# Thread: Problem with exponential integral

1. ## Problem with exponential integral

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.

2. 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 $\displaystyle e^{4x}=\left( e^{2x}\right)^2$ Then set $\displaystyle u=e^{2x}$

3. Originally Posted by TheEmptySet
Hint $\displaystyle e^{4x}=\left( e^{2x}\right)^2$ Then set $\displaystyle u=e^{2x}$
What would du be? 2?

4. Originally Posted by Archduke01
What would du be? 2?
What is the derivative of $\displaystyle e^{\lambda x}$?

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

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

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

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

Then in terms of differentials you get $\displaystyle du=...$
$\displaystyle 2e^{2x}$!

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

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

... How can I proceed though?

6. Originally Posted by Archduke01
$\displaystyle 2e^{2x}$!

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

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

... How can I proceed though?
Now use a trigonometric substitution $\displaystyle u = 8\tan{\theta}$. Make note that $\displaystyle du = 8\sec^2{\theta}\,d\theta$.

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

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

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

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

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

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

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

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

$\displaystyle = \frac{1}{8}\arctan{\frac{e^{2x}}{8}} + C$.

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

Also note that $\displaystyle \theta = \arctan{\frac{u}{8}}$.
Thank you for the detailed solution. But why did you choose 8?

8. 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

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

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

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

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