My failed attempts to the integral of sqrt(sin^2(x))

[@blackpenredpen]

Try this one next: kzbin.info/www/bejne/o36pe4iQdq2bnJI

[@prasdyandisamramadhan3358]

11:31 sir, u make mistake, isn't (1/cos x) . sin x = tan x ??

[@jotajaviergonzalezgarcia7504]

Feyneman?

[@abdelrahmanayman8415]

I used the trig identity Cos(2x)=1-2sin²(x) And then u sub cos2x And i got -√cos²(x)

[@abdelrahmanayman8415]

I might have done smth wrong

[@forgetittube5882]

Perhaps the simplest substitution: sin(x) = u … dx=du/cos(x)=du/sqrt(1-u^2)

[@ele2735]

One way I found it: sqrt(sin^2x) = sin^2x/sqrt(sin^2x) = 1/sqrt(sin^2x) - cos^2x/sqrt(sin^2x) = sqrt(sin^2x)/sin^2x - [(1/2)cotx][2sinxcosx/sqrt(sin^2x)] Integrate the part with square bracket expressions using integration by parts differentiating (1/2)cotx and integrating 2sinxcosx/sqrt(sin^2x) by using u=sin^2x, du=2sinxcosxdx You are left with -cotxsqrt(sin^2x) and two integrals of the leftover expression sqrt(sin^2x)/sin^2x but with opposite sign, leaving finally just the result -cotxsqrt(sin^2x) +C

[@沈博智-x5y]

nice. u did it. obviously there is a bit of a hole when sin(x) = 0 since you expressed sqrt(sin^2(x)) = sin^2(x)/sqrt(sin^2(x)) (so undefined at sin(x) = 0) but we are just trynna get it to be the same as wolframalpha (i guess w/o piecewise manipulation) and there we have it.

[@blackpenredpen]

Wow! I think this works and it’s exactly what I am looking for. I will come back to this comment latter to read more. Big thanks!!

[@brian554xx]

@@blackpenredpen i'm looking forward to seeing this on a whiteboard. i struggle with plaintext math.

[@_skyslayer]

@@blackpenredpen you should also check my method, unfortunately my comment got lost in the stack. I found a method that works for any integral { sqrt(f(x)^2) } dx without introducing absolute value in the process...

[@klmcwhirter]

@@blackpenredpen Dude! Your determination and humility is a breath of fresh air for me. This world needs more math profs like you. As I think back at the profs I had (shudder). If I had you I just might have stuck with math instead of transitioning my career to Comp Sci. I applaud you for making videos like this! Those of us who have been on the planet for some time recognize that, of course, none of us can know it all. Thanks for being a trend setter - helping math students see first hand what teamwork is all about.

[@hamdansaad3406]

For the second method multiplying 1/cos(x) and 1/2(sin(x)sqrt(sin²(x)) gives you 1/2(tan(x)sqrt(sin²(x)) not cot(x)

[@utuberaj60]

Yes, dear blackpenredpen- you should be writing uv=tanx.sqrt sin^2x and NOT cot x....

[@Kero-zc5tc]

How can it give tan(x)? Isn’t the sin and cos being multiplied on the denominator? Isn’t tanx = sinx divided by cosx?

[@Kero-zc5tc]

Do you mean 1/2 multiplied by sinx then, sorry I commented before getting up to that part

[@lantami1199]

@@Kero-zc5tc Yes, that's what they wrote

[@Kero-zc5tc]

Thanks

[@hvok99]

I love this kind of video - Showing a problem that you are not totally certain about and authentically presenting your ideas and challenges. I would love to teach more like this in my class and see how students engage. It made for a very engaging hook.

[@pierreabbat6157]

If you graph -√(sin²(x))cot(x), you get a graph with discontinuities at 0, π, 2π, etc., which is not the integral of √(sin²(x)). The integral has the discontinuities fixed by adding a staircase function.

[@沈博智-x5y]

yep. i saw this algebraically when i tried to shoehorn -cot(x)|sin(x)| into the solution when solving the integral of |sin(x)| piecewise in order to shoehorn the soln (when doing piecewise), multiplication of sin(x)/sin(x) is necessary, and this is not possible with sin(x) = 0 i.e. at x = any integer multiple of pi. so yeah, -cot(x)|sin(x)| is not the most accurate form of the antiderivative of sqrt(sin^2(x)). nice.

[@domahidipeter6092]

I dont get it. Sqrt(sin(x)^2) always positive, but -cot(x)*sqrt(sin(x)^2) have negative values

[@沈博智-x5y]

@@domahidipeter6092 just because a function is always positive does not guarantee it's antiderivative to always have positive values take x^2 which is never negative integrate it you get x^3/3 + C But last time i check for x < 0, it is negative.

[@r-prime]

​@@沈博智-x5y by integral comparison test, if f(x)>0, and a 0 also. So the first comment is correct. [Conversely, if a > b then the integral < 0, explaining the negative on the integral of x²]

[@沈博智-x5y]

@@r-prime sure, if we're talking about definite integrals. but the comment wasnt necessarily stating definite integrals., merely it was stating a singular point rather than an interval. i.e. for some values of x, the anti-derivative has negative outputs. (this is fine). ---- so if sqrt(sin^2(x)) was represented as a definite integral then the antiderivative -cot(x)sqrt(sin^2(x)) can be used on the condition that the integral is split every integer multiple of pi (due to the holes at sin(x) = 0) [and this would lead to the 'positive values case'] but again, i dont think thats what the original commentor who had doubts about the 'answer' -cot(x)|sin(x)| was talking about.

[@沈博智-x5y]

11:36, isn't it tan(x) instead of cot(x)? 1/cos * sin = tan

[@eduardodocarmoszadkowski4163]

I think so

[@JohnnyLaw3134]

Yeah, it's tan x

[@simonvh7092]

He didn't have the right board up :(

[@blackpenredpen]

Yes u r right. That was my mistake

[@ideatronzvvvicunit9476]

You are really a great person with even greater petsonality. This video is a proof of this fact , which is more important than proving that integration problem.

