Integral of sqrt(1+tan(x))

[@LS-Moto]

I feel so honored :) ... Hello to everyone and warm greetings from Belgium 😀😀😀

[@chirayu_jain]

Hi Lars 👋😃

[@LS-Moto]

@@chirayu_jain Hey... ✌😄

[@blackpenredpen]

Here’s the man!

[@LS-Moto]

@@blackpenredpen 😀

[@Cat-yz1tk]

hey lars hope you are doing well

[@blackpenredpen]

U, w, v

[@chirayu_jain]

And x 😉

[@ayonbarua8949]

Bro why are you so cool? Pls ans.

[@ayoobbhat9180]

Ur Amazing bro From KASHMIR

[@andreiplesa1518]

how about this calculus with this formula (in degrees) limit for x approach for x pi=x\2*sin (360\(x+2)) \ sin((x\4)*(360\(x+2))) I want to know how approach you can get even with a calculator (I tried this formula and works as much as you dont round the numbers (less the calculator to do)) and approach as much as bigger as x is (I sugggest to began aproach with x= 100 to get first digit 3 and for second you need x= apraxch 1000 so almost 10 to the power of 3 plus number of digit you want (for smaller values of diggits ) gets the most approach you can get by hand and calculator

[@mokouf3]

And if I'm the one doing this, I will not use w and v, instead writing down the algebraic twin Something like: d(u-sqrt(2)/u), d(u+sqrt(2)/u)

[@meedonexus]

I got A+ in calculus 21 years ago Integration was my favorite game but after these years I totally lost my skill I love math more than any other science but unfortunately I left studying it to have better job in engineering to gain more money Math is the science that all inventions based on it Love math

[@annevanderbijl3510]

cool story :)

[@samirelzein1978]

i found them back in my job :))

[@aashsyed1277]

@@annevanderbijl3510 hello

[@annevanderbijl3510]

@@aashsyed1277 hahahh hi ive seen you many times before

[@aashsyed1277]

@@annevanderbijl3510 :0

[@VibingMath]

What a show man! And also big congrats to Lars for his winning the cancer finally! You two are awesome integral-fighter and cancer-fighter respectively 😎

[@blackpenredpen]

Thank you!!! Yay!! We are all happy!

[@aashsyed1277]

@@blackpenredpen & you are so awesome!!!!!!!

[@abhinavbhutada9b484]

Dabloo?

[@aLumpOfParticles]

when you changed the W to the X you wrote + instead of - btw check the answer via derivative :D

Came here just to say that hahahah

[@RoyEduworks]

kzbin.info/www/bejne/lYPagn99jtpor80

[@solidwaterslayer]

16 minute int just to fuk up at the end lol

[@nite5963]

Nearly as painful as if he’d have forgotten the + C

[@ayoubachak01]

there is a misstake sir !!! when you were substituting u in the w expression you puted a (+) insted of a (-) I love your work

[@GayAnnabeth]

now prove this by differentiating

[@justabunga1]

The 379th Hero you can split into products as tan^2(x)*tan(x)=(sec^2(x)-1)tan(x)=tan(x)sec^2(x)-tan(x). The integral will come out to be (tan(x))^2/2+ln(abs(cos(x)))+C. You might end up the answer in terms of secant as (sec(x))^2/2-ln(abs(sec(x)))+C since it end up a different constant and using properties of logarithms.

[@RoyEduworks]

kzbin.info/www/bejne/lYPagn99jtpor80

[@subhadeepsarkar5606]

Justin Lee lol

[@alhassanelkossei8481]

Please check the answer by differentiation.

[@tomatrix7525]

Those who are womdering why he used hyperbolic cot instead of hyperbolic tan, it is because with tan there is a limitation, namely |x| 1 so that limitation works here

[@seb538_]

Do the proof that sqrt(2) is irrational in under sqrt(2) minutes!

[@RoyEduworks]

kzbin.info/www/bejne/lYPagn99jtpor80

[@bharatipatel5076]

Can be done in 2 mins

[@mokouf3]

​@@bharatipatel5076 He means "square root of 2" minutes, shorter than 2 minutes.

[@chhromms.8138]

@@bharatipatel5076 it's about 84,85 seconds

[@ernestschoenmakers8181]

The other method is partial fraction decomposition where you divide u^4-2u^2+2 by u^2-au+b. After working this out you'll get: a=sqrt(2+2sqrt(2)) and b=sqrt(2).

[@blackpenredpen]

Did you pause & try? Also, check out my 100 integrals where I first mentioned Lars. kzbin.info/www/bejne/mpjQZWBpYq6padU

[@supriyajyoti22]

Math tricks for any competition plz....

