For anybody having trouble understanding integration by parts, you can remember it by keeping in mind it's just the reverse of the product rule for derivatives. That is, the product rule says (f(x)g(x))' = f(x)g'(x) + f'(x)g(x) But taking the anti-derivative of both sides you get f(x)g(x) = ∫f(x)g'(x) + ∫f'(x)g(x) + c If you shift it around you then get the integration by parts formula ∫f(x)g'(x) = f(x)g(x) - ∫f'(x)g(x) + c So in the video, for instance, you have f(x) = sin(x) and g'(x) = cos(x) and go from there.
So why didn't Ben just use u=sin(x), du=cos(x)dx and write the integral as §udu = (u^2)/2 ? It bypasses the extra bits of integration by parts. @sparksmath
@@mydroid2791 That's the other two techniques in the middle, which goes off of the chain rule, rather than the product rule. He's generalized it to include power rule, where he's written u^1 = (sin(x))^1, du/dx = cos(x), and equivalently u^1 = (cos(x))^1, du/dx = -sin(x), except that he went about it in a roundabout way by pretending not to know the coefficient in front. That coefficient can always be determined, it's just the reciprocal of 1 more than the power 'u' is raised to; so here it's u^1, thus the constant is 1/(1+1) = 1/2. That should be familiar, being just the power rule for integration. Then a minus sign which instead comes from du/dx. (If du/dx itself had a different coefficient than -1, for example maybe 3/2, you should remember to apply the reciprocal of that coefficient, 2/3 rather than 3/2 for example, because the du/dx is telling you what you need, not necessarily what you have. Most rigorously, you do the trick where you multiply and divide by some number at the same time, such that, after moving those coefficients around, you see exactly du/dx on the inside. When the integration is finished, it will happen that you're left with just the reciprocal of what you needed for du/dx [multiplied along with the coefficient from the power rule and anything you might've started with].)
I actually went through the third ring of Dijarmha and reached the eternal light of Nurrzhenzhis by watching this on LSD. opening my mind is an understatement.
i have given this "conundrum" in every calculus class i taught. Most students tended not to have a problem with it, but rather they were entertained. The few who struggled afterwards seemed like something finally clicked. By this i don't mean my students were all aces in calculus, just that they were a little more clear on the meaning of indefinite integrals. happy to see this "conundrum" appear on Numberphile.
What I like about this particular example is that the meaning of the constant becomes obvious. (Replace (cos x)² with 1 - (sin x)² and expand out, and you get the other integral with a +1 term - the (c + 1) terms group into a new constant.) It's something I wish I had been taught instead of having to figure it out, although figuring out the meaning of the +c was interesting in itself.
I think it would have been informative to also discuss that since cos(2x) = cos²(x) - sin²(x) and cos²(x) + sin²(x) = 1, then you can work out that cos(2x) = 2cos²(x) - 1 = 1 - 2sin²(x), so that the 3 answers are indeed identical up to a constant [ (-1/4)cos(2x) is (-1/2)cos²(x) plus a constant and is (1/2)sin²(x) plus some other constant, both absorbed in the undefined constants in the answers]. It may also have helped to label the constants c1, c2, c3 and c4, to explicitly say they are not necessarily the same (they are not) though that would have given away the punchline.
Yeah, I was doing that in my head a bit during the video and was waiting for this reveal. Felt incomplete without it! But the graphing was nice too. :)
Right, and it might also lead to a confusion in definite integration, "there's no C so which method is right!" But since the functions really are the same up to a constant, any will work with a definite integral where the constant sort of cancels out.
you can get the same effect by integrating functions like 1/(4x) in two different ways: First by writing it as 1/4 * 1/x, which integrates to 1/4 * ln|x| + c, or secondly by a linear substitution, then giving 1/4 * ln|4x| + c. Of course, the trick is again in the constant, because ln|4x|=ln(4)+ln|x|
I would've used t/u substitution, since it may be the faster. Realising they're the derivative of each other so you change sin(x) for t and cos(x)dx for dt then integrate and rewrite, and you end up with t²/2 + C = sin(x)²/2 + C
@@genessab yeah, I thought so, but I just thought it was a bit more convoluted than necessary, though it still is nice to have that thought process, "normal" t/u substitution are nice to think about too
@@mr.prometheus3320 they are for sure a special case, yeah. Actually, when you have integral(udu) it's always gonna be u²/2 + C. Since you can use integration by parts to see that indefinite integral(udu) = u² - indefinite integral(udu) indefinite integral(udu) = u²/2 It's fun to see that now
What I want to know is why Ben didn't show that the answers were equivalent by more than one method? Between the double-angle formula for cos and the Pythagorean identity, you can show that the three different expressions only differ by their constants (which is easier than sketching the three graphs to do on paper).
This actually happened in my first year calc class. This problem was on a test and different students got these various answers, and were very confused when they got their tests back with points taken off. Eventually our professor graphed them to see they were the same.
I remember one teacher showing it the other way around. He wrote a trigonometric function and differentiated it. Then he wrote a completely different function, differentiated it, and to our amazement, they turned out to have the same derivative. He then proved using some trigonometric identities that they differ by a constant.
Method 5 Put sinx = t (now diffrentiate both side wrt x) Cosx =dt/dx Cosxdx=dt Now just put the value of sinx = t & cosxdx = dt To get integral of t dt Which is on integrating (t^2)/2 + c .......(1) Now put back the value of t (= sinx) into the eqn (1) So the answer is ((Sinx)^2)/2 + c Same ans as in method 4. But simpler than integration by part....
Numberphile going back to its roots! This is outstanding! You should have one of these a week, your audience would learn so much! Thank you for doing this one.
I specifically avoided learning trig identities when I took calculus. When you posed the problem I jumped to u-substitution with u = sinx, and when you hinted that different methods may give seemingly different answers my next approach was integration by parts.
I like a video that challenges my intuition. I know there’s multiple ways to integrate, but it hadn’t occurred to me that the functional expression might be different (though in effect equivalent). Very cool!
It wasn't until grad school that I actually understood what an indefinite integral actually is: the preimage of an operator (the derivative operator) in a function space. That is, an indefinite integral tells you all the things that you apply the derivative operator to in order get the integrand. We write the result as a function, but it's actually a set of functions indexed by (in this case) the real numbers.
this is why I fell in love with integrations, finding new and tricky functions to integrate and differentiate to check the answer! Introduction to calculus made me appreciate Mathematics big time :)
Holy cow, ever since i first saw the notation bit mentioned at 7:20 I've never heard anyone else ever comment on it but it was a strong memory for me. It's very validating to see it mentioned here.
That's why you should write the areafunction as definite integral: A(x) = ∫ _0^t f(t) dt Because then, the constant 'C' is vanishing away, and you get the correct answer.
If f(x) = eˣ, then the ∫₀ˣ eᵗ dt = eˣ-1. By comparison, c=-1, so the lower limit should be the inverse antiderivative evaluated at 0 (i.e. F⁻¹(0)) in general.
@@adiaphoros6842 You're right. What I wrote is not always correct. Because there are functions you can't evaluate at x=0. F(x)= 1/x, F(x)=1/x², F(x)=1/(x+x²+..), F(x)=e^(1/x), etc... There are better ways to handle this constant.
