Problem with exponential integral

**Archduke01** · February 24th 2010, 07:39 PM

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.

**TheEmptySet** · February 24th 2010, 07:49 PM

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 $u=e^{2x}$

**Archduke01** · February 24th 2010, 07:51 PM

Originally Posted by TheEmptySet

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

What would du be? 2?

**TheEmptySet** · February 24th 2010, 07:57 PM

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 $du=...$

**Archduke01** · February 24th 2010, 08:03 PM

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?

**Prove It** · February 24th 2010, 08:14 PM

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 $\theta = \arctan{\frac{u}{8}}$ .

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

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

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

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

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

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

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

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

**Archduke01** · February 24th 2010, 08:19 PM

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?

**Prove It** · February 24th 2010, 08:24 PM

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).

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