[@healthygamer8192]

I don't understand how you keep so motivated doing math.

[@user-wu8yq1rb9t]

I love the hidden symmetry in this integral! Great ... I just enjoyed. Thank you so much ❣️

[@masonholcombe3327]

when you went from u to x when integrating, you did + instead of - for u! all good though, nice job!!:)

[@maskedman8368]

youtube must encourage these typeof educative channels

[@QuantumHistorian]

Way easier to start off with the substitution 1 + tan(x) = cos^2(u) and then do some simple trig until you can integrate by partial fraction. Gives a MUCH nicer answers too: 1/sqrt(2) ln[sqrt(2) + sqrt(1+tan(x)) / sqrt(2) - sqrt(1+tan(x))] + c

[@leponpon6935]

Keep making more of these amazing videos! The world needs more of this!

[@AmooBaktash]

A sign error happened at 16:34! Note that w = u - sqrt(2)/u.

[@edusoto91]

The polynomial P = u^4 + 2 u^2 + 2 factors over the reals (the only irreducible polynomials over R are linear or quadratic with negative discriminant). Here is a factorization P = (x^2 - a x + b) ( x^2 +a x + b) where a = sqrt(2sqrt 2 + 2) and b = sqrt 2 Once you compute this, the integral is straightforward.

[@ernestschoenmakers8181]

Yeah i did it this way, doing a long division by using u^2-au+b as the division factor.

[@rashmigupta6227]

Your change of face expression at 2:41 😂😂

[@JulesvanPhil]

Very nice video :-) But you did a mistake in the last line: when resubstituting the w you wrote a plus instead of a minus :D

[@itsviv1]

Thanks very much. I was stuck in integration of similar kind, and your videos did provide me with a solution.

[@genocider5868]

This is such a big brain math play to make two integrals this way

[@nchoosekmath]

Wow, this one does not even involve special function in the answer. But the steps were really long. Nice video!

[@leonardocampigli8320]

Noticed that too

[@Mernusify]

You could also write the answer with tanh^(-1). The differentiation for this is MONSTROUS (and fairly tedious), but it's doable.

[@Reallycoolguy1369]

I agree, I thought the choice between tanh^(-1) and coth^(-1) was based on the domain and since it's an indefinite integreal it's arbitrary. And maybe (1/2)ln|(1+x)/(1-x)| would be best since its domain includes all real numbers except +/- 1. It's not as fun as the inverse hyperbolic functions though

[@faizahmed7907]

7:37 Who else remembered Arthur??

[@rubikscuber1114]

U r a cool teacher🤟👍 Greetings from India

[@3420undertaker]

Do it for Lars!

[@blackpenredpen]

That would be a cool harshtag! #doitforLars

[@pablorestrepodiaz8520]

Please check the answer by diferentietion :)

[@martincurley8107]

Hi BlackPenRedPen! Didn't know if you noticed that in the inverse tan of your answer you switched plus for minus the answer should have been: 1/sqrt(2sqrt(2)-2)*arctan([1/sqrt(2sqrt(2)-2)](sqrt(1+tanx)-2/sqrt(1+tanx)))-1/sqrt(2sqrt(2)-2)*arctanh([1/sqrt(2sqrt(2)-2)](sqrt(1+tanx)+2/sqrt(1+tanx)))+C. But that was an awesome job. Integrals can be tricky, but you do an amazing job.

[@avdylkrasniqi4687]

14:03 should be minus. Much respect for you!

[@herlysqr1650]

Imagine fail just for that.

[@ansper1905]

13:20 can't we do the second integral using partial fractions?

[@egillandersson1780]

I did not find it. This was a very difficult one ! I prefer your aswer to that of wolframalpha, which goes needlessly to the complex world.

[@abhishekchakraborty2316]

I didn't see that the video was 16 mins long and i thought this seems like an easy integral. Three mins into my attempt and i realized this integral is out of my league. So i watched the whole video. I am fascinated by the simplicity with which you explained everything.

[@mehmeteminconkar2590]

Proce by differentiation def of derivative and epsilon delta

[@arpwable]

Why did you choose coth^-1 rather than tanh^-1? Both can be differentiated to the form you need, right?

[@chetansanap3398]

Plz solve integral of sin theta^2

[@justabunga1]

It’s hard to tell without the use of parentheses. If you meant the integral of sin(x^2), then it’s non-elementary but will come out as the answer of the sine Fresnel integral as S(x)+C. If you meant the other way as the integral of (sin(x))^2, then you have to change the identity as (1-cos(2x))/2. Doing so will get the answer to be x/2-sin(2x)/4+C.