Method 1 and Method 3 are equivalent because *cos 2x = 1 - 2 sin² x* so *-1/4 cos 2x + c = 1/2 sin² x + (c - 1/4) = 1/2 sin² x + c'* Method 2 and Method 3 are equivalent because *sin² x + cos² x = 1* so *-1/2 cos² x + c = 1/2 sin² x + (c - 1/2) = 1/2 sin² x + c''*
Yeah. I was also confused as to why did not use trig identities after his integrals to simplify, just like he did in one of his methods. The video title and mood, initially made the different answers seem paradoxical.
To answer one of the questions at the end, about which methods students would use: in the AP program, a Calc AB student would probably use method 2 or 3 (a basic example of ‘u sub’), and a Calc BC student would probably use method 4, integration by parts. Most American students in my experience would look at the trig identity used in method 1 and just nod their head sheepishly while thinking, “yea that thing exists but I’ll never think to use it in that way.”
Agreed. It's been decades since I was in a Calc class, but I'm pretty sure I'd have done it then the same way I did it today: as a basic u substitution. u = sin(x); du = cos(x)*dx; I = Int(u*du). But then, I never could be bothered with memorizing trig identities.
Though on the notation thing, something Geogebra did that would probably have given more of a hint to students is that it called the constant "c1", rather than just c - which would reinforce the idea that all of the answers have some constant c applied to them, but it is not necessarily the *same* constant between different answers.
You need to make your next video shedding light on the trig identities that explain how to convert between the 3 or 4 very different looking formulas, given specific C values.
For me, it was a revelation when I learned about complex numbers, and the fact that: e^ix = cos x + i sin x Now it easier to generate the angular formulae and differentiate them.
Method 3 & 4 show constant as the same value c. But methods 1 and 2 have different solutions so the constant can't be c, choose other values e.g. a and b. That makes it easier to explain and understand
It doesn't matter how many times I see it, the fact that a trig function squared is jsut another trig function (plus some shifts and frequency change) never ceases to be unintutitive
I think method 2 and 3 are both "the best" since it's basically a substitution, one of the more useful methods for solving integrals. That being said, I went for partial integration and I'm glad it appeared in the video!
I would’ve first thought u substitution similar to the second method shown. And this kinda thing happens all the time another example is the integral of tan(x)*sec^2(x) you can use tangent or secant for the substitution variable
There is actually another method, and it’s the easiest in my opinion You can simply consider the cos(x)dx as a differential of a trigonometric function which is sin(x) so, cos(x)dx=d(sin(x)) Integral(sin(x)d(sin(x))) = (sin^2(x)/2) + C And you get the same result as in the last methods
That's method 3 in the video, just phrased slightly differently. "cos(x)dx = d(sin(x))" is the same as saying "cos(x) = d/dx(sin(x))". Usually you'd just refer to this as u substitution, using u = sin(x), du = cos(x)dx. It's all the same thing, just the chain rule in reverse.
My immediate reaction to the different answers were that you can add any constant, and any constant is just c = c(sin^{2}(x) + cos^{2}(x) ). Moreover, the expanded version of cos2x is cos^{2}(x) - sin^{2}(x) so in the end we just have lots of sin^{2} and cos^{2}.
@@zachb1706 It can because that x gets 'cancelled out' by a trig identity, which is that the term there equals 1. It's the same as saying c = c(x/x) (except without the minor division by zero issue that has).
Two other ways (not necessarily easier or harder, but different enough): 1. Complex Integration 2. Tangent half-angle substitution In more detail: 1. If you don't mind complex numbers, you can use Euler's identity, along with the fact that e^(cx) is easy to integrate. Using the two easy identities e^(ix) = cos x + i sin x e^(-ix) = cos x - i sin x We get: cos x = (e^(ix) + e^(-ix))/2 sin x = (e^(ix) - e^(-ix))/2i So sin x cos x = ((e^(ix) - e^(-ix))/2i) * ((e^(ix) + e^(-ix))/2) = (e^(2ix) - e^(-2ix))/4i Integrating this gives (e^(2ix)/2i + e^(-2ix)/2i)/4i + C = -(e^(2ix) + e^(-2ix))/8 + C This is yet another solution that "looks different" (but ultimately the same, though we're using complex numbers now). Unsurprisingly, you can rewrite it into one of the already-found forms by noticing that cos 2x = (e^(2ix) + e^(-2ix))/2, so the result is actually just -(1/4) cos(2x) + C. The advantage of this is that it works for all polynomial expressions involving sin x and cos x. This technique can be extended to cover all rational expressions involving sin x and cos x. 2. The so-called tangent half-angle substitution takes care of basically all trig integrations "automatically/mechanically". The idea is to use the substitution t = tan(x/2). This gives us: sin x = 2t/(1 + t^2) cos x = (1 - t^2)/(1 + t^2) dx = 2/(1 + t^2) dt This turns any rational expression involving only trig functions into a pure rational function, which can then be integrated in the standard way, e.g., by doing partial fraction stuff. A lot of the time, this technique is harder to do by hand than other techniques, but the advantage is that it works for all of them, and can be automated.
I think the average calc student would be more likely to use integration by parts because that stuff would be fresh in their mind. Trig identities not so much unless they're especially studious!
I'm not sure whether this is already well known, but the double angle identities are trivially easy to rederive if you know Euler's Formula: exp(iθ) = cos(θ) + i sin(θ) exp(2iθ) = cos(2θ) + i sin(2θ) exp(2iθ) = exp(iθ) · exp(iθ) = (cos(θ) + i sin(θ))^2 = cos(θ)^2 + 2i sin(θ) cos(θ) - sin(θ)^2 Now equate the two sides: cos(2θ) + i sin(2θ) = cos(θ)^2 + 2i sin(θ) cos(θ) - sin(θ)^2 Hence: cos(2θ) = cos(θ)^2 - sin(θ)^2 sin(2θ) = 2 sin(θ) cos(θ) I'm sure it varies from person to person, but I find it much easier to remember Euler's Formula than the angle identities so this is how I remember them! You can easily extrapolate for the angle sum identities as well.
I learned 2 and 3 as u-substitution. You can write sin x as u. Then you take du/dx = cos x => du = cos x dx, which gives integral of u du = 1/2u^2 = 1/2 sin^2 x.
I particularly like giving students the indefinite integral of tan(x)*[sec(x)]^2 for a similar reason: it can be found via u-substitution in two different ways that give two very different-looking answers, and you can reconcile those differences using trigonometric identities and the constant of integration
3 ways. u=tan(x), du=sec²(x)dx You get the integral of u du u=sec(x), du=sec(x)tan(x)dx You get the integral of u du u=½sec²(x), du=tan(x)sec²(x)dx You get the integral of du
Both method 2 and method 3 are going to save you at some point (3 works like a charm if you have both exponential and trigonometric functions at once). My favorite way of writing it would be ∫sinxcosxdx= =(1/2)∫2sinx(sin)'(x)dx =(1/2)∫((sinx)^2)'dx =(1/2)(sinx)^2+c though, because it happens in an uninterrupted line.