[@markbrightwell3536]

A piecewise solution shows that the generic antiderivative of |sin(x)| is discontinuous at multiples of pi. Each continuous piece of the discontinuous antiderivative has a separate integration constant. Only a very special choice of the integration constants gives a continuous antiderivative.

[@Silvar55x]

11:21 It's only the coolest calculus teachers that can make the whiteboard erase itself out if respect, just by tapping it. (Kudos for an amusing video cut.)

[@philipal7170]

Correction: at 11:40 it's tan(x), not cot(x).

[@laroete]

Your videos make me happy, i havent done any math since high school and my grades werent perfect but i just love working through stuff with you in my mind

[@danielspivak3926]

The integral of a non-negative function cannot be periodic (unless the non-negative function is 0). If you look at the right side closely, it is not continuous, so that's why its derivative is the left side, but the integral of the left side is not the right side.

[@JavSusLar]

Exactly what I was thinking, but couldn't express it in a mathematically proper way. Thanks.

[@tomhejda6450]

That's why I gave a dislike. Also, if anyone wondered where the problem is in the derivative of the Wolfram Alpha result: You cannot ignore handling of discontinuities. What BPRP is doing is non-mathematical technical usage of derivative rules without any understanding of what they mean.

[@TomFarrell-p9z]

Yes. You need to limit the range of x, or include a term like 2*pi*n where n are the number of cycles of the sin wave. This really is more natural as a definite integral rather than an anti-derivative.

[@leeyc0]

Googling integral abs(sin(x)) seems finds the correct answer, which is 2 * floor(x/pi) - cos(x)sgn(sin(x))

[@General12th]

​@@tomhejda6450I'm reasonably certain he has a PhD in mathematics. If you think this video is evidence that he doesn't understand the basics, try to build a case and get his degree stripped from him.

[@zhabiboss]

Please i want you to make the formula for a^x + bx + c = 0 🙏

[@matthewtallent8296]

You got what you wished for 😎

[@John-cl8iv2]

where?@@matthewtallent8296

[@Pharoom]

If you formulate sqrt(sin(x)^2) as (sqrt(sin(x)^2)/sin(x))*sin(x) and do integration by parts, then you get the correct result. Not sure if it's technically allowed, though, since the first part is not continuous.

[@Fematika]

I don't understand why continuity would be needed here? So long as they are continuous except on a measure zero subset it should be fine.

[@willie333b]

Integral calculator says you just pull the absolute value out of the integral

[@Bo0mber]

I'm pretty sure sqrt(sin(x)^2) formulates as sqrt(sin(x)^2/sin(x)^2)*sin(x), which indeed should not be allowed, since we get sqrt(1)*sin(x), implying that sqrt(sin(x)^2)=sin(x) (obviously not true)

[@tomhejda6450]

​@@FematikaFor integration by parts you of course need continuity for some bits. Certainly an integral of a continuous function is continuous, so the bit you pull out of the integral must be continuous! And exactly the result by Alpha is simply WRONG because it's discontinuous, so if BPRP arrived at it, he would have been wrong.

[@lazarussevy2777]

@@tomhejda6450 I don't think sqrt(sin^2(x)) is exactly continuous, though. There those sharp points every π, and wolfram alpha's solution is discontinuous and undefined every π. You might want to fact check me on that though.

[@satindra.r]

You can rewrite |sin(x)| as sin(x)*sgn(sin(x)), since the derivative of sgn(sin(x)) is 0 when we use integration by parts the second bit becomes 0 and we get -cos(x)*sgn(sin(x)) which can we written as (-cos(x)*|sin(x)|)/sin(x) which is equal to |sin(x)|*-cot(x)

[@GreenMeansGOF]

I wonder if this is too related to the piecewise solution he is avoiding

[@vladislavanikin3398]

This. But it's interesting how Wolfram Alpha does it, because I remember I've once asked it about the integral of |sin(x)|dx and it used signum in the answer, but here it doesn't seem to recognize that √(x²)=|x| and uses something else.

[@Joey47600]

For the final attempt, you can simplify sin^-1(u) to sin^-1(cos(x)) and then to sin^-1(sin(x+pi/2)), which is just x+pi/2

[@elizabethsusanlibra]

That doesn't work since x+ pi/2 is not in the domain of arcsin, rather it would simplify to pi/2 - x instead.

[@Joey47600]

@@elizabethsusanlibra Thank you

[@BassemFanari]

Here's how to solve it ✅ • Step 1: Multiply and divide by √sin²(x): ∫ √sin²(x) (√sin²(x)/√sin²(x)) dx = ∫ sin²(x)/√sin²(x) dx • Step 2: Isolate a single sin(x): = ∫ sin(x) (sin(x)/√sin²(x)) dx • Step 3: Apply IBP: u = sin(x)/√sin²(x) , dv = sin(x) du = 0 dx , v = -cos(x) ∫ u dv = uv - ∫ v du ∫ √sin²(x) dx = (sin(x)/√sin²(x)) (-cos(x)) - ∫ 0 dx = (sin(x)/√sin²(x)) (-cos(x)) + C • Step 4: Again as Step 1, multiply and divide by √sin²(x): = (sin(x)√sin²(x)/sin²(x)) (-cos(x)) + C • Step 5: Cancel a single sin(x) from both numerator and denominator: = (√sin²(x)/sin(x)) (-cos(x)) + C = √sin²(x) (-cot(x)) + C.

[@profismate]

@blackpenredpen , please consider sqrt(sin² x) = |sin(x)| and that integral answer can be written like - sgn(sin(x))cos(x). Solving this way is easier!

