \frac{dy}{dx}=\sinh(x+\ln2)$, and so $\displaystyle \sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\sin h^2(x+\ln2)}$, and since $\displaystyle \cosh^2x-\sinh^2x=1$, $\displaystyle 1+\sinh^2x=\cosh^2x$, so
$\displaystyle \sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\sin h^2(x+\ln2)}$
$\displaystyle =\sqrt{\cosh^2(x+\ln2)}$
$\displaystyle =\cosh(x+\ln2)$, as the hyperbolic cosine is always positive, and you should be able to integrate that.

--Kevin C.

4. I dont think ive ever worked with an hyperbolic function like

cosh(x+ln2)

how would i go about starting to integrate such a function? do i make a substiution.
Ive never worked with these types before?

5. would i make a sub. $\displaystyle u=x + ln2 \ du=dx$

then i would get

my antidervitive is ---> $\displaystyle sinh(x + ln2) + c$

6. Originally Posted by zangestu888

$\displaystyle L = \int\ {\sqrt{1 + (e^x -1/4e^x)^2}}$

my problem is the integral under the square root, i tried substitution i tried parts i couldnt get anything, any clues on how to start off this integral. Thank You
You need to do first some of algebra!

$\displaystyle 1+{{\left( {{e}^{x}}-\frac{1}{4{{e}^{x}}} \right)}^{2}}=1+{{e}^{2x}}-\frac{1}{2}+\frac{1}{16{{e}^{2x}}}={{\left( {{e}^{x}}+\frac{1}{4{{e}^{x}}} \right)}^{2}}.$

Extract the square root and you'll get a very easy integral to solve.

7. how is that easier? am not sure how to solve it do i make a sub?

8. got it thanks!