Question

what is the best way to integrate 1/sqrt(e^(2x)-25)?

Answer 1

\[\int\limits_{}^{}\frac{1}{ \sqrt{e^{2x}-25}} dx\]
Hmmm... so we are looking at this...

Answer 2

double substitution?

Answer 3

u = 2x , du = 2dx
\[\frac{1}{2}\int \frac{1}{\sqrt{e^u-25}}du\]

Answer 4

\[5 \sec(\theta)=e^x \text{ this might work too }\]

Answer 5

And then...

p = e\(^u\) , dp = e\(^u\)du \(\implies\) du = \(\frac{dp}{e^u}\)
\[\frac{1}{2}\int \frac{1}{\sqrt{p-25}}\cdot \frac{dp}{p}\]

Answer 6

\[1/2\int\limits1/(\sqrt(p)(p+25)) dp\]

Answer 7

i don't understand the step between your last response and this... I understand up to this point.

Answer 8

Thanks for your help, btw.

Answer 9

Freckles, how are you doing that substitution?

Answer 10

\[5 \sec(\theta)=e^x \\ 5 \sec(\theta) \tan(\theta)d \theta=e^x dx \text{ differentiated both sides } \\ \text{ since } e^x=5 \sec(\theta) \text{ we have } e^x \tan(\theta) d \theta=e^x dx \\ \text{ dividing both sides by } e^x \text{ gives } \\ \tan(\theta) d \theta=d x \\ \int\limits_{}^{} \frac{ \tan(\theta) d \theta}{\sqrt{(5 \sec(\theta))^2-25}}=\int\limits \frac{ \tan(\theta) d \theta}{\sqrt{25 \sec^2(\theta)-25}}\]

Answer 11

\[=\int\limits \frac{\tan(\theta)}{\sqrt{25} \sqrt{\sec^2(\theta)-1}} d \theta \]
I think this can be finished from here pretty easily

Answer 12

My way is nearly the exact same way as freckles' but he just used trig subs to reduce the number of variable substitions in order to simplify the problem.

Answer 13

substitutions*