## anonymous 4 years ago Improper integral from 0 to infinity ( e^x/e^2x+) dx

1. anonymous

could you solve that problem please guys?

2. anonymous

$\int\limits_{0}^{\infty} \frac{e ^{x}}{e ^{2x}+?????}dx$

3. anonymous

sorry 3

4. 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 $\lim_{b \rightarrow \infty} \frac{\sqrt{3}}{3} \tan^{-1}(\frac{e^x}{\sqrt{3}})|_0^b$

5. myininaya

$\frac{\sqrt{3}}{3}\lim_{b \rightarrow \infty}(\tan^{-1}(\frac{e^b}{\sqrt{3}})- \tan^{-1}(\frac{1}{\sqrt{3}}))$

6. myininaya

$\frac{\sqrt{3}}{3}(\frac{\pi}{2}-\tan^{-1}(\frac{1}{\sqrt{3}}))$ $\frac{\sqrt{3}}{3}(\frac{\pi}{2}-\frac{\pi}{6})$

7. myininaya

$\frac{\sqrt{3}}{3}(\frac{3 \pi - \pi }{6})=\frac{\sqrt{3}}{3}(\frac{2 \pi}{6})=\frac{\sqrt{3}}{3}(\frac{\pi}{3})=\frac{\sqrt{3} \pi}{9}$

8. anonymous

Thank you so much but I don't understand it

9. myininaya

Which part?

10. anonymous

I don't know all of it I don't get it

11. anonymous

why don't you use the substitution in the first step

12. anonymous

?

13. myininaya

i did use substitution in the first step

14. 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

15. anonymous

Thank You so much Myininaya You alway help me that is really kind of you

16. myininaya

17. anonymous

To be honest Myininaya I don't understand it :(

18. myininaya

Do you understand the sub that used?

19. anonymous

I don't know why you used (tan and sec)

20. anonymous

for it

21. 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 $3(\frac{1}{3}(\sqrt{3} \tan(\theta))^2+1)=3(\frac{1}{3}(3 \tan^2(\theta)+1)=3(\tan^2(\theta)+1)$ $=3\sec^2(\theta)$

22. 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)

23. anonymous

other form of intregration: $\int\limits_{0}^{\infty}\frac{e ^{x}}{3+(e ^{x})^{2}}=\int\limits_{0}^{\infty}\frac{e ^{x}}{3\left[ 1+ (\frac{e ^{x}}{\sqrt{3}})^{2} \right]}=\frac{1}{\sqrt{3}}\int\limits_{0}^{\infty}\frac{e ^{x}}{\sqrt{3}\left[ 1+ (\frac{e ^{x}}{\sqrt{3}})^{2} \right]}=\frac{1}{\sqrt{3}}\tan^{-1} (\frac{e ^{x}}{\sqrt{3}})+C$

24. 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

25. anonymous

Thank You guys some much I understood right now thank you again