[@ilias-4252]

hope you are joking

[@kokainum]

F(x) = 2* floor(x/PI) - cos(x-PI*floor(x/PI)) + C, best to see this by drawing the plot. First look at F(x) at (0,PI), this should be something like 1-cos(x) + C. We see that f(x) is cyclic so you're gonna get increase of 2 at every distance of PI and then you need to get the same shape and also you can shift whole plot up and down and we get it from C. The formula from Wolfram Alpha works at each of intervals (k*PI, (k+1)*PI) separately, but it fails to describe whole function because you need to use these parts and shift them each by different C to glue them into one continuous function. I mean as I said function needs to grow 2 per each PI (it can look like it grows linearly but with small oscillations, it has a trend) and the formula from Wolfram Alpha is bounder and has values from (C-1, C+1). It's because you need to choose different C for each interval to get the final formula. I leave it as excercise to prove details of my formula but I gave you a sketch already. :) Edit: I guess you can change -cos(x-PI*floor(x/PI)) by -cos(x) * sgn(sin(x))=(kinda but not really because cot(x) explodes) -cot(x) * abs(sin(x)). Still -cos(x) * sgn(sin(x)) looks better and is defined everywhere so F(x) = 2*floor(x/PI) - cos(x) * sgn(sin(x)) + C and I think it's the prettiest of the formulas. The one with cot explodes at points like k*PI so I don't like it, but besides those points it works.

[@MichaelRothwell1]

Very nice! I found a similar solution using the floor function to fix the jump discontinuities, but yours is neater.

[@AbouTaim-Lille]

Notice that the concept 'integrable does not imply differentiability or even continuity in the Riemannian Integral. As of Lebesgue integral this concept is much larger. For example the function f(x)= |x| which is not differentiable at zero, can be written as f(x) = { X if x>0 , -x if x0 , -½x²+C if x

[@bartekabuz855]

the result has holes in domain at x=k*pi but sqrt(sin^2(x)) does not. I think that's the reason you cannot arrive at answer

[@barco1ml]

I can almost get there with u = sin(x) and dx = du/cos(x). Then x = arcsin(u) and so cos(x) = sqrt(1-u^s) so it becomes integral(sqrt(u^2/(1-u^2)),u) which gives the answer as -sqrt(sin^2(x)) ⋅ sqrt(cos^2(x))/sin(x) which is almost the WA answer, but not quite...

[@mtaur4113]

|sin| is a piecewise function, integrate sin and -sin on different intervals. To get a continuous antiderivative on all reals, you have to pick "+C" as a step function sewing the two pieces of antiderivative together.

[@Bhuvan_MS]

He mentioned that he didn't want to do it that way.

[@mtaur4113]

​​@@Bhuvan_MSwell, a continuous antiderivative on the real domain exists, and this discontinuous function in the video isn't it. But sure, integrating by parts with absolute values in cognito is cuter, so there's that.

[@arielhaslam4249]

Early to blackpenredpen, nothing could be better :)

[@patrikhallsjo]

Might not the the way forward that you are looking for, but this is quite simple if you write it as complex exponents and then integrate them.

[@CliffChafin]

You are integrating a positive definite expression. The result must be increasing. It should be a stair step function plus a cosine of the remainder. I don't know how nicely you can do this with elementary functions. abs value can be written with squares and square roots but I think you are into the land of infinite trig series here.

[@sergeygaevoy6422]

Abs() isn't an analytic function so I think it is the problem. This fact produces these discontinuities apparently. That way 1 / (1 + x *x) diverges when abs(x) >= 1 despite no apparent problems (but we have (i^2 = (-i)^2 = -1 of course). Abs() isn't infinitely differentiable and integrable. It doesn't behave "well".

[@ahmedtharwat9870]

Multiply by sin(x)/sin(x) Then D sqrt(sin^2(x))/sin(x) And Int sinx And the new integration will equal 0 Try it

[@neel18_18]

I have a trick I = int of √(sin²x)dx remove the square root and inside square to obtain I = int of sinx•dx = -cos(x) +c. And the answer is given as -√(sin²x)•cotx +c = -sinx•cotx +c = -sinx•(cosx/sinx) + c = -cosx + c.

[@kennethalfredo5532]

Hey, I have an idea (Consider this as alternative method) Why don't we write sqrt(sin² x) as sqrt(1/(csc² x)) since csc x = 1/(sin x) so that sin x = 1/(csc x). Then using the trigonometric identity csc² x = cot² x + 1 Then we can write the sqrt(sin² x) as 1 over sqrt(cot² x + 1). This is interesting because we can assume that maybe the result will still have the cot x term. From there, we can use trigonometric substitution and it will yield a beautiful result. Let cot x = tan y, then we have -csc² x dx = sec² y dy (There is "-" there, I think we are on the right track). We can use the trigonometric identity once more csc² x = cot² x + 1 Then, we will obtain -(cot²x + 1) dx = sec² y dy Remember that cot x = tan y, so cot² x = tan² y, so we can rewrite it as -(tan² y + 1) dx = sec² y dy dx = -(sec²y/(tan²y + 1)) dy Hey, wait a second. We know that tan²y + 1 = sec²y from trigonometric identity. So in the end, we yield dx = -dy. Now we can write the integral of [1/sqrt(cot²x + 1)] dx as the integral of [1/sqrt(tan²y + 1)] • -dy or the minus of the integral of [1/sqrt(tan²y + 1)] dy. Let's try simplify the 1/sqrt(tan²y + 1). 1/sqrt(tan²y + 1) = 1/sqrt(sec²y) = 1/sec y = cos y (since sec y = 1/cos y so that cos y = 1/sec y) So we will have the minus of the integral of cos y dy. Wow, another simple integral form. Thus, the integral of it will be -sin y + c Now, we need to change it back to x term. However, we only have cot x = tan y, but we need sin y. Then rewrite tan y as cot x over 1. So we have a triangle with sqrt(cot²x + 1) as the length of the hypotenuse. Now we can write the integral result as - (cot x)/sqrt(cot² x + 1) + C Now we have the result similar to what wolfram alpha want, but it missing the sqrt(sin²x). Don't worry, we just need simplified the sqrt part. This we will have - (cot x)/sqrt(cot² x + 1) + C = - (cot x)/sqrt(csc² x) + C = - sqrt(sin²x) cot x + C (QED) How wonderful is that. This is my way to beat Wolfram Alpha I guess 😅.

