Question

Integrate e^3x(1+e^2x)^1/2

Answer 1

Here is one approach,
\[\large\rm \int\limits e^{3x}\sqrt{1+e^{2x}}~dx\quad=\int\limits \left(e^x\right)^2\sqrt{1+\left(e^x\right)^2}~e^x~dx\]From there you can make the substitution,\[\large\rm \tan \theta= e^x\]

Answer 2

Works out well enough,
you have to apply Secant Reduction Formula at one point.

Answer 3

There's probably some way to do Integration by Parts,
I'm just not seeing it.

Answer 4

I tried integration by parts but you would keep on getting increasing power of e^x.

Answer 5

Have you considered using substitution, two times. First substitute \[u=e^x\]Then substitute \[u=tan(s)\]

Answer 6

And then you get something where you need to use the reduction formula: \[\int sec^m(s)ds=\frac{sin(s)sec^{m-1}(s)}{m-1}+\frac{m-2}{m-1}\int sec^{-2+m}(s)ds\]

Answer 7

@zepdrix I never studied reduction formulas before i don't get why we choose our substitution for e^x to be tanx. I used your substitution and it worked well

Answer 8

Are you sure that the first thing is \(e^{3x}\), and not \(e^{2x}\)?
That would be indeed much easier:)

Answer 9

here you have \(\sqrt {1 + blah^2}\), and so you want to sort that bit out right away

the identity \(\tan^2 x + 1 = \sec^2 x\) is prolly the most obvious way to get rid of that radical

Answer 10

so if it was \(\sqrt{1- e^{2x}}\) you'd look at the cos sin identity

Answer 11

\[\int\limits_{}^{}\tan x \tan x^2\sqrt{secx^2}dx=\int\limits_{}^{}\tan x(\sec^2x-1)secxdx=\int\limits_{}^{}\tan xsec^3xdx-\int\limits_{}^{}tanxsecx\]

Answer 12

i got my answer as secx^3/3-secx

Answer 13

it should come out as \(\int \; \sec^5 \theta - \sec^3 \theta \; d\theta\)

then those reductions things kick in

Answer 14

Ya I think you made a boo boo somewhere, hmm..\[\large\rm \color{royalblue}{e^x=\tan \theta},\qquad\qquad\qquad \color{orangered}{e^x~dx=\sec^2\theta~d \theta}\]Gives us,\[\rm \int\limits\limits (\color{royalblue}{e^x})^2\sqrt{1+(\color{royalblue}{e^x})^2}~\color{orangered}{~e^x~dx}\qquad=\qquad\int\limits\limits (\color{royalblue}{\tan \theta})^2\sqrt{1+(\color{royalblue}{\tan \theta})^2}~\color{orangered}{\sec^2 \theta~d \theta}\]Which simplifies to,\[\large\rm \int\limits \tan^2 \theta \sec^3 \theta~d \theta\]I don't think you ended up with enough secants in yours.

Answer 15

I know it is not the case, by why isn't like this?
\(\color{#000000 }{ \displaystyle \sqrt{\sec^2\theta}=\color{red}{\left|\color{black}{\sec \theta}\right|} }\) ?

That would (but doesn't) go according to the definition of the absolute value, i.e. \(\color{#000000 }{ \displaystyle \sqrt{\varphi^2}=\left|\varphi \right| }\)

Answer 16

@SolomonZelman i agree, i see that all the time with trig subs

Answer 17

\(\color{#000000 }{ \displaystyle \sqrt{1+\tan^2\theta~}=\sqrt{\sec^2\theta}=\color{red}{\left|\color{black}{\sec \theta}\right|} }\)

maybe something ehre that makes sec\(\theta\) undefined for negatives(?)

Answer 18

Well I remember when doing First-Order ODE's,
for integrating factor you didn't have to be too picky,\[\large\rm \mu=e^{\int\limits p(x)dx}\]We didn't have to be careful and do something like this,\[\large\rm e^{\int\limits\limits 2x~dx}\quad=e^{x^2+c}\]because we were looking for `a suitable substitution`,
not a family of substitutions or anything.
We just needed one that worked.
We just didn't need the +c.

Maybe that is what's going on here?
We just care about the principal root because ummmm
Hmm no -_- hmm

Answer 19

Err not substitution :) But integrating factor*
blah, whatever, i failed to make any kind of point there anyway lol

Answer 20

http://math.stackexchange.com/questions/1118400/trig-substitution-why-can-we-ignore-the-absolute-value

Answer 21

Refer to the tiff attachment.

Answer 22

Woops rob :( The x's be in the exponent, e^{3x} not e^3x

Answer 23

A detailed solution from WolframAlpha through Mathematica v9 is attached.