1. ## Trignometric Integrals

$\displaystyle \int \sqrt{x^2 - 4} dx$

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

'e' is for Euler's number.

I'm having a serious problem with these. I just can't get it.
Can you also show the steps at how you arrived at the answer.
Thank you so much!

2. To the first you can do: $\displaystyle \sec \theta=\dfrac{x}{2}$ and to the second $\displaystyle u=e^x$

3. Originally Posted by Nacho
To the first you can do: $\displaystyle \sec \theta=\dfrac{x}{2}$ and to the second $\displaystyle u=e^x$
i ended up with
$\displaystyle 4 \int \frac{sin^2\theta}{cos^3\theta} d\theta$

i dont know how else to go from there

4. Originally Posted by larryboi7
i ended up with
$\displaystyle 4 \int \frac{sin^2\theta}{cos^3\theta} d\theta$

i dont know how else to go from there
$\displaystyle 4 \int \frac{sin^2\theta}{cos^3\theta} d\theta = 4\int tan^2(\theta) sec(\theta) d\theta$
$\displaystyle = 4 \int (sec^2(\theta)-1)sec(\theta) d\theta = 4 \int sec^3(\theta) d\theta - 4 \int sec(\theta) d\theta = ....$

Can you complete it ?
Use integration by parts for the first one.

For the second integral in your question:
Its $\displaystyle e^2x$ or $\displaystyle e^{2x}$ ?

5. the details of the first are given in a previous post of yours "competing the integration"

6. Originally Posted by Calculus26
the details of the first are given in a previous post of yours "competing the integration"
Normally I'd close one of the threads but in this instance it's too difficult to straighten things out so I'm leaving both threads open.

Please post replies here that are only related to $\displaystyle \int \frac{e^x}{1+e^{2x}} dx$. Replies to the other one should be posted here: http://www.mathhelpforum.com/math-help/calculus/128484-completing-integration.html