[@yoylecake313]

can’t we just do sqrt(sin²x) = |sin(x)|

[@agent3667]

I can't figure out how to integrate but the derivative is almost the integral (kinda expected since the derivative of sinx is almost the integral of sinx). If we consider sqrt(sin²x) to be |sinx|, and the derivative of |x| can be thought of as |x|/x so the derivative of |sinx| can be (|sinx|/sinx)*cosx = |sinx|cotx which is almost the same, it just needs the negative sign. I couldn't think of another way of getting the answer wolfram alpha gave though.

[@Bo0mber]

A really solid attempt for someone who can't integrate! There is a problem though: derivative is not almost like integral, they are, in fact, inverse, integral of derivative gives you the original function. So no, you can't differentiate istead of integrating. Also, yes, integral and derivative of sin(x) can be considered similar, but check this out: integral arctan(x) = arctan(x) *x - 1/2 * ln(1 + x^2) derivative arctan(x) = 1/(1 + x^2)

[@SkorjOlafsen]

Related: have you ever talked about integrating sign(sin x)) and the surprise of functions like arccos(cos x) ? Seems like a fun topic.

[@adayah2933]

sgn(sin x) doesn't have an antiderivative

[@SkorjOlafsen]

@@adayah2933 sure it does, piecewise. It's obvious what the area has to be between any a and b (taken in finitely many pieces). The fun is finding a function with the right derivitive everywhere but nπ.

[@philipal7170]

The solution: make the integral to be: sinx(sqrt(sin^2(x)))/sinx, then integrate by parts by integrating sinx, so it'll become -cosx, which will give -cotx(sqrt(sin^2(x))) minus the integral of - cosx times the derivative of sqrt(sin^2(x)), which is equal to 0, so the integral will be equal to a constant, hence we get the answer -cotx(sqrt(sin^2(x))) + C

[@_skyslayer]

goal: ∫ √(f(x)²) dx rewrite it as: ∫ √(f(x)²) dx = ∫ [√(f(x)²) / f(x)] * f(x) dx integrate by parts taking: D = √(f(x)²) / f(x) I = f(x) we will get: ∫ [√(f(x)²) / f(x)] * f(x) dx = ∫ I dx * D - ∫ { ∫ I dx * D'(x) } dx after some manipulation we can show that D'(x) = 0 which means that: ∫ [√(f(x)²) / f(x)] * f(x) dx = ∫ I dx * D - 0 = ∫ f(x) dx * √(f(x)²) / f(x) + C so: ∫ √(f(x)²) dx = ∫ f(x) dx * √(f(x)²) / f(x) + C for f(x) = sin(x) we will get: ∫ √(sin(x)²) dx = ∫ sin(x) dx * √(sin(x)²) / sin(x) + C = - cos(x) / sin(x) * √(sin(x)²) + C = - √(sin(x)²) * cot(x) + C

[@_skyslayer]

Here are the details on how to show D'(x) = 0: 1) Simplify notation for readability: d/dx [√(f(x)²) / f(x)] = d/dx [√(f²) / f] 2) Use chain rules: ... = [2 f f' / (2 √(f²)) * f - √(f²) * f' ] / f² = [f² f' / √(f²) - √(f²) * f' ] / f² 3) Multiply the first term by √(f²) / √(f²): ... = [f² f' * √(f²) / (√(f²) * √(f²)) - √(f²) * f' ] / f² = [f' * √(f²) - f' * √(f²) ] / f² = 0

[@dimosbachlas8203]

In your second way you can write sin²x/cos²x as tan²x+1-1 and the integral=integral of (tanx)'sqrt(sin²x) - integral of sqrt(sin²x) and we have to just calculate the integral of (tanx)'sqrt(sin²x) with DI method i tried it and it works

[@adrified9352]

The way you do math is so elegant

[@kmsbean]

at 13:00, we need to get the co-efficient of -sqrt(sin2x)cotx from 1/2 to -1. observation: sin2x(sqrt(sin2x)) = sin(3/2)x, aka u*sqrt(u) = u^1.5 if u=sin2x. if we can get that intergral to equal (3/2) sqrt(sin2x)cotx then we're good

[@TranquilSeaOfMath]

Nice example of the problem solving process. It isn't easy to show this as an instructor.

[@TramNguyen-pk2ht]

|x| < 0

[@alessandroarmenti5562]

To keep advancing on the third attempt you can treat arcsincosx as one side of a right triangle as one of the proofs of the derivative of arctanx

[@rikschaaf]

You already solved it in reverse, so you know how you have to change the sqrt(sin(x)^2). Turn it into sqrt(sin(x)^2)*(csc(x)^2-cot(x)^2) and continue working your way backwards.

[@GoldSleezy]

By the double angle formula: (Sinx)^2 =[1/2sin(x/2)cos(x/2)] ^ 2 Let u = sin 0.5x Du = 0.5 cos0.5x dx The remainder is self explanatory

[@Bhuvan_MS]

Isn't it called Half Angle formula?

[@yatharthshrivastava846]

As easy and simple as that: √f² = |f| = f * sgn(f) Use then Integration By Parts, or D/I method. Differentiate sgn(f) and integrate f. You must know that d/dx [sgn(f(x))] = 0, if f(x) ≠ 0. Thus, the integral reduces to: Integral(f) * sgn(f) Or, if f(x) ≠ 0, Integral(f) / sgn(f) Note: sgn(f) = f / |f| = |f| / f, if f(x) ≠ 0. A high school student from India tried to solve it as easy as possible.

