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Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1show full steps if possible o_o

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0Ready to work together on this?

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0This one's going to be a bit tricky, I'm afraid.

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1lol i know its pretty rediculous

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0Okie dokie. Let's do this. I'll give you a hint: The thing is in the form of a trig identity.

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0Can 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

galacticwavesXX
 3 years ago
Best ResponseYou've already chosen the best response.0don't you have to find the antiderivative of the function?

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1well i know its arctan its just the steps involved inbetween to find the answer

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1im just not sure if its u sub or something thats why i wana see it with full steps

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1@galactic yes you have to find the antiderivative saying integral is another way of saying anti derivative

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0You 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.

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0If 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.

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1shouldnt the 1/2 go on the top and turn into 2arctan(2x)?

galacticwavesXX
 3 years ago
Best ResponseYou've already chosen the best response.0@uber i wanted to ask how comes we are using the trig identities. is it because it is one over the function??

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0@Blisk: Well I mean it's originally just (1/(2(tan^1(2x)). So you pull the 1/2 to the left.

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0@galactic; Look at the derivative of tan^1.

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1how did you get 1/(2tan^1(2x))?

daveboyback
 3 years ago
Best ResponseYou've already chosen the best response.0Can Someone Help Me After They finish doing this guy!! :D

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0@blisk: I chain ruled it.

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1i didnt know you could chain rule backwards

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0Oh shoot! You're right! I think I made up a rule then. :(

galacticwavesXX
 3 years ago
Best ResponseYou've already chosen the best response.0isnt the derivative of tan^1......sec^2(x)tan^1?!?

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1the deriv of tan^1 is 1/x^2+1

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1the deriv of tan = sec^2(x) and the deriv of sec is sectan

galacticwavesXX
 3 years ago
Best ResponseYou've already chosen the best response.0your right i confused myself sorry!!!

abtrehearn
 3 years ago
Best ResponseYou've already chosen the best response.1Draw a right triangle having legs of lengths 1 and 2x. Place the unit length segment om the positive xaxis with one end at the origin O. Erect the perpendicular to the xaxis 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.

Blisk
 3 years ago
Best ResponseYou've already chosen the best response.1would 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?
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