I really liked how using different approaches leads to different looking answers that all correct. The easiest method, for me, was u-substitution (let u=Sin(x) then du=Cos(x)dx). Using Trigonometric identities, the results can be shown as equal to each other. Now, showing that would be a great trig review for the students, though they might not think so. 😁
This is weirdly nostalgic for me... in 1982 I had an interview for a place at Nottingham University studying maths. In the interview they asked me to integrate something a little trickier: e^x.sin(x).cos(x) - I ended up doing it by parts, but then applying the "by parts" method again. It was a pretty tense experience! I did get an offer in the end but I chose to go to Southampton instead (I think I just wanted to live by the sea!). Every time I see Brady's videos at Nottingham I always wonder how many of the people on screen would have been my teachers, had I chosen differently... obviously not the younger ones, but still.
I wish we lived in a world where educators at lower levels weren't afraid of complex numbers. They make most of trigonometry _very_ straightforward, after all: sin x = (e^ix - e^-ix)/2i, cos x = (e^ix + e^-ix)/2, so multiply all those things together (including e^x) and you get (e^(1+2i)x - e^(1-2i)x)/4i right away. While it will require some later manipulation, this is very easy to integrate as it is.
You can even get two more "different" results, if when using integration by parts, you substitute -1/4 cos(2x) or -1/2 cos^2(x) for the integral of sin(x)cos(x) at the end.
what this also means is that we can change 1/4 cos(2x) to -1/2 sin²(x) to 1/2 cos²(x) just by adding constants maybe there exists out there another family of functions who each look different, but you can get one from the other by adding (the same?) constant
This is a classic. What we're doing here is *not* integration, it's antidifferentiation*. Since antidifferentiation results in a one-parameter family of functions, we have to compare results as one-parameter families, not as singular functions. Basically, we have different additive constants, and we can write each of those constants in terms of any of the other constants. On the other hand, if we do *actual* integration, we will get the same value from all of the methods (assuming you're careful about how limits go through substitutions). * I'm being opinionated about terminology here. The term "indefinite integration" is definitionally equivalent to "antidifferentiation", while "definite integration" is what I simply call "integration". I find that making a clearer distinction between the nature of these operations helps students keep them in mind as separate things, which happen to be nicely related thanks to FTC.
Yeah I learned this back in calculus when I got a different answer than the answer sheet, so I graphed them on Geogebra and noticed exactly what this video explains.
With the first method you can get -1/4 ((cos(x))^2 - (sin(x))^2) + C or -1/4(1-2(sin(x))^2) + C = 1/2(sin(x))^2 - 1/4 + C. Also you can get -1/4 (2(cos(x)^2) - 1) + c = -1/2(cos(x))^2 +1/4+ C. So I guess the first method gives the most complete answer since it provides the answer of the other three methods.
My instinct was "u substitution," u = sin(x), du = cos(x) dx. Thus u^2/2 + c, or sin(x)^2/2 + c. When I noticed the different answer, I was like, "ah, but that's the same answer shifted by a constant." And then I realized that was the point of the video.
This was a cool video! Reminds me of how 1/(x+1) and x/(x+1) has the same derivative, for the same reason. And it freaked me out the first time I saw it
1/(x+1) and x(x+1) don't have the same derivative. 1/(x+1) and -x/(x+1) do, and the reason (I know you know, but just to complete your idea for other readers) is that: 1/(x+1) = (1+x-x)/(x+1) = [-x + (x+1)] / (1+x) = -x/(1+x) + (x+1)/(1+x) = -x/(x+1) + 1, which means that they are almost the same function except shifted one unit vertically one from the other, and hence they have the same slope for any value of x.
once you show all 4 results it's quite obvious that there are trigonometric identities that tie them together. Specifically, the cosine double angle formula [cos(2x)=2cos²x-1] and the Pythagorean identity [sin²x + cos²x = 1].
Just another technique to add to the ones presented: since sinx cosx dx = sinx d(sinx), we can make a variable substitution y=sinx and integrate for y: Int(ydy) becomes 1/2y^2+C and finally, replacing back y, 1/2 (sinx)^2 + C. And of course, we can also go the other way around with -d(cosx)...
Really enjoyed this video. What was not explained is why C the constant of integration disappears if the integral has limits which might have been helpful. More fundamentally where C comes from.
It's fairly simple algebra to see why the constant disappears in a definite integral. You're subtracting the values of the same integral evaluated at two different points: (f(a)+c)-(f(b)+c), so the two constants cancel. I find the easiest way to see why you need the constant in the first place is that when you differentiate, all constant terms vanish. So when you integrate you have to put them back - but you don't know what they were; the information is lost.
I insta-changed variables, sin(x)=s, ds=cos(x), answer is s²/2 ie answer 3 and 4. I found his explanation of "methods" 2 and 3 a bit confusing/shortcut-ish, it's nice to change variables explicitly and show what's happening.
Note to Ben: There is no problem with using sin²(x), sin³(x) et cetera because the meaning is unmistakable. Of course, the issue is when we see sin-¹(x) and there is confusion about whether this is (sin(x))-¹ = 1÷sin(x) or the inverse trig function, arcsine of x. The problem is entirely the result of using sin-¹(x) for the inverse sine function when the better, unmistakable expression is arcsin(x).
In some contexts, given a function f(x), notation like f²(x) refers to f(f(x)) and f³(x) refers to f(f(f(x))), where the exponent right after the function but before the argument refers to _repeated composition._ So the meaning very well could be mistakable. I suspect Ben holds this view (but I haven't asked to be fair!) After all, we have a designated location to denote exponentiation already. Why allow exponentiation to take _two_ possible locations, including the only sensible location for repeated composition. By the way, under the repeated composition view, f-¹(x) is unmistakably the inverse function of f(x). I find this a better option than saying we should write arcf(x).
I used Leibniz method on this one and got (-1/4)*cos(2x). Defined F(t) = integral( sin(x)*sin(x + t*Pi/2) )dx, then found that F'' + (Pi/2)^2*F(t) =0, solved for F(t) and evaluated F(1) = integral( sin(x)*sin(x + Pi) ) dx = integral( sin(x)*cos(x)) the desired integral
In beam theory, when integrating from rotation to displacement, then twice to shear and finally to bending, the need to carefully establish what the constant is at each integration.
If you actually put in limits, let's say 0 to pi/2, all three give the same answer, 0.5. Yellow curve: 0.5 - 0 = 0.5. Red curve: 0.25 - (-0.25) = 0.5. Blue curve: 0 - (-0.5) = 0.5. Here is an interesting twist. Consider the fact that cos(2x) = [cos(x)]^2 - [sin(x)]^2 if you add method 2 to either method 3 or 4, you end up with -1/2cos(2x) which is simply 2I from method 1. Using the graph, If you look at the sum of the two, i.e., the superposition you will immediately see it is 2I.
Yet another version, very similar to 2&3 (maybe you'd say the same only different) that I was also taught. substitute y = sin(x) Then dy = cos(x)dx I = (sin(x))(cos(x)dx) = y dy = y^2/2. Substituting sin(x) back in for y I=(sin(x))^2/2 And identically for cos(x) to get the other result.
The notation of putting a number after fuction name normaly means apply the function that many times, but it is customary to use it with trigonometric functions to raise that function to the power, because those are not usually applied one after other. But of course they are in navigation and in survey where you can see sin(sin(x)) and others.
