From a teacher’s perspective, A is a student who made a careless mistake. C seems like a student who either froze/ran out of time or just forgot and wrote the only formula they could remember. D knows the antiderivative of rational functions is associated with the natural log function, so they have the big picture but not the details. B is pretty rough. Unless the student forgot several trig identities, this shows they either didn’t properly do a u-sub and/or don’t know the chain rule.
The +c is not needed in most cases as mathematicians know that the function stands for an equivalent class of many functions all with the same derivative. In my calculus class we did not need to write that unnecessary +c.
@@AtzenGaffiit matters in differential equations when the problem has an initial or boundary condition. But even then, the auxiliary steps in any calculation don’t need a constant because it cancels in the last integration. Most integrals don’t need a constant of integration so it’s a needless complication in general.
His channel and Dr. Peyam are helpful for learning basics and intermediate university level maths. If you want something even more challenging, check out Math 505. I can't even understand some of the videos from Math 505. Watching multiple math channels really help.
Here is another one: Anti derivative of lnx is 1/x. I see this mistake a lot in Integration by parts where lnx is the g'. For example an integral like x^5*lnx...Calculus 2! If a student only forgot +C, well that is still something I can live with...kind of...
@@epikherolol8189 -- It's xln(x) - x + C. 1) You need the "+ C." 2) the logarithm is a function, so the argument should be inside grouping symbols. That goes for the original poster as well.
Well that’s not certainly true. Here, you are assuming that x is a number. d/dx (x^x) Rewrite x^x as e^xlnx Find the derivative of e^xlnx with respect to x We get (e^(xlnx))*(1+lnx) x^x(1+lnx) is the real answer
I learnt how to change sin^2x and cos^2x into cos(2x) from cos(2x)=2cos^2(x)-1 So I would argue that in B you could actually use sin^2(x)=1-cos^2(x) And substitute cos^2(x)=(cos(2x)+1)/2
In my honest opinion, C is the worst mistake. Missing the constant of integration is forgivable because they at the very least did the bulk of the computation correctly. B is the second worst because they forgot chain rule but, it pales in comparison to C where they literally forget the concept of derivative and integral all together and mismatch them. Forgetting the chain rule is one thing because you at least still know the other rules but, forgetting the difference between integral and derivative is a whole new level of stupidity from that. D is more forgivable because I see someone getting tripped up by that easily (at least early on) because, it looks so similar to something completely different that they would be tempted to say it even if it’s far off. It shows they at the very least can differentiate between integral and derivative and can do other types of integrals correctly which is better than B and C but still worse than A because with A the whole computation was correct it was just they forgot the plus C.
If you're talking about 6:44, you don't need to include the constant at intermediate steps in integration by parts. You can if you prefer, and you'll see that most of the time, it will just cancel out anyway, and be absorbed in a master constant of integration. Most of the time, we just keep it simple by letting the constant be zero. There some applications where you'll prefer a non-zero constant of integration at intermediate steps, in integration by parts. It is usually for the regrouper stops, where we have either a log or inverse trig being differentiated, and an algebraic function being integrated. You can strategically assign a constant other than zero, to cancel out part of your regrouped integrand in the next step.
It's a constant, because when you do a derivation of a constat it becomes 0, and when you do an integration you can't know if it have an constant before or dont, so you just put and +C Well you can calculate the valor of the C but you need additional values
The first 3 mistakes are straightforward, the last mistake is that average student is not going to see that trigonometric substitution is needed to get the antiderivative.
Then it will equal -sinxcosx-int(-cos^2(x)dx)=-sinxcosx+int(cos^2(x)dx) And doing it twice will yield -sinxcosx+cosxsinx+int(sin^2(x)) which will lead to the original integral Proof: attempt integration by parts and that will happen
To solve you have to use trig identity: cos2x=1-2sin^2(x) 2sin^2(x)=1-cos2x sin^2(x)=(1-cos2x)/2 Then sub into integral and you can solve it from there
A) reminds me of "don't forget the limit notation" but worse, since you're not getting an exactly correct answer it is probably due to laziness or time anxiety, but i bet it is a very commom mistake, so it is not really that bad B) you're assuming that the ∫ f(x)^N dx = (f(x)^N+1)/N+1 + C i think most students would do that mistake because they haven't learned about the chain rule or u-sub, which i guess it's fine for beginners, but it can definitely hurt someone C) i don't know how you would do that type of mistake normally, but i think the main reason why that happens is time anxiety (which i have aswell) or a memory glitch if not, then... i think you should go to the basics again D) well... it's the same as B), most likely from beginners, but if not, you definitely skipped alot of trigonometry classes. i will go with D) because well... if you remember cos(x) and sin(x) and a few identities related to them, then you have basically memorized 90% of the trigonometry formulas
Regarding A: When I was student-teaching, one of the other teachers said she could always tell when a student's previous math teacher was a man because their work was so sloppy (or something along those lines), and leaving out the "limit" notation was one of those topics.