[@yatharthshrivastava846]

Easy generalisation, but leads to discontinuous graph, can be corrected by staircase function.

[@yatharthshrivastava846]

Actually this was taught to us at school.

[@nfergie19]

since you can do (uv)' = u'v+uv'. you can reverse the step to get int(u'v+uv') = int(u'v) + int(uv') = uv - int(uv')+ int (uv') = uv. and voila.

[@sandglass9928]

bro, if you multiply by secxtanx on the top and the bottom, and you integrate by parts, integrate secxtanx and derivate the rest, you could reach the result, it also works if you multiply by senx on the top and the bottom and integrating by parts, integrate senx and derivate the rest. I think that only using this two functions and equivalents it works.

[@TomFarrell-p9z]

Looks like there are several good answers via trig substitutions in the comments. One thing to point out in general: If you already have a solution (as from Wolfram Alpha), and you take the derivative, as you did, to verify that it gives you the integrand, you then have multiple lines of equivalent functions. You can try integrating any of those to see if they are easily solvable. For example, I would have tried the last two lines at 4:04. Before doing that, I would have also tried plotting a numerical solution to each side (of course, that would be for the definite integral version) to see what insight it might yield. For example, the left hand side appears not to be periodic, while the right hand side (from Wolfram) does, and the numerical solution may explain what's going on. But, I'm an engineer, not a mathematician. Another thought: Recognizing sqrt(sin(x)^2) = |sin(x)| = sin |x|, you might try the Taylor expansion and then integrating term by term. That yields an infinite sum which would be difficult to resolve back into trig functions, but it would take care of the discontinuities.

[@ceezan8085]

I found it out thid way, We can multiply the integrand by cosec(x) sin(x) which becomes cosec(x) sin(x) √sin²(x). Now if we integrate by parts by differentiating cosec(x)√sin²(x) and integrate sin(x), the differenriation (by rationalizing the denominator and all) will turn out to be 0. So we can get the final result -cosec(x)cos(x)√sin²(x) which is equivalent to -cot(x)√sin²(x).

[@sylowlover]

sqrt(sin²(x))=abs(sin(x)) then do it piecewise for every interval of the form [kπ,(k+1)π]. No need for explicit expressions.

[@wavingbuddy3535]

ok so, integral( sqrt( sin^2(x) ) ) = integral( abs( sin(x)) ) = sin(x)/( abs( sin(x) ) * integral( sin(x) ) = - ( cos(x) )( sin(x) ) / abs( sin(x) ) = -cos(x)sin(x)/ sqrt(sin^2(x)) = -cos(x)sin(x)sqrt(sin^2(x))/ sin^2(x) = -cos(x)sqrt(sin^2(x))/ sin(x) = -cot(x)sqrt(sin^2(x)) and dont forget the +C

[@sergeygaevoy6422]

Did you try u = sin(x) ? It will eliminate sin() altogether. Or abs(x) = x * sign(1/x)? And then to x < 0 and x > 0 separetedly.

[@沈博智-x5y]

Honestly. I dunno how to get to the answer without using piecewise. Here is how I get it piecewise (without just 'differentiating the goal anti-derivative and calling it a day') \int sqrt(sin^2(x)) dx = \int |sin(x)| dx = { \int sin(x) dx when sin(x) >= 0, \int (-sin(x)) dx when sin(x) < 0} = { - cos(x) + c1 when sin(x) >= 0, cos(x) + c2, when sin(x) < 0} = { -cot(x)sin(x) + c1 when sin(x) > 0 (multiply by sin(x)/sin(x)) so sin(x) cannot equal 0 , cot(x)sin(x), when sin(x) < 0} = {-cot(x)|sin(x)| + c1 when sin(x) > 0 , -cot(x)(-sin(x)), when sin(x) < 0} = {-cot(x)|sin(x)| + c1 when sin(x) > 0, -cot(x)|sin(x)|, when sin(x) < 0} = -cot(x)|sin(x)| + c = -cot(x)sqrt(sin^2(x)) + c things glitch out when sin(x) = 0, because cot(x) is undefined when sin(x) = 0

[@沈博智-x5y]

so maybe we can't reach it 'without piecewise' because it isn't valid for sin(x) = 0. but the original integral and the integrand allows sin(x) = 0 just fine.

[@justinbrentwood1299]

Btw, arcsin(cos(x))= pi/2 +/- x + 2pi(n) for all integers n

[@tomctutor]

I was thinking of subs sin(x) with complex function exp(u) -> sqrt(sin^2(x)) = sqrt(e^2u) but Wolfram Alpha gives integral {sqrt(e^(2u))} = sqrt(e^(2u)) -> sin(x) because its treating u as a real! I'm sure that some complex substitution might work but I wouldn't trust Wolfram to do that correctly!

[@Gman6365]

One online calculus site uses this approach: The integral of |sin(x)| is simply sin(x)/|sin(x)| multiplied by the integral of sin(x), giving c - sin(x)cos(x)/|sin(x)| Multiplying top & bottom by |sin(x)| gives the Wolfram answer. However, there is an alternative: sin^2(x)=(|sin(x)|)^2 so sin(x)/|sin(x)| = \sin(x)|/sin(x) You now have |sin(x)|/sin(x) multiplied by the integral of sin(x), giving c - |sin(x)|cot(x)

[@adnantasin485]

Sir, I found the answer to be 1/2 tan(x) √sin^2(x) - 1/2 √sin^2(x) cot(x) - 1/2 √sin^2(x) cosec(x) sec(x). I differentiated it and it was √sin^2(x).

