# Arclength of exp from 0 to 1

1. Dec 1, 2009

### Samuelb88

1. The problem statement, all variables and given/known data
Find the arc length of y=e^x, from [0,1].

2. Relevant equations

3. 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. :)

Last edited by a moderator: Dec 1, 2009
2. Dec 1, 2009

### clamtrox

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

3. Dec 1, 2009

### HallsofIvy

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".

4. Dec 1, 2009

### 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

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

?