U are using the term tan^-1 to talk about the inverse function of tg x. While there is a simple notation for it as arctg X. Ur writing is confusing and makes people think that tg^-1 x = ctg x = cos x/ sin X.
It's convention that trig^-1(x) means the inverse trig function, since the superscript -1 means inverse function in general, when used on a function name. Superscripts on a trig function names ONLY mean exponents, when they are positive numbers. We have completely different names for reciprocal trig functions, which are different than inverse trig functions. Yes, I agree that the superscript -1 is misleading, and it is better to use either the atrig or arctrig. The notation of trig^-2 (x) is even more misleading and mysterious. Should we have a different notation? Yes, we should. There are three meanings that a superscript number could mean on a function name, and you have to know from context which one it is: 1. Exponent, such as sin^2(x) 2. Derivative order, as we use for Taylor series. Usually, you use apostrophes, until you get to the 4th derivative, in which case you use Roman numerals. I prefer to enclose these in brackets, when it's a variable, such that f^[n] (x) means the nth derivative. 3. Iteration degree. This is the number of times you compose a function with itself. An inverse function is a special case of an iteration degree of -1.
@@carultch I would have completely agreed with you if there was no special symbol specified for the inverse function of tg (x) which is arctg (x) . It is like using the -1 power to refer to the inverse function of e^x while it is written as ln x. And it would have been very confusing if you write (e^x)^-1 in order to talk about the logarithmic function.
@@AbouTaim-Lille Historically, natural log was discovered before the number e, and before exponential functions in general, even though modern math classes teach it the other way around. Thus, there never was a direct notation for inverse exponential before the term logarithm was coined. Instead, the reverse happened. There was a function called antilog(x) that used to be what e^x was called before mathematicians made the connection that it had anything to do with exponentials, and before Euler's number was discovered.
@@AbouTaim-Lille I generally avoid the superscript -1 for inverse trig, and will always either call it arctrig or atrig. It makes me laugh when I see the arc prefix applied to hyperbolic trig, since the inverse to hyperbolic trig functions, has nothing to do with arcs. The a could stand for area or anti.
In Soviet Russia, C is the equivalent letter of S, and they even call it "es". C in Russian always sounds like the C in Cindy. I would guess that Soviet textbooks used +K as their default name for the constant of integration, to make it more intuitive.
So I see you never move some function under derivitive. Is it forbiden in you country? Sudv = uv - Svdu. Sarctg(x)dx = x*arctg(x) - Sx* d arctg(x) = x*arctg(x) - S(x/(1+x^2))dx= x*arctg(x) - S (dx^2)/(1+x^2) = x*arctg(x) - ln(1+x^2)/2 +c This form seems more preatty, clean and understandable for me. Instead of all your tables and useless rules
Mistake A: Almost correct, missing +C, you are just not careful enough. Far from the worst. Mistake B: forgetting the chain rule, please, go back and learn how to differentiate first. Also learn how to use double / half angle formulae. You missed 2 essential skills, that makes this mistake really bad. Mistake C: it is integration, not differentiation, only "C1*e^x + C2*e^-x" or "C3*cosh(x) + C4*sinh(x)" satisfy integral = derivate + C. Go back and learn how to integrate by parts. Really bad mistake, nearly as bad as B. Mistake D: Hey, I guess that the common integral table has shown the answer already, do you even remember that? Or, do you even remember cos²(θ) + sin²(θ) = 1 can be divided by cos²(θ) to get 1 + tan²(θ) = sec²(θ)? If you manage to mess this simple question up after you have ever studied about that, I really don't know why. This mistake is the worst one for me, even worse than B.
My point in Mistake C can be proved with this: ∫ydx = dy/dx + C Let u = ∫ydx, y = du/dx, dy/dx = d²u/dx² u = d²u/dx² + C d²u/dx² - u = -C Consider u = u_p + u_c, u = u_p if C = 0 Using auxiliary equation a²-1 = 0, a = ±1 u_p = C1*e^x + C2*e^-x Obviously, u_c = C u = C1*e^x + C2*e^-x + C As y = du/dx, y = C1*e^x - C2*e^-x I had made mistakes in that point, and I have corrected it.