This reminds me of a (digital) math quiz we had at school. One question was about integrating something like -1/sqrt(1-x^2), I answered: -arcsin(x) +c but the "correct" answer was arccos(x)+c so I lost a point for it. -arcsin(x) and arccos(x) are the exact same functions just ofset in the y-axis
I think I've figured this out using trig identities: I'll use cos^2 x and sin^2 x to mean (cos x)^2 and (sin x)^2 start with the 2 identities cos^2 x + sin^2 x = 1 cos^2 x - sin^2 x = cos 2x add them 2 cos^2 x = 1 + cos 2x cos^2 x = (1/2)*(1 + cos 2x) subtract them 2 sin^2 x = 1 - cos 2x sin^2 x = (1/2)*(1 - cos 2x) expand and rearrange a) (1/2)cos 2x = cos^2 x - (1/2) b) (-1/2)cos 2x = sin^2 x - (1/2) a)*(-1/2) and b)*(1/2) (-1/4)cos 2x = (-1/2)cos^2 x + (1/4) (-1/4)cos 2x = (1/2)sin^2 x - (1/4) joining these together (-1/4)cos 2x = (-1/2)cos^2 x + (1/4) = (1/2)sin^2 x - (1/4)
Since this is an indefinite integral we are calculating the anti derivative so the derivative of all 3 of those functions will all be the same since vertical shifts do not change the slope at any point.
What I might have added on as an explanation is a more general interpretation of the relationship between derivatives and integrals. By performing an indefinite integral, you're essentially starting with the rate of change and asking what function has that (the original) function as its rate of change. Since a rate of change doesn't care about where you start, there has to be a degree of uncertainty (i.e. the constant at the end).
I'd say that the thing to know (other than remembering your constants when integrating) is that CYCLIC functions (like sine and cosine) can get REALLY WEIRD.
I would have used precisely none of these methods! Method 5: u substitution! Let u = sin(x). Then du = cos(x) dx. This then becomes ∫ u du = 1/2 u² + C = 1/2 (sin(x))² + C.
I always thought that +c was inconvenient and slightly imprecise notation. It hides the fact that the solution is a family of functions (+c makes it look like it's one specific function that you don't know exactly), while also causing annoying algebra inaccuracies that are irrelevant (like the c that should technically be c/2 in this video). I think it's better to think of the answer as a coset of the constant functions, which would look like f(x)+ℝ or f(z)+ℂ (depending on whether you are talking about a real or complex function). Then you can multiply by scalars, and not have to change/redefine your added constant, because the whole set of numbers (which is a field) is fixed under multiplication by a (non-zero) scalar. Plus, writing it like this is exactly in line with the notation you'd use for an element in the group of functions under pointwise addition, modulo the subgroup ℝ or ℂ of constant functions. Indefinite integration is a group homomorphism into this group.
@@BlueEyesWhiteTeddy you could write it using the full notation for a set, but I'm using the notation of cosets to simplify this. Like how in group theory you would write g+H to represent the coset {g+h for h in H}, i.e. add g to everything in H to get the new set g+H. This is common notation across all of mathematics to represent the elements of a structure quotiented by a substructure.
Should have been strictly correct had he used different constants in each integral (C1, C2, etc...). At the end of the day as you say, is that once you get one specific solution for the antiderivative (without the c), then shifting up or down that solution by an arbitrary amount will also be a solution, and sometimes (many times) functions f and g that are just the same shifted up or down one from the other can be written as something other than f=g+c, like f=1/(1+x), if we shift it down by 1 unit we get 1/(1-x)+1 = (taking common denominator (1/x) ) [1-(1+x)]/(1+x) = (cancelling +1 and -1) -x/(1+x), which at first sight doesn't look to be the same as 1/(1+x) just shifted 1 unit, but it is.
Seen this many times before. All are equivalent, with a different value for the constant, that is determined by trig identities (like sin^2 + cos^2 = 1 or cos2x = cos^2(x) - sin^2(x) )
For anybody having trouble understanding integration by parts, you can remember it by keeping in mind it's just the reverse of the product rule for derivatives. That is, the product rule says
(f(x)g(x))' = f(x)g'(x) + f'(x)g(x)
But taking the anti-derivative of both sides you get
f(x)g(x) = ∫f(x)g'(x) + ∫f'(x)g(x) + c
If you shift it around you then get the integration by parts formula
∫f(x)g'(x) = f(x)g(x) - ∫f'(x)g(x) + c
So in the video, for instance, you have f(x) = sin(x) and g'(x) = cos(x) and go from there.
I haven't worked with integral for 20 year. Thanks for reminding. It's easier for me to remember this process than the rule itself.
@@loose4bet 40 years here. But it's amazing how much sense it makes even with a thick coat of rust on my technique.
So why didn't Ben just use u=sin(x), du=cos(x)dx and write the integral as §udu = (u^2)/2 ?
It bypasses the extra bits of integration by parts.
@sparksmath
@@mydroid2791 That's the other two techniques in the middle, which goes off of the chain rule, rather than the product rule.
He's generalized it to include power rule, where he's written u^1 = (sin(x))^1, du/dx = cos(x), and equivalently u^1 = (cos(x))^1, du/dx = -sin(x), except that he went about it in a roundabout way by pretending not to know the coefficient in front.
That coefficient can always be determined, it's just the reciprocal of 1 more than the power 'u' is raised to; so here it's u^1, thus the constant is 1/(1+1) = 1/2. That should be familiar, being just the power rule for integration.
Then a minus sign which instead comes from du/dx.
(If du/dx itself had a different coefficient than -1, for example maybe 3/2, you should remember to apply the reciprocal of that coefficient, 2/3 rather than 3/2 for example, because the du/dx is telling you what you need, not necessarily what you have.
Most rigorously, you do the trick where you multiply and divide by some number at the same time, such that, after moving those coefficients around, you see exactly du/dx on the inside. When the integration is finished, it will happen that you're left with just the reciprocal of what you needed for du/dx [multiplied along with the coefficient from the power rule and anything you might've started with].)
I never thought about integration by parts that way. But it makes total sense.
Ben is the unsung hero of this channel. Dude is incredible at opening your mind about math we're already familiar with
He is brilliant. A super communicator and always such enthusiasm 😊
I actually went through the third ring of Dijarmha and reached the eternal light of Nurrzhenzhis by watching this on LSD. opening my mind is an understatement.
@@aceman0000099 I’ve watched a lot of these on nutmeg; so, I can *_KINDA_* feel, where you’re coming from 😵💫.
@Zach Gates True and beautifully motivated.
Yes, Love love Ben!
i have given this "conundrum" in every calculus class i taught. Most students tended not to have a problem with it, but rather they were entertained. The few who struggled afterwards seemed like something finally clicked. By this i don't mean my students were all aces in calculus, just that they were a little more clear on the meaning of indefinite integrals.
happy to see this "conundrum" appear on Numberphile.
What I like about this particular example is that the meaning of the constant becomes obvious. (Replace (cos x)² with 1 - (sin x)² and expand out, and you get the other integral with a +1 term - the (c + 1) terms group into a new constant.) It's something I wish I had been taught instead of having to figure it out, although figuring out the meaning of the +c was interesting in itself.