[@RoyEduworks]

@@justabunga1 kzbin.info/www/bejne/lYPagn99jtpor80

[@justabunga1]

Niraj Roy :Motivational and Teacher you’re doing the infinite nested square root derivative. He already did that in the video to show work.

[@chetansanap3398]

@@RoyEduworks isko Hindi kaise smjega

[@chetansanap3398]

@@justabunga1 yeah,my doubt was first case sin(x^2),got it

[@isaacmedina9962]

what an insane integral!!!!

[@qu2k458]

from South Africa! Hugs fan of yours! hope to be as profiecient as you are someday! much love.

[@larissa8232]

I love how you go and just say "pause the video and try this first :)"

[@larissa8232]

have to say that i felt a little emotional at the beginning of the video, I'm also an engineer student and fighted cancer exactly 3 months ago, everything is fine now but anyways I'm in quarantine watching Big Integrals playlist AGAIN hah it's just sooo good. Cheers to Lars

[@egohicsum]

easy peasy lemon sqeezy for that one. holy moly

[@maskgamings19]

imagine in a test, this question comes and you forgot to write the C at the end.🙂

[@heldertvillegasjaramillo6343]

I did tried using integración by parts multiple times and some regular substitution in the middle, i got to a point were i had de integral of sqrt(tanθ), that's when i stopped because i know that isn't pretty (or is way too pretty, depending on how messed up you are).

[@siddharthamondal4346]

let 1+tanx = u^2 sec^2xdx = 2udu so the original integral becomes (u.2udu)/(1+u^2) then simplifying it we get 2du - 2du/(1+u^2) so we get 2u-2tan^-1(u) 2*(sqrt(1+tanx)) - 2*tan^-1(sqrt(1+tanx)) Won't this be easier? Or am I wrong somewhere?

[@backyard282]

Hey bprp, I checked your apparel, why doesn't your "for every ϵ > 0" t shirt have the rest of the limit definition on the back side? thanks :)

[@blackpenredpen]

It was requested by someone who just wanted that to be in the front. And the good thing is the cost is cheaper if it’s just one side print, too.

[@davisnganga6266]

Tricky one without knowing with substitution to use.

[@achrafbaiz5287]

W=U - note + sqrt2/u

[@coolguy4989]

Hey! you should check the answer by differentiating it! just to be sure you got it right

[@mr.cabbage4428]

🗣️DIVIDEE BY YOU SQUARE🔥

[@rarewc3uploader]

Hello blackpenredpen, May I ask, what is the limit of the expression "(W(x)/ln(x))^x" as x approaches a sideways 8 (infinity)?

[@IbraheemMatanmi]

the correct factor in the denorminator is {u-(1/u)}²-4

[@IbraheemMatanmi]

please kindly disregard my above comment i wasn't thinking very well thanks so much for sharing this video you're highly appreciated sir

[@alse72]

Check the answer via differentiation

[@bopaliyaharshal2399]

Mistake 14.04 -- minus ave

[@hassansameh8960]

Check by Differentiation😎😎😎

[@mattwik7467]

Love you ❤️❤️

[@DebarghyaBasak_hi_every_one]

Hey, Blackpenredpen! I was trying to find the area under the graph of y=|[x^3]| from x=-2 to x=3. [.]- Greatest Integer Function and |.|-Modulus. Result involves a series of summation of the cube roots of first n integers. How do I go about solving this? Because the answer in my text book is given in the form of a numerical value. I really enjoy all of your integral videos, btw!

[@JUANAMPIE]

Eso si es de gánster, muy buen video siempre es genial ver el nivel hard de estos ejercicios

[@وريانهاد]

In 13:14 you make a little mistake . You should write tanh-1(v) not coth-1(v) !!! I'm right ???

[@infamous992]

very helpful! thx a lot!

[@absolutezero9874]

Thank you for your videos. But did you see my question on integration of x^(x^2)? Is it integrable? Thank you

[@blackpenredpen]

I don’t think it has a nice answer

[@prollysine]

Hi bprp, w=(u-(sqrt2/u)), the good thing is, the arctg() argument is not a typo because you wrote a + sign there?

[@pinchus2714]

Check answer via derivative.

[@dork8656]

I am confused on how hyperbolic integrals work. I instead use (1/2a)log((x + a)/(x - a)) on 1/(x^2 - a^2)

[@nvapisces7011]

Btw it is ln not log. I didnt learn hyperbolic functions yet but based on what i know about them, they have the same shape as ln. However, they have different domains. U just use them because they are more convenient instead of spending time doing more trig sub or partial fractions

[@serbanhoban1517]

@@nvapisces7011 You can say log if you want. Please read this first paragraph here on Wikipedia en.m.wikipedia.org/wiki/Natural_logarithm#:~:text=The%20natural%20logarithm%20of%20a,is%20implicit%2C%20simply%20log%20x.

