anonymous
  • anonymous
Improper integral from 0 to infinity ( e^x/e^2x+) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
could you solve that problem please guys?
anonymous
  • anonymous
\[\int\limits_{0}^{\infty} \frac{e ^{x}}{e ^{2x}+?????}dx\]
anonymous
  • anonymous
sorry 3

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myininaya
  • myininaya
\[\int\limits_{}^{}\frac{e^x}{e^{2x}+3} dx\] \[\text{ Let } e^x=\sqrt{3} \tan(\theta) => e^x dx= \sqrt{3} \sec^2(\theta) d \theta\] \[\tan(\theta)=\frac{e^x}{\sqrt{3}}\] Make right triangle. Label that triangle based on what you are given. \[Opp=e^x, Adj=\sqrt{3} => Hyp=\sqrt{e^{2x}+3}\] So we have \[\int\limits_{}^{}\frac{ \sqrt{3} \sec^2(\theta) d \theta}{3 \tan^2(\theta)+3} \] \[\sqrt{3} \int\limits_{}^{}\frac{\sec^2(\theta)}{3(\tan^2(\theta) +1)} d \theta\] \[\frac{\sqrt{3}}{3} \int\limits_{}^{}\frac{\sec^2(\theta)}{\sec^2(\theta)} d \theta= \frac{\sqrt{3}}{3} \theta+C\] \[=\frac{\sqrt{3}}{3} \tan^{-1}(\frac{e^x}{\sqrt{3}})+C\] So we have \[\lim_{b \rightarrow \infty} \frac{\sqrt{3}}{3} \tan^{-1}(\frac{e^x}{\sqrt{3}})|_0^b\]
myininaya
  • myininaya
\[\frac{\sqrt{3}}{3}\lim_{b \rightarrow \infty}(\tan^{-1}(\frac{e^b}{\sqrt{3}})- \tan^{-1}(\frac{1}{\sqrt{3}}))\]
myininaya
  • myininaya
\[\frac{\sqrt{3}}{3}(\frac{\pi}{2}-\tan^{-1}(\frac{1}{\sqrt{3}}))\] \[\frac{\sqrt{3}}{3}(\frac{\pi}{2}-\frac{\pi}{6})\]
myininaya
  • myininaya
\[\frac{\sqrt{3}}{3}(\frac{3 \pi - \pi }{6})=\frac{\sqrt{3}}{3}(\frac{2 \pi}{6})=\frac{\sqrt{3}}{3}(\frac{\pi}{3})=\frac{\sqrt{3} \pi}{9}\]
anonymous
  • anonymous
Thank you so much but I don't understand it
myininaya
  • myininaya
Which part?
anonymous
  • anonymous
I don't know all of it I don't get it
anonymous
  • anonymous
why don't you use the substitution in the first step
anonymous
  • anonymous
?
myininaya
  • myininaya
i did use substitution in the first step
myininaya
  • myininaya
for this one you don't need to draw the right triangle i did anyways because i didn't know how the antiderivative was gonna turn out but my sub came first
anonymous
  • anonymous
Thank You so much Myininaya You alway help me that is really kind of you
myininaya
  • myininaya
If you want to ask more questions about this problem, don't be afraid to.
anonymous
  • anonymous
To be honest Myininaya I don't understand it :(
myininaya
  • myininaya
Do you understand the sub that used?
anonymous
  • anonymous
I don't know why you used (tan and sec)
anonymous
  • anonymous
for it
myininaya
  • myininaya
Ok lets look at it for a second we have... \[\int\limits_{}^{}\frac{e^x}{e^{2x}+3} dx\] \[\int\limits_{}^{}\frac{e^x}{(e^{x})^2+3} dx\] Can you recall the trig identity: \[\sin^2(x)+\cos^2(x)=1\] There was also: \[\tan^2(x)+1=\sec^2(x)\] (if you divide that first trig identity I wrote by cos^2(x)) So he have \[(e^x)^2+3 \text{ \in the denominator }\] We want to figure out a sub that we can make so we can write this as one term well we just said: \[(\tan(x))^2+1=(\sec(x))^2\] Doesn't this look similar to: \[(e^x)^2+3\] (well at least the left hand side of that equation anyways, right?) How can we use this identity here: What if I wrote : \[\frac{3}{3}(e^x)^2+3=3(\frac{1}{3}(e^x)^2+1)\] What if we made the sub \[e^x=\sqrt{3} \tan(\theta)\] so we will be able to use the trig identity that we wrote \[3(\frac{1}{3}(\sqrt{3} \tan(\theta))^2+1)=3(\frac{1}{3}(3 \tan^2(\theta)+1)=3(\tan^2(\theta)+1)\] \[=3\sec^2(\theta)\]
myininaya
  • myininaya
since \[\tan^2(p)+1=\sec^2(p)\] \[x^2 + a \text{ try something with tan( )}\] since \[\tan^2(p)=\sec^2(p)-1\] \[x^2 - a \text{ try something with sec( )}\] since \[\cos^2(p)=1- sin^2(p)\] \[a - x^2 \text{ try something with sin( )} \] --- Now these are really the only three trig sub you need (you can use the others but you only need these 3)
anonymous
  • anonymous
other form of intregration: \[\int\limits_{0}^{\infty}\frac{e ^{x}}{3+(e ^{x})^{2}}=\int\limits_{0}^{\infty}\frac{e ^{x}}{3\left[ 1+ (\frac{e ^{x}}{\sqrt{3}})^{2} \right]}=\frac{1}{\sqrt{3}}\int\limits_{0}^{\infty}\frac{e ^{x}}{\sqrt{3}\left[ 1+ (\frac{e ^{x}}{\sqrt{3}})^{2} \right]}=\frac{1}{\sqrt{3}}\tan^{-1} (\frac{e ^{x}}{\sqrt{3}})+C\]
myininaya
  • myininaya
So we had that first one I mentioned \[x^2+a\] We can even say instead of this that we have \[ (f(x))^2+a\] and since we have this we use tan( ) in our sub
anonymous
  • anonymous
Thank You guys some much I understood right now thank you again

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