I think it would have been informative to also discuss that since cos(2x) = cos²(x) - sin²(x) and cos²(x) + sin²(x) = 1, then you can work out that cos(2x) = 2cos²(x) - 1 = 1 - 2sin²(x), so that the 3 answers are indeed identical up to a constant [ (-1/4)cos(2x) is (-1/2)cos²(x) plus a constant and is (1/2)sin²(x) plus some other constant, both absorbed in the undefined constants in the answers].
It may also have helped to label the constants c1, c2, c3 and c4, to explicitly say they are not necessarily the same (they are not) though that would have given away the punchline.
Yeah I know it's "mathy" but we shouldn't skip the trig identities to show that all three are indeed the same.
Yeah, I was doing that in my head a bit during the video and was waiting for this reveal. Felt incomplete without it! But the graphing was nice too. :)
Right, and it might also lead to a confusion in definite integration, "there's no C so which method is right!" But since the functions really are the same up to a constant, any will work with a definite integral where the constant sort of cancels out.
Yeah when I show this to calc 1 students I stress that they are all equal up to a constant.
That's really helpful, thank you!
Fantastic video. I’ve actually been struggling with this integral for an upcoming test. Very timely!
Glad to be of service!
Destin doing a level maths???
well i could have used it a few month back when i had my calculus exam!
??.
integration is an interesting area
ha
ha
Nc hahaha and derivative is a pretty interesting slope
That about sums it up well
Galois theory is an interesting field
I love Ben Sparks' videos
Ben, is this your burner account?
@@numberphile If it is, it's so well disguised that I don't recognise it :)
I am not Ben Sparks, and so far have never claimed to be.
@@platypi_otbs ...which is exactly what Ben Sparks would say if you were him, or he were you.
@@platypi_otbs bro, he's just joking.
you can get the same effect by integrating functions like 1/(4x) in two different ways: First by writing it as 1/4 * 1/x, which integrates to 1/4 * ln|x| + c, or secondly by a linear substitution, then giving 1/4 * ln|4x| + c. Of course, the trick is again in the constant, because ln|4x|=ln(4)+ln|x|
It wouldn't be a Ben sparks numberphile video without geogebra somewhere in it
I would've used t/u substitution, since it may be the faster. Realising they're the derivative of each other so you change sin(x) for t and cos(x)dx for dt then integrate and rewrite, and you end up with t²/2 + C = sin(x)²/2 + C
My first thought was the same! I kept watching video thinking, oh surely the next method will be a u-substitution, and then it never came!
Well his methods for 2/3 were in fact u substitution, just an informal way of doing it.
@@genessab yeah, I thought so, but I just thought it was a bit more convoluted than necessary, though it still is nice to have that thought process, "normal" t/u substitution are nice to think about too
Yeah! I thought of u-subs right away too. I puzzled with it for a bit and wondering if they're describable as a special case of parts where u = v.
@@mr.prometheus3320 they are for sure a special case, yeah. Actually, when you have integral(udu) it's always gonna be u²/2 + C. Since you can use integration by parts to see that
indefinite integral(udu) = u² - indefinite integral(udu) indefinite integral(udu) = u²/2
It's fun to see that now
What I want to know is why Ben didn't show that the answers were equivalent by more than one method?
Between the double-angle formula for cos and the Pythagorean identity, you can show that the three different expressions only differ by their constants (which is easier than sketching the three graphs to do on paper).
Because that wasn't the point of the video
This actually happened in my first year calc class. This problem was on a test and different students got these various answers, and were very confused when they got their tests back with points taken off. Eventually our professor graphed them to see they were the same.
I remember one teacher showing it the other way around. He wrote a trigonometric function and differentiated it. Then he wrote a completely different function, differentiated it, and to our amazement, they turned out to have the same derivative. He then proved using some trigonometric identities that they differ by a constant.
Method 5
Put sinx = t (now diffrentiate both side wrt x)
Cosx =dt/dx
Cosxdx=dt
Now just put the value of sinx = t & cosxdx = dt
To get integral of
t dt
Which is on integrating
(t^2)/2 + c .......(1)
Now put back the value of t (= sinx) into the eqn (1)
So the answer is
((Sinx)^2)/2 + c
Same ans as in method 4. But simpler than integration by part....
That's method 3.. :)
Numberphile going back to its roots! This is outstanding! You should have one of these a week, your audience would learn so much! Thank you for doing this one.
I really love this guy because he also uses visualization in a great way which can sometimes really clarify why something is the case.
Oh nice you can do this by parts, substitution, trig identity and recognition, I'm definitely going to have to use this question in future
This video came at the right time
I just completed some exercises on indefinite integration.
I specifically avoided learning trig identities when I took calculus. When you posed the problem I jumped to u-substitution with u = sinx, and when you hinted that different methods may give seemingly different answers my next approach was integration by parts.
I like a video that challenges my intuition. I know there’s multiple ways to integrate, but it hadn’t occurred to me that the functional expression might be different (though in effect equivalent). Very cool!
It wasn't until grad school that I actually understood what an indefinite integral actually is: the preimage of an operator (the derivative operator) in a function space. That is, an indefinite integral tells you all the things that you apply the derivative operator to in order get the integrand. We write the result as a function, but it's actually a set of functions indexed by (in this case) the real numbers.
it never made sense to me that it's taught as a separate special thing in high school, where you wouldn't dare teaching about function spaces.
Thanks Ben, great video to share with my IB class.
this is why I fell in love with integrations, finding new and tricky functions to integrate and differentiate to check the answer! Introduction to calculus made me appreciate Mathematics big time :)
Holy cow, ever since i first saw the notation bit mentioned at 7:20 I've never heard anyone else ever comment on it but it was a strong memory for me. It's very validating to see it mentioned here.
That's why you should write the areafunction as definite integral: A(x) = ∫ _0^t f(t) dt
Because then, the constant 'C' is vanishing away, and you get the correct answer.
If f(x) = eˣ, then the ∫₀ˣ eᵗ dt = eˣ-1. By comparison, c=-1, so the lower limit should be the inverse antiderivative evaluated at 0 (i.e. F⁻¹(0)) in general.
@@adiaphoros6842
You're right. What I wrote is not always correct.
Because there are functions you can't evaluate at x=0.
F(x)= 1/x, F(x)=1/x², F(x)=1/(x+x²+..), F(x)=e^(1/x), etc...
There are better ways to handle this constant.
For me the go-to solution was to notice that (sin x)' = cos x, the rest follows nearly instantly: ∫ sinx cosx dx = ∫ sinx d(sinx) = sin²x/2 + c
We can slightly change 3rd method:
try y=sinx
dy/dx=cosx
dy=cosxdx
Sinx*cosx*dx=y*dy
Integral (ydy) =y^2/2 + C = 0.5*(sinx)^2+C
Method 1 and Method 3 are equivalent because *cos 2x = 1 - 2 sin² x* so *-1/4 cos 2x + c = 1/2 sin² x + (c - 1/4) = 1/2 sin² x + c'*
Method 2 and Method 3 are equivalent because *sin² x + cos² x = 1* so *-1/2 cos² x + c = 1/2 sin² x + (c - 1/2) = 1/2 sin² x + c''*
That's what I missed in the video: you can use the trig identities again to actually show how the functions are equivalent
Yeah. I was also confused as to why did not use trig identities after his integrals to simplify, just like he did in one of his methods. The video title and mood, initially made the different answers seem paradoxical.
