Blisk 4 years ago integral(1/((2x)^2+1)dx

1. Blisk

show full steps if possible o_o

2. uber1337h4xx0r

Ready to work together on this?

3. uber1337h4xx0r

This one's going to be a bit tricky, I'm afraid.

4. Blisk

lol i know its pretty rediculous

5. uber1337h4xx0r

Okie dokie. Let's do this. I'll give you a hint: The thing is in the form of a trig identity.

6. uber1337h4xx0r

Can you tell me the trig identities for cos^-1 tan^-1 and sin^-1 first? After that, I think you'll know exactly what the answer is :D

7. galacticwavesXX

don't you have to find the antiderivative of the function?

8. Blisk

well i know its arctan its just the steps involved inbetween to find the answer

9. Blisk

im just not sure if its u sub or something thats why i wana see it with full steps

10. Blisk

@galactic yes you have to find the antiderivative saying integral is another way of saying anti derivative

11. uber1337h4xx0r

You know your "X" is 2x. Because of that, you must chain rule it. When you do the derivative of 2x, you get 2. Pull that two out, and because it's on the bottom, it turns to 1/2. So 1/2 tan^-1 (2x) + c should be the answer. But double check with other people to be safe.

12. uber1337h4xx0r

If my wording was weird, I meant that normally it'd be just tan^-1 if x^2 was there. But we have (2x)^2 instead of just (x)^2.

13. Blisk

shouldnt the 1/2 go on the top and turn into 2arctan(2x)?

14. galacticwavesXX

@uber i wanted to ask how comes we are using the trig identities. is it because it is one over the function??

15. uber1337h4xx0r

@Blisk: Well I mean it's originally just (1/(2(tan^-1(2x)). So you pull the 1/2 to the left.

16. uber1337h4xx0r

@galactic; Look at the derivative of tan^-1.

17. Blisk

how did you get 1/(2tan^-1(2x))?

18. daveboyback

Can Someone Help Me After They finish doing this guy!! :D

19. Blisk

that sounds wrong lol

20. daveboyback

ok i know :D

21. uber1337h4xx0r

@blisk: I chain ruled it.

22. daveboyback

Sory

23. uber1337h4xx0r

d/dx (2x) = 1

24. Blisk

i didnt know you could chain rule backwards

25. uber1337h4xx0r

Oh shoot! You're right! I think I made up a rule then. :(

26. galacticwavesXX

isnt the derivative of tan^-1......sec^2(x)tan^-1?!?

27. Blisk

the deriv of tan^-1 is 1/x^2+1

28. Blisk

the deriv of tan = sec^2(x) and the deriv of sec is sectan

29. galacticwavesXX

your right i confused myself sorry!!!

30. Blisk

31. abtrehearn

Draw a right triangle having legs of lengths 1 and 2x. Place the unit length segment om the positive x-axis with one end at the origin O. Erect the perpendicular to the x-axis at A(1, 0) and place the other leg on that perpendicular with its lower end at (1, 0). The other end of that vertical leg is at B(1, 2x). Lrt t be angle AOB. Then tan(t) = 2x/1 = 2x, and (sec(t))^2 dt = 2 dx =>dx = (sec(t))^2/2 dt. If we substitute into the integral, we get(1/2)(sec(t))^2 dt/(1 + (tan(t))^2. and we end up integrating dt/2, so that simplifies to t/2. Getting back to original variable x, we have (1/2) Arctan(2x) + C for the final answer.

32. Blisk

would this work too? S = integral S1/((2x)^2+1) dx u = 2x du = 2 dx => 1/2 du = dx S(1/u^2+1)(1/2) du 1/2S(1/u^2+1) du 1/2 tan^-1(u) Plug U back in 1/2 tan^-1(2x) is that also correct abtre?

33. abtrehearn

Yes it is.