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.

Printable View

- Feb 24th 2010, 07:39 PMArchduke01Problem with exponential integralhttp://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.

- Feb 24th 2010, 07:49 PMTheEmptySet
- Feb 24th 2010, 07:51 PMArchduke01
- Feb 24th 2010, 07:57 PMTheEmptySet
- Feb 24th 2010, 08:03 PMArchduke01
- Feb 24th 2010, 08:14 PMProve 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 $\displaystyle \theta = \arctan{\frac{u}{8}}$.

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

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

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

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

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

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

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

$\displaystyle = \frac{1}{8}\arctan{\frac{e^{2x}}{8}} + C$. - Feb 24th 2010, 08:19 PMArchduke01
- Feb 24th 2010, 08:24 PMProve It
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).