Calculating the Arc Length of y=e^x from 0 to 1

[Samuelb88]

Homework Statement

Find the arc length of y=e^x, from [0,1].

Homework Equations

The Attempt at a Solution

s = \int_0^1 (1 + e^2^x)^(^1^/^2^)dx

I let t = e^x, dt=e^xdx; therefore dt/t=dx

s = \int_1^e \frac{(1+t^2)^(^1^/^2^)}{t}\right) dx

Let t = tanT, dt = sec^2(T)dT (T for theta)

s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{sec^3T}{tanT}\right)dT

s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{1/cos^3T}{sinT/cosT} \right) dT

s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{1}{sinT}\right) * \frac{1}{cos^2T} \right)dT

Where do I go from here?

Any help would be greatly appreciated. :)

 

[clamtrox]

A substitution t = sinh y might be a bit easier. It takes some work too though.

 

[HallsofIvy]

Homework Statement

Find the arc length of y=e^x, from [0,1].

Homework Equations

The Attempt at a Solution

s = \int_0^1 (1 + e^2^x)^(^1^/^2^)dx

I let t = e^x, dt=e^xdx; therefore dt/t=dx

s = \int_1^e \frac{(1+t^2)^(^1^/^2^)}{t}\right) dx

Let t = tanT, dt = sec^2(T)dT (T for theta)

s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{sec^3T}{tanT}\right)dT

s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{1/cos^3T}{sinT/cosT} \right) dT

s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{1}{sinT}\right) * \frac{1}{cos^2T} \right)dT

Where do I go from here?

Any help would be greatly appreciated. :)

Since that is an odd power of sin(T), a standard technique is to multiply both numerator and denominator by sin(T) to get
s= \int_{\frac{pi}{4}}^{arctan(e)}\frac{cos^2(T)}{sin^2(T)} sin(T)dT
= \int_{\frac{pi}{4}}^{arctan(e)}\frac{cos^2(T)}{1- cos^2(T)} sin(T)dT

Now let u= cos(T) so du= -sin(T)dt and the integral becomes
\int_{\frac{\sqrt{2}}{2}}^{\frac{1}{\sqrt{1+e^2}}}\frac{u^2}{1- u^2} du
Which you can do by "partial fractions".

 

[Donaldos]

u=\sqrt{1+e^{2x}}

{\rm d}u=\frac{e^{2x}}{\sqrt{1+e^{2x}}}=\frac{u^2-1}{u} {\rm{d}x}

and

s={\int\limits_{\sqrt{2}}^{\sqrt{1+e^2^}}} {\frac{u^2}{u^2-1^}} \quad {\rm d}u

?