To answer one of the questions at the end, about which methods students would use: in the AP program, a Calc AB student would probably use method 2 or 3 (a basic example of ‘u sub’), and a Calc BC student would probably use method 4, integration by parts. Most American students in my experience would look at the trig identity used in method 1 and just nod their head sheepishly while thinking, “yea that thing exists but I’ll never think to use it in that way.”
Agreed. It's been decades since I was in a Calc class, but I'm pretty sure I'd have done it then the same way I did it today: as a basic u substitution. u = sin(x); du = cos(x)*dx; I = Int(u*du). But then, I never could be bothered with memorizing trig identities.
Ben Sparks has got to be my favorite Numberphile regular.
Though on the notation thing, something Geogebra did that would probably have given more of a hint to students is that it called the constant "c1", rather than just c - which would reinforce the idea that all of the answers have some constant c applied to them, but it is not necessarily the *same* constant between different answers.
You need to make your next video shedding light on the trig identities that explain how to convert between the 3 or 4 very different looking formulas, given specific C values.
Agreed. It would also be great to see the double angled formula proved by using the unit circle.
For me, it was a revelation when I learned about complex numbers, and the fact that:
e^ix = cos x + i sin x
Now it easier to generate the angular formulae and differentiate them.
Method 3 & 4 show constant as the same value c. But methods 1 and 2 have different solutions so the constant can't be c, choose other values e.g. a and b. That makes it easier to explain and understand
2:30 Sadly; there’s no Chain Rule, for integration. At least, not an elementary Chain Rule.
It doesn't matter how many times I see it, the fact that a trig function squared is jsut another trig function (plus some shifts and frequency change) never ceases to be unintutitive
I think method 2 and 3 are both "the best" since it's basically a substitution, one of the more useful methods for solving integrals.
That being said, I went for partial integration and I'm glad it appeared in the video!
I would’ve first thought u substitution similar to the second method shown. And this kinda thing happens all the time another example is the integral of tan(x)*sec^2(x) you can use tangent or secant for the substitution variable
There is actually another method, and it’s the easiest in my opinion
You can simply consider the cos(x)dx as a differential of a trigonometric function which is sin(x) so, cos(x)dx=d(sin(x))
Integral(sin(x)d(sin(x))) = (sin^2(x)/2) + C
And you get the same result as in the last methods
That's method 3 in the video, just phrased slightly differently. "cos(x)dx = d(sin(x))" is the same as saying "cos(x) = d/dx(sin(x))". Usually you'd just refer to this as u substitution, using u = sin(x), du = cos(x)dx. It's all the same thing, just the chain rule in reverse.
That's literally what he did for method 3. He just went through all the steps for why it becomes half the square.
@@stanleydodds9 Exactly
My immediate reaction to the different answers were that you can add any constant, and any constant is just c = c(sin^{2}(x) + cos^{2}(x) ). Moreover, the expanded version of cos2x is cos^{2}(x) - sin^{2}(x) so in the end we just have lots of sin^{2} and cos^{2}.
C is a constant. Which means it can't have an x term in it
@@zachb1706 It can because that x gets 'cancelled out' by a trig identity, which is that the term there equals 1. It's the same as saying c = c(x/x) (except without the minor division by zero issue that has).
Two other ways (not necessarily easier or harder, but different enough):
1. Complex Integration
2. Tangent half-angle substitution
In more detail:
1. If you don't mind complex numbers, you can use Euler's identity, along with the fact that e^(cx) is easy to integrate. Using the two easy identities
e^(ix) = cos x + i sin x
e^(-ix) = cos x - i sin x
We get:
cos x = (e^(ix) + e^(-ix))/2
sin x = (e^(ix) - e^(-ix))/2i
So
sin x cos x
= ((e^(ix) - e^(-ix))/2i) * ((e^(ix) + e^(-ix))/2)
= (e^(2ix) - e^(-2ix))/4i
Integrating this gives
(e^(2ix)/2i + e^(-2ix)/2i)/4i + C
= -(e^(2ix) + e^(-2ix))/8 + C
This is yet another solution that "looks different" (but ultimately the same, though we're using complex numbers now). Unsurprisingly, you can rewrite it into one of the already-found forms by noticing that cos 2x = (e^(2ix) + e^(-2ix))/2, so the result is actually just
-(1/4) cos(2x) + C.
The advantage of this is that it works for all polynomial expressions involving sin x and cos x. This technique can be extended to cover all rational expressions involving sin x and cos x.
2. The so-called tangent half-angle substitution takes care of basically all trig integrations "automatically/mechanically". The idea is to use the substitution
t = tan(x/2).
This gives us:
sin x = 2t/(1 + t^2)
cos x = (1 - t^2)/(1 + t^2)
dx = 2/(1 + t^2) dt
This turns any rational expression involving only trig functions into a pure rational function, which can then be integrated in the standard way, e.g., by doing partial fraction stuff.
A lot of the time, this technique is harder to do by hand than other techniques, but the advantage is that it works for all of them, and can be automated.
That is true, but if you're at that level (so probably year 13) then one would hope that you know both of:
cos²(x)+sin²(x)=1
cos²(x)-sin²(x)=cos(2x)
I think the average calc student would be more likely to use integration by parts because that stuff would be fresh in their mind. Trig identities not so much unless they're especially studious!
and never forget the trig unity, sin² x + cos² x = 1 which can be used to show that methods 2 and 3 are the same, just off by a constant
I'm not sure whether this is already well known, but the double angle identities are trivially easy to rederive if you know Euler's Formula:
exp(iθ) = cos(θ) + i sin(θ)
exp(2iθ) = cos(2θ) + i sin(2θ)
exp(2iθ) = exp(iθ) · exp(iθ)
= (cos(θ) + i sin(θ))^2
= cos(θ)^2 + 2i sin(θ) cos(θ) - sin(θ)^2
Now equate the two sides:
cos(2θ) + i sin(2θ) = cos(θ)^2 + 2i sin(θ) cos(θ) - sin(θ)^2
Hence:
cos(2θ) = cos(θ)^2 - sin(θ)^2
sin(2θ) = 2 sin(θ) cos(θ)
I'm sure it varies from person to person, but I find it much easier to remember Euler's Formula than the angle identities so this is how I remember them!
You can easily extrapolate for the angle sum identities as well.
I learned 2 and 3 as u-substitution. You can write sin x as u. Then you take du/dx = cos x => du = cos x dx, which gives integral of u du = 1/2u^2 = 1/2 sin^2 x.
