1. ## Integral of e^sec^2x

Hello,

I need help evaluating this integral, which i cannot seem to get. It is part of a bigger question(ordinary differential equations)

The integral is

$\displaystyle \int_(e^{sec^2(x)})$

2. Originally Posted by mkelly09
Hello,

I need help evaluating this integral, which i cannot seem to get. It is part of a bigger question(ordinary differential equations)

The integral is

$\displaystyle \int_(e^{sec^2(x)})$
Let me start by saying first that - I don't know. Because this feels like $\displaystyle e^{u^2}$ which has no known anti-derivative. And if we were to consider parts, it doesnt look good.

What about The taylor series for $\displaystyle e^x$. We could replace x with sec^2x?

I don't know. Just some thoughts.

3. Originally Posted by mkelly09
Hello,

I need help evaluating this integral, which i cannot seem to get. It is part of a bigger question(ordinary differential equations)

The integral is

$\displaystyle \int_(e^{sec^2(x)})$
post the original problem in its entirety

4. OK, original problem is an ordinary differential equation.

[tex] y' + ytanX = y^2 [/mATH]

So this is a bernouilli differential equation with a=2

I set $\displaystyle u(x) = y^{1-a} = 1/y$

So it becomes linear ordinary differential equation by re-writing in terms of u like so:

$\displaystyle u' - (tanx)u = -1$

This is now a linear ODE with $\displaystyle f(x) = -tanx$ and $\displaystyle r(x) = -1$

Well i don't know this Latex code very well but im going to give it a try:

The general solution for a linear ODE: $\displaystyle u(x) = \dfrac{-\int {e^{\int{f(x)dx}} * r(x)dx} + c}{e^{\int{f(x)dx}}}$

ill try to figure out this latex code but i gotta run for now, i hope you can all see how this leads me too my original problem posted above.

$\displaystyle \int e^{-sec^2(x)}$

5. Originally Posted by mkelly09
OK, original problem is an ordinary differential equation.

$\displaystyle y' + ytanX = y^2$

So this is a bernouilli differential equation with a=2

I set $\displaystyle u(x) = y^{1-a} = 1/y$

So it becomes linear ordinary differential equation by re-writing in terms of u like so:

$\displaystyle u' - (tanx)u = -1$

This is now a linear ODE with $\displaystyle f(x) = -tanx$ and $\displaystyle r(x) = -1$

Well i don't know this Latex code very well but im going to give it a try:

The general solution for a linear ODE: $\displaystyle u(x) = \int {(e^{\int {f(x)dx})(r(x))dx} + c$

ill try to figure out this latex code but i gotta run for now, i hope you can all see how this leads me too my original problem posted above.

$\displaystyle \int e^{-sec^2(x)}$
ok, so there's your mistake. i assume you thought $\displaystyle \int \tan x ~dx = \sec^2 x + C$. This is false. $\displaystyle \int \tan x ~dx = \ln | \sec x | + C$. what you had before was the derivative.

so, your integrating factor is $\displaystyle \exp \left( - \int \tan x ~dx \right) = \exp ( - \ln |\sec x|) = \cos x$

so your differential equation becomes: $\displaystyle \cos x \cdot u' - \sin x \cdot u = - \cos x$

now you should be able to finish up. see how posting the original problem helps?

6. Originally Posted by Jhevon
ok, so there's your mistake. i assume you thought $\displaystyle \int \tan x ~dx = \sec^2 x + C$. This is false. $\displaystyle \int \tan x ~dx = \ln | \sec x | + C$. what you had before was the derivative.

so, your integrating factor is $\displaystyle \exp \left( - \int \tan x ~dx \right) = \exp ( - \ln |\sec x|) = \cos x$

so your differential equation becomes: $\displaystyle \cos x \cdot u' - \sin x \cdot u = - \cos x$

now you should be able to finish up. see how posting the original problem helps?

Great, thanks so much for your help. This of course did the trick!