[@watsonjunior85]

Check the answer

[@hafeezullahrahoojo9972]

Sir, have you uploaded the video of riemann sum proof??

[@Simplement724]

Whats the best way to write the integral of tan^3(x) or are they no better writings of the answer?

[@angelmendez-rivera351]

the chessmate tan(x)^3 = tan(x)·tan(x)^2 = tan(x)·[sec(x)^2 - 1]. Use linearity, and now you can consider the integrals of tan(x)sec(x)^2 and the integral of tan(x). The integral of tan(x) is given by ln|sec(x)| + C(x), where C(x) is an arbitrary piecewise step function with discontinuities whenever x = πn + π/2 for some integer n. The integral of tan(x)sec(x)^2 can be calculated by letting u = tan(x) => du = sec(x)^2 dx, which gives the integral of u with respect to u, which is equal to u^2/2 + C, or tan(x)^2/2 + C. Adding the integrals results in tan(x)^2/2 - ln|sec(x)| + C(x). Is there a better way to write the antiderivative? Other than switching tan(x)^2 for sec(x)^2, which is permitted because they differ by a constant, no, there is not a better way, as far as I am concerned.

[@Simplement724]

@@angelmendez-rivera351 alright thank you i wrote it with sec instead of tan but i know some people said they wrote it with ln(cos) instead of ln(sec) wanted To know if it was any better or worse lol

[@ernestschoenmakers8181]

@@Simplement724 Well lnIcos(x)I = -lnIsec(x)I

[@ayoobbhat9180]

Becoz of u i fall in love in MATHS from KASHMIR

[@O_Capivara]

You should do the diferenciation

[@GreenMeansGOF]

Check the derivative!!!

[@ppereztorres]

Please make a video checking this via derivative. Do it for Lars

[@neilgerace355]

13:11 backwards writing For The Win

[@Sg190th]

more substitutions than the pokemon move

[@jrli5952]

I paused the video To take my shoes off

[@akshatahuja2523]

Really I had done EXACTLY same

[@mokouf3]

Really similar to integrating sqrt(tan(x)), making algebraic twin again!

[@jbitddpggp]

on the 4th line do we get inverse cot instead of natural logarithm (the 2nd integral)? maybe i am confused but the derivative of inverse cot is -1/(1+x^2) not 1/(1-x^2)

[@nicolasgoubin]

Nice Lars ! Gg for having fought against that shitty cancer !!! This will turn into an old bad memory now ;)

[@martindolak2293]

It was great and understandable although I would recommend not using the inverse hyperbolic functions. Their usage is severely limited and they can be rewritten using natural logs which seems to be a better idea, especially considering that calc 2 students dont really use those things. But apart from that, it was a pleasant experience as always :)

[@itsviv1]

Logarithmic term as solution represents only the real part of the solution, whereas inverse hyperbolic function gives complete solution without neglecting imaginary terms. For the sake of completeness, I think it's better to write ans in inverse hyperbolic function.

[@edwardhudson815]

@@itsviv1 dont absolute value signs make it better

[@jaskiraatshah9445]

Ohhmmgggg please then differentiate the answer back to the question 😭 That would be lovely 😅😅😅

[@sujalkoirala3675]

How long should I try to solve an integral before giving up?

[@mbarekouazragh9982]

Error in minute 12 and 5 sec ! Taking out 1/sqrt(2 sqrt(2)-2) should be 1/(2 sqrt(2)-2) same for 1/(2 sqrt(2)+2)

[@brucefrizzell4221]

I like Scott Joplin. Who is the musician ?

[@mathswithpana]

My bro I see a mistake for w substitution . Put - not +. Or maybe I can't see properly but I think you forgot to subtract not addition

[@arghadipnandi5012]

Sir where you put the main function of 'w' there will be a minus sign

[@cecilhenry9908]

Brutal!!!!!!!!

[@chillforever6164]

# Teacher of Chirayu Jain #!

[@SartajKhan-jg3nz]

'It wasnt painful at all...' - BPRP

[@edwardhudson815]

lost all hope bro used the formula for the derivate of the inverse hyperbolic cotangent

[@zyadchoukri1654]

Amazing

[@user-fy2hp]

I think that you have a mistake, in the first part of the answer we have sqrt(1+tan(x))-sqrt(2÷(1+tan(x))) not a positive sign

[@wayneosaur]

That is a hairy answer... Does it ever make sense to convert the integrand to a Taylor series and just integrate the resulting (infinite degree) polynomial?