My Calculus professor used to say "differentiare humanum est, integrare diabolicum"
I particularly like giving students the indefinite integral of tan(x)*[sec(x)]^2 for a similar reason: it can be found via u-substitution in two different ways that give two very different-looking answers, and you can reconcile those differences using trigonometric identities and the constant of integration
3 ways.
u=tan(x), du=sec²(x)dx
You get the integral of u du
u=sec(x), du=sec(x)tan(x)dx
You get the integral of u du
u=½sec²(x), du=tan(x)sec²(x)dx
You get the integral of du
would be great a followup video explaining how to get from one solution to another using only trigonometric identities
use laplace transforms (laplacian of derivative) to more directly get the proper integral of any function, just have your laplace tables completed
Interesting wrap-up at the end. Despite having never learned calculus, that part was insightful.
sin²x is poor notation because it suggests sin(sin(x))
Game of integration constant and trigonometry. Expansion of Cos(2x)=2cos²x-1=1-2sin²x
Numbers concerted into integration constant.
i agree, sin² x is terrible notation
Both method 2 and method 3 are going to save you at some point (3 works like a charm if you have both exponential and trigonometric functions at once). My favorite way of writing it would be
∫sinxcosxdx=
=(1/2)∫2sinx(sin)'(x)dx
=(1/2)∫((sinx)^2)'dx
=(1/2)(sinx)^2+c
though, because it happens in an uninterrupted line.
I really liked how using different approaches leads to different looking answers that all correct. The easiest method, for me, was u-substitution (let u=Sin(x) then du=Cos(x)dx). Using Trigonometric identities, the results can be shown as equal to each other. Now, showing that would be a great trig review for the students, though they might not think so. 😁
Yeah to me too. Although the half angle tangent substitution would have been fun too lol.
This is weirdly nostalgic for me... in 1982 I had an interview for a place at Nottingham University studying maths. In the interview they asked me to integrate something a little trickier: e^x.sin(x).cos(x) - I ended up doing it by parts, but then applying the "by parts" method again. It was a pretty tense experience! I did get an offer in the end but I chose to go to Southampton instead (I think I just wanted to live by the sea!). Every time I see Brady's videos at Nottingham I always wonder how many of the people on screen would have been my teachers, had I chosen differently... obviously not the younger ones, but still.
I wish we lived in a world where educators at lower levels weren't afraid of complex numbers. They make most of trigonometry _very_ straightforward, after all: sin x = (e^ix - e^-ix)/2i, cos x = (e^ix + e^-ix)/2, so multiply all those things together (including e^x) and you get (e^(1+2i)x - e^(1-2i)x)/4i right away. While it will require some later manipulation, this is very easy to integrate as it is.
@@therealax6 I really like that approach. Thanks!
You can even get two more "different" results, if when using integration by parts, you substitute -1/4 cos(2x) or -1/2 cos^2(x) for the integral of sin(x)cos(x) at the end.
what this also means is that we can change 1/4 cos(2x) to -1/2 sin²(x) to 1/2 cos²(x) just by adding constants
maybe there exists out there another family of functions who each look different, but you can get one from the other by adding (the same?) constant
do you mean like tan and cotan, and sec and cosec?
@@vsm1456 maybe even outside the trig functions!
Ben is the best math demonstrator on this channel. Period.
This is a classic. What we're doing here is *not* integration, it's antidifferentiation*. Since antidifferentiation results in a one-parameter family of functions, we have to compare results as one-parameter families, not as singular functions. Basically, we have different additive constants, and we can write each of those constants in terms of any of the other constants.
On the other hand, if we do *actual* integration, we will get the same value from all of the methods (assuming you're careful about how limits go through substitutions).
* I'm being opinionated about terminology here. The term "indefinite integration" is definitionally equivalent to "antidifferentiation", while "definite integration" is what I simply call "integration". I find that making a clearer distinction between the nature of these operations helps students keep them in mind as separate things, which happen to be nicely related thanks to FTC.
let u = sinx
cant believe they forgot substitution
Yeah I learned this back in calculus when I got a different answer than the answer sheet, so I graphed them on Geogebra and noticed exactly what this video explains.
With the first method you can get -1/4 ((cos(x))^2 - (sin(x))^2) + C or -1/4(1-2(sin(x))^2) + C = 1/2(sin(x))^2 - 1/4 + C. Also you can get -1/4 (2(cos(x)^2) - 1) + c = -1/2(cos(x))^2 +1/4+ C. So I guess the first method gives the most complete answer since it provides the answer of the other three methods.
My instinct was "u substitution," u = sin(x), du = cos(x) dx. Thus u^2/2 + c, or sin(x)^2/2 + c. When I noticed the different answer, I was like, "ah, but that's the same answer shifted by a constant." And then I realized that was the point of the video.
This was a cool video! Reminds me of how 1/(x+1) and x/(x+1) has the same derivative, for the same reason. And it freaked me out the first time I saw it
An excellent example, but do you need to use -1/(x+1) and x/(x+1) (or vice versa), for them to have the same derivative?
1/(x+1) and x(x+1) don't have the same derivative. 1/(x+1) and -x/(x+1) do, and the reason (I know you know, but just to complete your idea for other readers) is that:
1/(x+1) = (1+x-x)/(x+1) = [-x + (x+1)] / (1+x) = -x/(1+x) + (x+1)/(1+x) = -x/(x+1) + 1,
which means that they are almost the same function except shifted one unit vertically one from the other, and hence they have the same slope for any value of x.
once you show all 4 results it's quite obvious that there are trigonometric identities that tie them together. Specifically, the cosine double angle formula [cos(2x)=2cos²x-1] and the Pythagorean identity [sin²x + cos²x = 1].
Just another technique to add to the ones presented: since sinx cosx dx = sinx d(sinx), we can make a variable substitution y=sinx and integrate for y: Int(ydy) becomes 1/2y^2+C and finally, replacing back y, 1/2 (sinx)^2 + C. And of course, we can also go the other way around with -d(cosx)...
Really enjoyed this video. What was not explained is why C the constant of integration disappears if the integral has limits which might have been helpful. More fundamentally where C comes from.
It's fairly simple algebra to see why the constant disappears in a definite integral. You're subtracting the values of the same integral evaluated at two different points: (f(a)+c)-(f(b)+c), so the two constants cancel. I find the easiest way to see why you need the constant in the first place is that when you differentiate, all constant terms vanish. So when you integrate you have to put them back - but you don't know what they were; the information is lost.
I did A level maths 53 years ago no maths since and I immediately saw it as integral y dy/dx dx and got y^2/2 where y= sinx.
The last method by parts can also be done by substitution
This is a brilliant observation. I learnt it when I was doing integration problems but getting the different answers for same problem.
I insta-changed variables, sin(x)=s, ds=cos(x), answer is s²/2 ie answer 3 and 4. I found his explanation of "methods" 2 and 3 a bit confusing/shortcut-ish, it's nice to change variables explicitly and show what's happening.
Note to Ben: There is no problem with using sin²(x), sin³(x) et cetera because the meaning is unmistakable. Of course, the issue is when we see sin-¹(x) and there is confusion about whether this is (sin(x))-¹ = 1÷sin(x) or the inverse trig function, arcsine of x.
The problem is entirely the result of using sin-¹(x) for the inverse sine function when the better, unmistakable expression is arcsin(x).
In some contexts, given a function f(x), notation like f²(x) refers to f(f(x)) and f³(x) refers to f(f(f(x))), where the exponent right after the function but before the argument refers to _repeated composition._ So the meaning very well could be mistakable. I suspect Ben holds this view (but I haven't asked to be fair!)