[@roulio7908]

i may be dumb but sqrt(sin²(x)) isnt it just |sin(x)| so we just want to multiply by 2 the integral of sin(x) which is -cos(x) so final ans -2cos(x) + C

[@MichaelRothwell1]

My first reaction on seeing this problem was that the antiderivative given by WolframAlpha is incorrect. One way to see this is by noting that the alleged antiderivative function is periodic (in fact with period π), whilst the integrand √(sin²x)=|sin x| is nearly always positive, so its antiderivative is strictly increasing. Then there is the problem of the jump discontinuity at multiples of π. However, I found that both problems can be fixed, as I show in my solution to finding the antiderivative below. To simplify things, I shall write the integrand as |sin x| instead of √(sin²x) So we are to evaluate ∫|sin x|dx. To illustrate the method, start with ∫|x|dx. We have two cases: x≥0: ∫|x|dx=∫x dx=½x²+c x

[@JonathanTot]

I think the problem is that the Wolfram Alpha answer is wrong. If we integrate sqrt[ (sin(x))^2 ], we will get a monotonically not-decreasing function. The WA answer =-cos(x) where sin(x)>0, cos(x) where sin(x)

[@CliffChafin]

Yes. I noticed that as well. Look up at my comment and find the real exact answer. I'm surprised how few people are noticing this.

[@hardubs]

As identified in the beginning, this integral is the integral of |sin(x)|. This is -cos(x) when sin is nonnegative and +cos(x) where sin is positive. Thus the integral is -sign(sin(x))cos(x). Our function is not integrable where sin(x) = 0, so wherever the integral exists it is true that sign(sin(x)) = |sin(x)| / sin(x) = sqrt(sin^2(x)) / sin(x). Therefore the integral is -[sqrt(sin^2(x)) / sin(x)]cos(x), i.e. sqrt(sin^2(x))(-cot(x)).

[@chrissmith4933]

Can you use Euler's identities to transform sin^2(x) into (cos2x -1)/2 via sinx = (e^ix - e^-ix)/2i and continue from there?

[@enderrequiem2176]

Mupltiply and divide by sinx. Then use integration by parts: integrate sinx and differentiate |sinx|/sinx

[@shaneruegg1402]

Let us simplify sqrt(sin^2(x)) to |sin(x)|, and we will remember that we can convert it back. int(|sin(x)|) = {sin(x)/|sin(x)|}int(sin(x)) {sin(x)/|sin(x)|} * -cos(x), + C We will then flip the fraction and still have the same expression. {|sin(x)|/sin(x)} * -cos(x), + C Rewrite the expression |sin(x)| * {-cos(x)/sin(x)}, + C Rewrite abs of |sin(x)| sqrt(sin^2(x)) * -cot(x), + C

[@zucazx]

One way I thought was doing sqrt(sin^2(x))=sin(x)*sgn(sin(x)), and I think that if we integrate a function times its sign, then we can put the sign outside and just integrate the function. We would need to prove this, but it is really correct to elementar functions. So we would have sgn(sin(x)) * I { sin(x) }, and so we have -cos(x)sgn(sin(x))=-cos(x)*|sinx|/sinx=-cot(x)*sqrt(sin^2(x)).

[@GreenMeansGOF]

I think he’s trying to avoid this since this is basically the piecewise solution

[@stumme]

Could we conclude or demomnstrate that : integral(abs(f(x)) dx) = abs(f(x)) / f(x) * F(x) ?

[@blackpenredpen]

Oh hey, that’s pretty interesting. I am not sure yet!

[@lechuu5683]

the last one the last step maybe you could substitute u = sinv so that then sqrt(1 - u^2) = cosv

[@iwilldefeatraymak2536]

∫|sinx|dx Multiply top and bottom by sin x ∫(|sinx|/ sinx)×sinx dx Since |sinx|/sin x is a constant we can bring it out the integral So now we have |Sinx|/sinx ∫sinx dx =- |sinx|/sinx ×cosx +c And cosx/sinx =cotx So the final ans is -|sinx|×cotx +c And u can write |sinx|as √sin²x

[@tobybartels8426]

0:25 : I'm just starting the video, so maybe you correct this, but don't try too hard to get the answer from Wolfram Alpha, because their answer is *wrong.* (Their function is not continuous; it's not an antiderivative of anything.)

[@tobybartels8426]

A correct antiderivative is 2[x/π]+cos(x) when [x/π] is odd, 2[x/π]−cos(x) when [x/π] is even. (Plus C to get all antiderivatives.)

[@tobybartels8426]

As for the method, I don't think you can get anywhere playing tricks with absolute values as square roots. You just have to break it into cases, depending on whether sin(x) is positive or negative, and then figure out how to stitch the pieces together to get a continuous function.

[@CliffChafin]

I noticed that this function has to increase steadily with x not be periodic.

[@tobybartels8426]

@@CliffChafin : Good point, since its derivative (the integrand) is strictly positive (except at isolated points, the multiples of π), it must be strictly increasing.

[@DelPhoxe]

I feel like this isnt as complicated as we think... You see... as |Number| in terms of real numbers actually meant to multiply by a '+' sign if the Number was +ve whereas by a '-' sign if the Number was -ve... Don't forget the signum function which can be used to adjust the sign (except at '0') So Integral of |sin(x)| is simply -cos(x) but with a exception to the variable +ve or -ve sign of sin(x) for which we multiply it by our signum function of sin(x), i.e, |sin(x)|/sin(x) making it our desired result... (constant of integration crying noises)

[@eduardoschardong4428]

The easiest way: sqrt(sin²x) = |sin x|, the integral wolfram offers is: -cos(x) sgn(sin(x)), from where -cos(x) sqrt(sin²(x))/sin(x)= -cot(x)sqrt(sin²(x))

[@offmano]

