Problem with exponential integral

• February 24th 2010, 07:39 PM
Archduke01
Problem with exponential integral
http://euler.vaniercollege.qc.ca/web...e4ff233941.png

I'm lost as to how to go about solving this. u = 4x and du = 4 dx but beyond that I'm stumped. That constant in the denominator confuses me.
• February 24th 2010, 07:49 PM
TheEmptySet
Quote:

Originally Posted by Archduke01

I'm lost as to how to go about solving this. u = 4x and du = 4 dx but beyond that I'm stumped. That constant in the denominator confuses me.

Hint $e^{4x}=\left( e^{2x}\right)^2$ Then set $u=e^{2x}$
• February 24th 2010, 07:51 PM
Archduke01
Quote:

Originally Posted by TheEmptySet
Hint $e^{4x}=\left( e^{2x}\right)^2$ Then set $u=e^{2x}$

What would du be? 2?
• February 24th 2010, 07:57 PM
TheEmptySet
Quote:

Originally Posted by Archduke01
What would du be? 2?

What is the derivative of $e^{\lambda x}$?

if $u=e^{\lambda x} \implies \frac{du}{dx}=\lambda \cdot e^{\lambda x}$

Then in terms of differentials you get $du=...$
• February 24th 2010, 08:03 PM
Archduke01
Quote:

Originally Posted by TheEmptySet
What is the derivative of $e^{\lambda x}$?

if $u=e^{\lambda x} \implies \frac{du}{dx}=\lambda \cdot e^{\lambda x}$

Then in terms of differentials you get $du=...$

$2e^{2x}$!

$1/2 \int 2e^{2x}/(e^{4x} + 64)$

$1/2 \int du/(u^2+64)$

... How can I proceed though?
• February 24th 2010, 08:14 PM
Prove It
Quote:

Originally Posted by Archduke01
$2e^{2x}$!

$1/2 \int 2e^{2x}/(e^{4x} + 64)$

$1/2 \int du/(u^2+64)$

... How can I proceed though?

Now use a trigonometric substitution $u = 8\tan{\theta}$. Make note that $du = 8\sec^2{\theta}\,d\theta$.

Also note that $\theta = \arctan{\frac{u}{8}}$.

$\int{\frac{1}{u^2 + 64}\,du} = \int{\frac{1}{(8\tan{\theta})^2 + 64}\,8\sec^2{\theta}\,d\theta}$

$= \int{\frac{8\sec^2{\theta}}{64\tan^2{\theta} + 64}\,d\theta}$

$= \int{\frac{8\sec^2{\theta}}{64(\tan^2{\theta} + 1)}\,d\theta}$

$= \int{\frac{\sec^2{\theta}}{8\sec^2{\theta}}\,d\the ta}$

$= \int{\frac{1}{8}\,d\theta}$

$= \frac{1}{8}\theta + C$

$= \frac{1}{8}\arctan{\frac{u}{8}} + C$

$= \frac{1}{8}\arctan{\frac{e^{2x}}{8}} + C$.
• February 24th 2010, 08:19 PM
Archduke01
Quote:

Originally Posted by Prove It
Now use a trigonometric substitution $u = 8\tan{\theta}$. Make note that $du = 8\sec^2{\theta}\,d\theta$.

Also note that $\theta = \arctan{\frac{u}{8}}$.

Thank you for the detailed solution. But why did you choose 8?
• February 24th 2010, 08:24 PM
Prove It
Quote:

Originally Posted by Archduke01
Thank you for the detailed solution. But why did you choose 8?

Because if you have an integral of the form

$\int{\frac{1}{a^2 + x^2}\,dx}$

you make the substitution $x = a\tan{\theta}$.

This is so that you can factor out the $a^2$ and then make use of the identity $1 + \tan^2{\theta} = \sec^2{\theta}$.

And since the derivative of $\tan{\theta}$ is $\sec^2{\theta}$, this also means that the $\sec^2{\theta}$ will be eliminated (as you will end up with them on the top and bottom).