After all, we have a designated location to denote exponentiation already. Why allow exponentiation to take _two_ possible locations, including the only sensible location for repeated composition. By the way, under the repeated composition view, f-¹(x) is unmistakably the inverse function of f(x). I find this a better option than saying we should write arcf(x).
@@MuffinsAPlenty At 7:35 he quickly says "it confuses with function notation" so this is probably what he meant.
I used Leibniz method on this one and got (-1/4)*cos(2x).
Defined F(t) = integral( sin(x)*sin(x + t*Pi/2) )dx, then found that F'' + (Pi/2)^2*F(t) =0, solved for F(t) and evaluated F(1) = integral( sin(x)*sin(x + Pi) ) dx = integral( sin(x)*cos(x)) the desired integral
In beam theory, when integrating from rotation to displacement, then twice to shear and finally to bending, the need to carefully establish what the constant is at each integration.
7:24 ".. which no-one argues with, but no-one writes." Hahaha
If you actually put in limits, let's say 0 to pi/2, all three give the same answer, 0.5. Yellow curve: 0.5 - 0 = 0.5. Red curve: 0.25 - (-0.25) = 0.5. Blue curve: 0 - (-0.5) = 0.5.
Here is an interesting twist. Consider the fact that cos(2x) = [cos(x)]^2 - [sin(x)]^2 if you add method 2 to either method 3 or 4, you end up with -1/2cos(2x) which is simply 2I from method 1. Using the graph, If you look at the sum of the two, i.e., the superposition you will immediately see it is 2I.
Yet another version, very similar to 2&3 (maybe you'd say the same only different) that I was also taught.
substitute y = sin(x)
Then dy = cos(x)dx
I = (sin(x))(cos(x)dx) = y dy = y^2/2.
Substituting sin(x) back in for y
I=(sin(x))^2/2
And identically for cos(x) to get the other result.
The notation of putting a number after fuction name normaly means apply the function that many times, but it is customary to use it with trigonometric functions to raise that function to the power, because those are not usually applied one after other. But of course they are in navigation and in survey where you can see sin(sin(x)) and others.
This reminds me of a (digital) math quiz we had at school. One question was about integrating something like -1/sqrt(1-x^2), I answered: -arcsin(x) +c but the "correct" answer was arccos(x)+c so I lost a point for it.
-arcsin(x) and arccos(x) are the exact same functions just ofset in the y-axis
I think I've figured this out using trig identities:
I'll use cos^2 x and sin^2 x to mean (cos x)^2 and (sin x)^2
start with the 2 identities
cos^2 x + sin^2 x = 1
cos^2 x - sin^2 x = cos 2x
add them
2 cos^2 x = 1 + cos 2x
cos^2 x = (1/2)*(1 + cos 2x)
subtract them
2 sin^2 x = 1 - cos 2x
sin^2 x = (1/2)*(1 - cos 2x)
expand and rearrange
a) (1/2)cos 2x = cos^2 x - (1/2)
b) (-1/2)cos 2x = sin^2 x - (1/2)
a)*(-1/2) and b)*(1/2)
(-1/4)cos 2x = (-1/2)cos^2 x + (1/4)
(-1/4)cos 2x = (1/2)sin^2 x - (1/4)
joining these together
(-1/4)cos 2x = (-1/2)cos^2 x + (1/4) = (1/2)sin^2 x - (1/4)
By parts is where I started. I didn’t know the other methods.
Since this is an indefinite integral we are calculating the anti derivative so the derivative of all 3 of those functions will all be the same since vertical shifts do not change the slope at any point.
It might be easier to change the identity. Sin(x)cos(x) = 0.5sin(2x). After changing it, just work from there.
Thanks for tossing a brick at cos^2(2)
I doubt that's anyone's biggest reason to hate maths, but it can't be helping.
I’ve always used U sub for these types of integrals. u= sinx du = cosxdx dx = du/cosx so the integral comes out to be u du.
*Ben:* ”Is sin(x) the same as cos(x)?”
*Me:* ”Well, in the case of an isosceles right triangle, yes 🧐.”
This can be thought of as proving certain trig identities using calculus (up to a constant).
Well, integration by parts is quite symmetrical in this example and thus can also result in 1/2cos(x)^2
Every one of these videos is another part of forgotten maths integrated into my head once again.
What I might have added on as an explanation is a more general interpretation of the relationship between derivatives and integrals. By performing an indefinite integral, you're essentially starting with the rate of change and asking what function has that (the original) function as its rate of change. Since a rate of change doesn't care about where you start, there has to be a degree of uncertainty (i.e. the constant at the end).
Çç
I'd say that the thing to know (other than remembering your constants when integrating) is that CYCLIC functions (like sine and cosine) can get REALLY WEIRD.
I would have used precisely none of these methods!
Method 5: u substitution!
Let u = sin(x). Then du = cos(x) dx. This then becomes ∫ u du = 1/2 u² + C = 1/2 (sin(x))² + C.
I always thought that +c was inconvenient and slightly imprecise notation. It hides the fact that the solution is a family of functions (+c makes it look like it's one specific function that you don't know exactly), while also causing annoying algebra inaccuracies that are irrelevant (like the c that should technically be c/2 in this video).
I think it's better to think of the answer as a coset of the constant functions, which would look like f(x)+ℝ or f(z)+ℂ (depending on whether you are talking about a real or complex function). Then you can multiply by scalars, and not have to change/redefine your added constant, because the whole set of numbers (which is a field) is fixed under multiplication by a (non-zero) scalar.
Plus, writing it like this is exactly in line with the notation you'd use for an element in the group of functions under pointwise addition, modulo the subgroup ℝ or ℂ of constant functions. Indefinite integration is a group homomorphism into this group.
shouldn't you make it
int f(x) dx = F(x) + c, c in R or C?
@@BlueEyesWhiteTeddy you could write it using the full notation for a set, but I'm using the notation of cosets to simplify this. Like how in group theory you would write g+H to represent the coset {g+h for h in H}, i.e. add g to everything in H to get the new set g+H.
This is common notation across all of mathematics to represent the elements of a structure quotiented by a substructure.
Should have been strictly correct had he used different constants in each integral (C1, C2, etc...). At the end of the day as you say, is that once you get one specific solution for the antiderivative (without the c), then shifting up or down that solution by an arbitrary amount will also be a solution, and sometimes (many times) functions f and g that are just the same shifted up or down one from the other can be written as something other than f=g+c, like f=1/(1+x), if we shift it down by 1 unit we get 1/(1-x)+1 = (taking common denominator (1/x) ) [1-(1+x)]/(1+x) = (cancelling +1 and -1) -x/(1+x), which at first sight doesn't look to be the same as 1/(1+x) just shifted 1 unit, but it is.
...substitution is taught *very* differently in the UK.
as a (pretty bad) A level student here with exams coming in less than 1 week
thank you for the accidental lesson in integration, very appreciated
2:46 - The constant c is already missing in the row with the squared brackets.
u=sinx du=cosxdx. Integral is u^2/2 + c. (sinx)^2 /2 + C
Seen this many times before.
All are equivalent, with a different value for the constant, that is determined by trig identities (like sin^2 + cos^2 = 1 or cos2x = cos^2(x) - sin^2(x) )
Can integrate using substitution too.