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BliskBest ResponseYou've already chosen the best response.1
show full steps if possible o_o
 2 years ago

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
Ready to work together on this?
 2 years ago

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
This one's going to be a bit tricky, I'm afraid.
 2 years ago

BliskBest ResponseYou've already chosen the best response.1
lol i know its pretty rediculous
 2 years ago

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
Okie dokie. Let's do this. I'll give you a hint: The thing is in the form of a trig identity.
 2 years ago

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
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
 2 years ago

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

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

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

BliskBest 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
 2 years ago

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
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.
 2 years ago

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
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.
 2 years ago

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

galacticwavesXXBest 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??
 2 years ago

uber1337h4xx0rBest 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.
 2 years ago

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
@galactic; Look at the derivative of tan^1.
 2 years ago

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

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

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
@blisk: I chain ruled it.
 2 years ago

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

uber1337h4xx0rBest ResponseYou've already chosen the best response.0
Oh shoot! You're right! I think I made up a rule then. :(
 2 years ago

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

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

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

galacticwavesXXBest ResponseYou've already chosen the best response.0
your right i confused myself sorry!!!
 2 years ago

abtrehearnBest ResponseYou've already chosen the best response.1
Draw 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.
 2 years ago

BliskBest ResponseYou've already chosen the best response.1
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?
 2 years ago
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