[@jakedanko7226]

What could dx/dy look like or mean in general?

[@bisakhbarman7344]

What's the integral of (x^(1/(1+x)) - x^(1/(1-x)))dx?

[@joluju2375]

Why are the graphs of sqrt(1+tan(x)) and 2x^2/(x^4 - 2x^2 + 2) so different ?They have the same integral.

[@ExTremeFlipper]

Joluju 😂 he did a u sub, thats why they are different

[@elementsslothdragon3216]

Quick question couldn’t you just substitute tan(x) for tan^2(θ). This would make this equation way more simple.

[@edwardhudson815]

you get sectheta as the the main thing but the dx is the problem

[@simonesora5573]

Why we can't made this integral by part and then we can call 1+tanx=u with du=dx/cos^2(x) I don't see why it doesn't work, i mean I find it much easier to do, so it must be wrong😅...(in the integration by part i considered 1 equals g'(x), and I derived sqrt(1+tanx)...). Any help?

[@angelmendez-rivera351]

Integration by parts does not work because you are forgetting the square root of the integrand. In this case, the function being differentiated is sqrt(1 + tan(x)), not 1 + tan(x). This yields a much different integrand than what you expected.

[@simonesora5573]

@@angelmendez-rivera351 I took that in mind, you get xsqrt(1+tanx) - 1/2 integral of (sec^2/sqrt(1+tanx))dx wich you can do by sustitution, ain't it ?

[@simonesora5573]

@@angelmendez-rivera351 Sorry, just found the error, i forgot an x in the integration by part, my mistake 🙌🏻

[@sirajkhalil6924]

Next: integral of cuberoot(1+tanx)

[@robertherbert8419]

what about its derivative ?

[@erikkonstas]

Let's see... so, we have f(x) = sqrt(1 + tan(x)), therefore f'(x) = sec(x)^2 / (2 * sqrt(1 + tan(x))). :P

[@MelonMediaMedia]

If only there was a tiny 2 next to the tan

[@fedefubbi5349]

can you do integral of sin^3x/(cos^3x+sin^3x) from 0 to pi/2? thanks sir in advance

[@mislavplavac6641]

The integral of f(x) from a to b is the same as the integral of f(a+b-x) from a to b. Could someone please correct me on this if it isn't always the case. But in this case it is so we'll let: I=integral from 0 to π/2 of sin³x/(sin³x+cos³x) and we'll also plugin 0+π/2-x into it so we get I=integral from 0 to π/2 of sin³(π/2-x) /(sin³(π/2-x) +cos³(π/2-x)). We notice that sin(π/2-x) =cos(x) and that cos(π/2-x) =sin(x) So we have sum the two I from which we get: 2I=integral from 0 to π/2 of (sin³x+cos³x) /(sin³x+cos³x) or the integral of 1dx So we get 2I = x [0 to π/2] or that 2I=π/2 in the end getting that the integral is π/4

[@fedefubbi5349]

@@mislavplavac6641 thank you!!

[@angelmendez-rivera351]

Goran Plavac Yes.

[@angelmendez-rivera351]

Fede Fubbi In general, for any arbitrary function f that is well defined on the real interval (0, 1), the integral from 0 to π/2 of f[sin(x)]/(f[sin(x)] + f[cos(x)]) with respect to x is equal to π/4. This uses the similar reasoning as the comments above. You can exploit the fact that sin(x) = cos(π/2 - x) and cos(x) = sin(π/2 - x) to prove this. In the special case of your integral, f(x) = x^3, but this could have been done with any other function with domain (0, 1). Even a discontinuous function would have done. There is also somewhat of similar trick with tan(x) and cot(x). For example, consider any function f : R -> R with the properties f(x)f(y) = f(xy), f(x^n) = f(x)^n, and f(1) = 1. Then the integral from 0 to π/2 of 1/(1 + f[tan(x)]) is the same as the integral from 0 to π/2 of 1/(1 + f[cot(x)]). Since cot(x), you can multiply numerator and denominator by f[tan(x)] to get f[tan(x)]/(f(1) + f[tan(x)]), and since f(1) = 1, f[tan(x)]/(1 + f[tan(x)]). Notice that this is identical to our original integral I, so if we add it to our original integral, then on the one hand, we have 2I, and on the other hand, we have the integral from 0 to π/2 of (1 + f[tan(x)])/(1 + f[tan(x)]) with respect to x, which is trivially equal to π/2. Thus, I = π/4.

[@fedefubbi5349]

@@angelmendez-rivera351 👌👌👌