At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
show full steps if possible o_o
Ready to work together on this?
This one's going to be a bit tricky, I'm afraid.
lol i know its pretty rediculous
Okie dokie. Let's do this. I'll give you a hint: The thing is in the form of a trig identity.
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
don't you have to find the antiderivative of the function?
well i know its arctan its just the steps involved inbetween to find the answer
im just not sure if its u sub or something thats why i wana see it with full steps
@galactic yes you have to find the antiderivative saying integral is another way of saying anti derivative
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.
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.
shouldnt the 1/2 go on the top and turn into 2arctan(2x)?
@uber i wanted to ask how comes we are using the trig identities. is it because it is one over the function??
@Blisk: Well I mean it's originally just (1/(2(tan^-1(2x)). So you pull the 1/2 to the left.
@galactic; Look at the derivative of tan^-1.
how did you get 1/(2tan^-1(2x))?
Can Someone Help Me After They finish doing this guy!! :D
that sounds wrong lol
ok i know :D
@blisk: I chain ruled it.
d/dx (2x) = 1
i didnt know you could chain rule backwards
Oh shoot! You're right! I think I made up a rule then. :(
isnt the derivative of tan^-1......sec^2(x)tan^-1?!?
the deriv of tan^-1 is 1/x^2+1
the deriv of tan = sec^2(x) and the deriv of sec is sectan
your right i confused myself sorry!!!
lol dont worry about it
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.
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?
Yes it is.