Guys, how I git gud at integral calculus. Time is running out until the end of the semester and my calculus finals, and this shit is killing me. I've spent 5 hours today just pondering over a single equation and I still can't figure it out. I've got other things to do but I can't move on until I understand these integration problems.
For those wondering, the equation in question is (1 / (g - k*v^2)) dv
Even a relatively simple expression like this has me absolutely stumped. What do?
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Trigonometric substitution
ok this one is easy
first factor out g, so you have 1/(g*(1-k/g*v^2))
now you substitute u=sqrt(k/g)*v, so du/sqrt(k/g)=dv
now you have some prefactor, lets call it A*1/(1-u^2)du. Now you rip it apart and say OK: 1/(1-u^2)=C1/(1+u)+C2/(1-u). Determine C1 and C2 such that the equality holds and now you only have to integrate 1+-u which is ln(1+-u). mix everything together, simplify and you should end up with something like prefactor*arctan(something)
you're welcome
The only way is to sit down and integrate like a madman, after you have integrated hundreds of exercises you will be set. There are no standard paths to integrate so you must get the intuition by grinding.
Not even close. I just worked it out. It's a sum of natural logs multiplied by a constant.
Now, on to the next ridiculously time-consuming integration question.
You can solve the problem with a trig sub, just not the one he mentioned. 1-x^2 = (sinx)^2
Which would then equal (cscx)^2, which is a trivial integral.
Lmfao, if you can't get at least a B in this class, you are unworthy of studying science.
Class is easy, just do your best to integrate then check it on your calculator if its a definite integral. If you got it wrong then find where you messed up. If its an indefinite integral then differentiate it to check it.
Maybe I am a brainlet, I don't understand trig substitution at all beyond the basic examples I've been taught in my course.
How what you wrote applies to the question in the OP is beyond me. In fact, how 1-x^2 = (sinx)^2 is beyond me.
not who you're asking, but...
1-x^2, substitute x=cosx, 1-cosx^2=sinx^2 because sinx^2+cosx^2=1
Probably because you just memorized trig cases, and never learned the technique.
I.e
I don't remember any of the "cases" for trig sub besides arctanx, but can easily work out any of the cases with the trig identities (which you should have memorized) and a right triangle. for some reason a lot of schools think memorizing the formulas is better than understanding the concept, but it's not.
AKA, go watch some professor Leonard videos. After you've got the few basic tools it just comes down to cranking out a shit ton of problems, which will translate into fluency on the exam.
Can you "prove" to me why the antiderivative of (secx) is ln |secx + tanx| + c ? Probably not if you've never seen the solution or have 15 minutes to spend on it. Simple shit like this that they tell you to just remember is a GREAT place to start practicing and honing your derivative/integral knowledge
Alright, advice much appreciated. I need to get trig identities learned.
use this site it shows all steps
integral-calculator.com
make up sentences or systems for each one, they're tough as all hell, but it CAN be done
source: 3rd time taking calc 1, hs got c, failed twice(including this semester)
I think if you've failed calc 1 three times you probably have no business taking it in the first place. Go back to algebra and come back when you're ready.
lmfao calc 2 is so fucking easy. the only thing that is kinda hard is the series stuff but u dont even need to read that shit. just learn the taylor expansion and power series stuff for a few functions and ur golden.
whatever you do, don't take calc advice from somebody who just failed calc 1 in their 3rd attempt, jesus christ senpai
OP, you know that you want the denominator to look like 1 - (something squared), so you factor out g.
Then you say, okay, what trig identity has 1 - (something1 squared) = (something2 squared)
Clearly you have to substitue (k/g)v^2 out for, you know, something1 squared.
So you make that substitution, and use your trig identity, and now all you have is some constants multiplied by some trig expression, which you integrate however you like, then work backwards through your substitution to get your answer in terms of your original variables, and you're done
idk it's honestly easier done than said; once you understand what's going on, problems at this level are just a boring series of arithmetic
practice more and make sure you remember your differentiation and integration rules for inverse sin functions, arctan is really common
gl on your final bro
That should be an inverse hyperbolic tangent
>arctan
unless k/g is negative to begin with, the solution will not involve arctan. This is because the substitution u = -v can't change the minus sign to a plus sign when the v term has an even power. It will be arctan (hyperbolic), but professors may not take that as an acceptable answer because they didn't teach it. The method they want you to use is factorizing the 1 - (sqrt(k)*v/sqrt(g))^2 expression into linear terms, then using partial fractions. The result will involve natural log.
Do loads of examples. Get frustrated. Het the shit out of it. Eventually you will get pretty good at it and have a kind of intuition about where to go with it. Some times they are just hard as fuck and you might not get the answer. But remember that people try to figure out shit like this for a living. But I doubt you'll come across anything that hard.
My bad, should be arctanh.
OP here, so I went and learned some trig-substitution, and it's great.
What's interesting though, is that the ODE which the equation in the OP is part of is SO much more difficult to solve if you use trig substitution to solve the integral, compared to the intuitive method. (). What was a simple algebraic manipulation using the intuitive method turns into a nightmare if you use trig substitution to get your f(v) = (g)t expression. I confirmed using wolfram alpha that they are indeed both correct, but I still haven't worked out v in terms of t after using trig substitution to get my expression.