Integrate ( |sinx| ) dx = y' + C y = - |sinx| - |sinx| cosx ( |sinx| )' = ------------------------ sinx ( -|sinx| )' = - |sinx| cotanx ( -|sinx| )' = - sqrt [ (sinx)^2 ] cotanx integrate ( |sinx| ) dx = y' + C integrate ( |sinx| ) dx = - sqrt [ (sinx)^2 ] cotanx + C because { -cosx , x>= 0 ( -|sinx| )' = { cosx , x=0 integrate( |sinx| ) dx = { cosx + C , x

[@hevado01]

Because of the stepwise nature of abs(sin(x)) should the integral not use discontinuous functions and become: -cos(x-pi*floor(x/pi))+2*floor(x/pi)

[@SirCumference31]

I found a way to do it. If you consider the derivative of -sqrt(sin^2 x)*cot(x) you get sqrt(sin^2 x). Therefore the integral of sqrt(sin^2 x) is -sqrt(sin^2 x)*cot(x) + C

[@godinhs]

@11:32 1/cosX x sinX = tanX, not cotX

[@thefallinglink]

Haven't tried this but since the "trick" to the derivative of the solution simplifying is the fact that (csc^2(x)-cot^2(x))=1, wouldn't it follow that the intuition or missing piece to solve the integral is to multiply by 1 in that same form?

[@tchntrn]

I have seen this before... this morning while going to the work. it was instagram broadcasting

[@blackpenredpen]

Yup 😆

[@wonghonkongjames4495]

Good Afternoon，Sir，we've got an idea and see if it could be of some help to this special integral: The 0 ÷ 0 =;1 Proof : Since (1÷2)*sinxcos²x

[@dylanoracle3742]

Integral of sqrt(sin^2 x) dx= Sqrt(1-cos^2x) dx Let cos x = sin t Integral of Cos t sqrt(sin^2 t) dt

[@gambini5777]

I am not a mathematican but the engineer in me would guess intuitivelly following solution: 2(x-x%pi)/pi+(1-cos(x%pi))+C. Can someone explain where I am wrong?

[@hevado01]

I came to something similar using floor(), but modulo should work as well to solve the problem that the primitive is stepwise periodic in intervals of pi

[@nanamacapagal8342]

how i got the answer from wolfram: The integral of sin(x) is -cos(x). Everyone knows that. What's pesky is that we have an absolute value. So we can attempt piecewise, then piece it back together. Whenever sin(x) is positive, the integral of |sin(x)| is -cos(x). But when sin(x) is negative, the integral is cos(x). You could say the answer is -cos(x) * sgn(sin(x)). sgn(x) can be written as |x|/x (at least in the reals) so now it's -cos(x) * |sin(x)|/sin(x). A little rearranging and you have |sin(x)| * (-cot(x)), or sqrt(sin²(x)) * (-cot(x)).

[@michelesiosti7461]

Int(sqrt(sin^2(x))dx)=Int(ABS(sin(x))dx)=-cos(x)(ABS(sin(x))/sin(x))+c=-cot(x)sqrt(sin^2(x))+c which is the solution given by WolframAlpha. This solution, however, is valid for individual pi intervals but if it is extended to larger intervals it does not give correct results for a definite integral. In fact it is Int(0,pi)(ABS(sin(x))dx)=[-cot(x)sqrt(sin^2(x))+c](0,pi)=2 while Int(0,2pi )(ABS(sin(x))dx)=4 different from [-cot(x)sqrt(sin^2(x))+c](0,2pi)=2. This is because this antiderivative is periodic and discontinuous in multiples of pi. To make it continuous you need to adjust the constant c for each interval pi. Must be c=2floor(x/pi).

[@holyshit922]

sqrt(sin(x)^2)=abs(sin(x)) Assume sin(x) > 0 Int(sin(x),x) = -cos(x) Multiply by sqrt(sin(x)^2)/sin(x) Int(sqrt(sin(x)^2),x) = -cos(x)*sqrt(sin(x)^2)/sin(x) + C

[@pacocastillo3278]

I solved it by doing the following: Multiply and divide by sin(x), so you get: sqrt(sin²(x))*sin(x)/sin(x) = [sqrt(sin²(x))/sin(x)]*sin(x) Next, do integration by parts. Differentiate the part in brackets [] and integrate sin(x). Now, since sqrt(sin²(x))=|sin(x)|, then sqrt(sin²(x))/sin(x) will always be equal to ±1, so its derivative is 0. And, of course, the integral of sin(x) is -cos(x). Therefore, the integral of sqrt(sin²(x))dx is going to be [sqrt(sin²(x))/sin(x)]*(-cos(x)), and remember that cos(x)/sin(x)=cot(x), so the integral is -sqrt(sin²(x))*cot(x)+C

[@ibraheemalani3584]

A possible solution might be to convert this into absolute value and rewrite it as a split function Assume the integral is I I=-cos(x)+c when sin(x)>0 I=cos(x)+c when sin(x)

[@derwolf7810]

I doubt that what wolfram says, really is in general (= outside of a limited intervall) the correct integral: If i didn't miss anything and F(x) := -sqrt(sin^2(x)) cot(x) + c really was correct, then the following should be true: integral from PI/4 to (2 PI + PI/4) of sqrt(sin^2(x)) dx = F(2 PI + PI/4) - F(PI/4) = -1/sqrt(2) - -1/sqrt(2) = 0. But shouldn't it be 4 (in case i calculated correctly - or at least something greater 0)?

[@CliffChafin]

So I believe this is the actual answer in terms of function you like: abs(sin(x))/sin(x)*(1-cos(x))-(4/Pi arctan(tan(x/2)))+2x/Pi

[@frankjohnson123]

I didn’t do it to completion, but if you try the tangent half angle substitution it seems much simpler.

[@maxrs07]

substituting from step 0 seems promising, ive tried (sinx)^2=u and got an interesting answer. PS why tf people in some places use (co)secant functions just use 1/sin smh

[@mokouf3]

Easy way: 2 case analysis ∫|sin(x)|dx = -cos(x)+C when sin(x)>=0 cos(x)+C when sin(x)

[@offmano]

Integrate ( |sinx| ) dx = y' + C y = - |sinx| - |sinx| cosx ( |sinx| )' = ------------------------ sinx ( -|sinx| )' = - |sinx| cotanx ( -|sinx| )' = - sqrt [ (sinx)^2 ] cotanx integrate ( |sinx| ) dx = y' + C integrate ( |sinx| ) dx = - sqrt [ (sinx)^2 ] cotanx + C because { -cosx , x>= 0 ( -|sinx| )' = { cosx , x=0 integrate( |sinx| ) dx = { cosx + C , x

[@mokouf3]

@@offmano Right.

[@moderator538]

Wouldn't the integral of csc^2(x)*sqrt(sin^2(x)) dx work? We would get: -cot(x)*sqrt(sin^2(x)) + integral of cot^2(x)*sqrt(sin^2(x)) dx which leads to the result of wolframalpha.

[@iamrepairmanman]

So I proved that xlnx-x+c was the integral of lnx because I figured it would work and guessed and got the test correct.

[@eduardochappa4761]

The problem with your antiderivative is that it is wrong without restrictions on the intervals in which you use it. The reason being is that it gives you the wrong value in the interval [pi/2, 3pi/2] (your formula gives you a value of 0 in that interval.) The reason why this does not work is that the cotangent function has a vertical asymptote at pi, which makes your formula a non-continuous function in the interval [pi/2, 3pi/2], which breaks your antiderivative, so if you want a formula like the one you are trying to produce, you can only do it in intervals where the sine function has constant sign, and in that case you can compute the integral as follows: sqrt(sin^2(x)) = |sin(x)| = sign(sinx)*sin(x), so when you integrate, you can put the sign(sin(x)) outside the integral sign (because it is constant), and then you get that integral of |sin(x)| = sign(sin(x))*integral(sin(x)) = sign(sin(x))*(-cos(x)) = -|sin(x))/sin(x)*cos(x) = -|sin(x))cot(x) = -sqrt(sin^2(x))*cot(x). You can use this idea to compute the antiderivative of other trigonometric functions.

[@hassanchoukrani4732]

I find your videos very interesting

[@jamesstrickland833]

-sqrt(sin^2x)*cotx = -cosx So the rhs is the integration of sinx controlling for the restricted values of the original problem's domain

[@jamesstrickland833]

In other words, rewrite the lhs as sinx and Integrate it to -cosx +c, then to maintain the restricted values of x, just rewrite -cosx into sqrt(sin^2x) times something.... which happens to be -sqrt(sin^2x)*cotx

[@jacopostortini7948]

How about you do it geometrically instead of analitically? The curves of |sinx| are quite related to the curves of sinx and if it were a defined integral we would be able to compute easily the area. I didn't try to do the calculation but it seems much more elegant

[@markoj3512]

That video is brilliant!

[@stevekilner5283]

Just rewrite the integrand as sqrt(sin^2(x))(csc^2(x)-cot^2(x)), then distribute to get two integrals. Use integration by parts for the first integral (the second will end up getting cancelled out) with u=sqrt(sin^2(x)) and dv=csc^2(x)dx. From there everything works out nicely. Try it out.

[@Ninja20704]

That does work but it is kind of cheating since we are just working backwards from the differentiation process which is how we know to do that first step. It would be a lot more challenging to do the integral without knowing the differentiation process.

[@stevekilner5283]

⁠@@Ninja20704That is very often how mathematics works. It’s not cheating but rather a common method of problem solving.

[@chhandaMondal-y4y]

@blackpenredpen i challenge you to find the answer of this question If the rectangle has a length of 4 and breadth 2, and from the intersection point of the two diagonals of the rectangle originate two line segments which extend till the perimeter connecting the perimeter to the point of intersection of the diagonals and there's an angle 130 in between them, find the section enclosed within the interior of the angle

[@chrisglosser7318]

Things like sin^-1 (cos (x)) can be evaluated (draw a right triangle)

[@davidspencer3726]

What about Feynmann's technique where you, I don't know, assume everything's a constant, differentiate it to zero then magic a rabbit out of a hat?

[@rotemlv]

This is my reasoning: Let's stay simple and remember that: int(sqrt(sin^2(x))) = { -cos(x), for x s.t. sin(x) >= 0 cos(x) , for else } now, we want a non piecewise function that gives this result, so use one of the sign functions: f1(x) = abs(x)/x f2(x)= x/abs(x) and multiply with the positive answer (in this case -cos(x)) And you'll get the desired result (-sqrt(sin^2(x))*cot(x)) Notice the natural double of this result plots identically: f1 result: This is the one you got from WA. f2 result: (sin(x) * -cos(x))/sqrt(sin^2(x)) As you'll see in the derivative they are identical. You can also prettify it to: -sin(2x)/(2*sqrt(sin^2(x))) Gotta add, it's much easier to do when you already know the answer (and don't play by the rules). (edit - forgot the 2 in the denom at the end)

[@juanbriceno3172]

Why this not taken into consideration when doing trigonometric substitution?

[@_skyslayer]

I think you might want to integrate by parts integral {sin(x) * sqrt(sin^2(x)) / sin(x) } dx, taking D = sqrt(sin^2(x)) / sin(x), I = sin(x). Logically it should work because sqrt(sin^2(x)) / sin(x) = |sin(x)| / sin(x) = sign(sin(x)). And d/dx (sign(sin(x))) = 0. You will get the correct result immidiately, because second part will be 0. You can apply this approach to any integral { sqrt(f(x)^2) } dx

[@xnaaloh4437]

Oh shit I made a comment abt this, i didn’t know sqrt(sin^2(x) / sin(x) had a name lol