Integral of sec(x) but without that trick!

[@blackpenredpen]

Red or black?

[@salociiin42]

Black

[@lucassssss4575]

Red

[@nischalada8108]

blackpenredpen black

[@buckeye49]

Green

[@pichass9337]

pen

[@pbj4184]

Ah this "mechanical" evaluation is much better than the "magical" original because this shows all the relevant motivation and isn't too hard either!

[@pablorestrepodiaz8520]

wow u love this video , its like magic when you demostrate that those 2 expression are the same

[@blackpenredpen]

: ) thanks

[@itszeen7855]

Love it when there is a new upload when I get home from school!

[@changenow9228]

100 differential equations please...thank you

[@ishaanivaturi2387]

This is so nice! I was curious about the same thing last year and ran through this same derivation too! Glad to know you're bringing it to everyone :)

[@blackpenredpen]

Yay glad that you like it!

[@VibingMath]

Thanks man! This technique is also applicable to integrate cosec x and it's more natural to do in this way rather than multiplying a complicated expression in the first place! I will let my students to watch this so they dont need to memorize that trick! Yay!

[@blackpenredpen]

Mak Vinci hahaha yea!! My students and even viewers get mad when they see that trick

[@VibingMath]

@@blackpenredpen I also got mad when first time seeing that trick lol

[@japotillor]

It's not an intuitive trick, I teach both. I also like using half angle forms too.

[@MrAznBoyWins]

you have a real heart of a teacher, love to see it

[@blackpenredpen]

Thank you!! Btw I have been teaching for years already : )

[@osmankanu1112]

The best sex in wold .

[@Nonsense_2021]

I'm in my first calculus class, and just today I was wondering how to integrate that thing. Coincidence or destiny? Great vid as always. Greetings from Honduras :D

[@blackpenredpen]

Fernando Sarmiento Hahahah must be destiny!!!

[@blackpenredpen]

Fernando Sarmiento and thank you!

[@Avaflyne]

¿A quien le importa de dónde saludes? Pobre pendejito ilegal

[@AloncraftMC]

@@Avaflyne mmm mejor calla. La proxima vez no seas tan toxico por favor.

[@Avaflyne]

@@AloncraftMC Mejor vete a ctpm

[@japotillor]

A way for me to remember it Is that one multiplies top and bottom by 1 which is (sec x)^2 - (tan x)^2, which factors to (sec x + tan x) (sec x - tan x) Eliminating the (sec x - tan x) terms, you get sec x + tan x :)

[@angelmendez-rivera351]

Joe Potillor Yes, this works perfectly fine too.

[@diegofnndiego9938]

I love how you explain every step

[@yashovardhandubey5252]

This was uploaded at 3 am in the morning in my country...... Still watched it

[@GreenMeansGOF]

I think it all comes down to already knowing the answer. The second way is more intuitive but the first way is faster.

[@blackpenredpen]

Yup!! Agree!!

[@jeffreys8199]

im new to this channel and i love how he smiles after doing a step like that shows how he loves maths😀 and boys girls find these kind of boys attractive bcz they like the men with dedication (kinda motivation for the boys)😂

[@jeffreys8199]

and somebody say what does the channel name mean??

[@JohnSmith-rf1tx]

CHEATER!!! :) You claimed you would do the integration without using the the trick of multiplying by (sec x + tan x)/(sec x + tan x) but that's exactly what you did in the process of converting your answer to the standard form. You may have dressed it up in different clothes, but it's still there! At 2:34, you first multiply by (cos x)/(cos x). Well, that's the same as multiplying top and bottom by (1/cos x). Then, at 8:20, you use (1 + sin x)/(1 + sin x). When you multiply those two together, you get exactly the (sec x + tan x)/(sec x + tan x) you claimed you wouldn't use.

[@ishaanivaturi2387]

It is more intuitive and logical than outright multiplying by secx + tanx, which is what most people do. I was mad first seeing this integral because I wouldn't have known to do that, so I ran through this same integration which is much more intuitive.

[@ssdd9911]

but the integration IS done without the multiplication

[@Metalhammer1993]

@@ishaanivaturi2387 yup i never converted it to standard result when i tried it myself. Did it several times and always got the nonstandars result and thought I'm too stupid xD

[@Ksh3104]

Wow that was one hell of a point u made there My opinion on this is that he technically finished integrating without 1+sinx and used 1+sinx to just make a point that those results are same

[@iabervon]

Looked at that way, the trick is that you do the problem the verbose and well-motivated way, and then combine everything you turned out to need into one mysterious step.

[@krukowstudios3686]

Nice one. I think it is a bit sad that we do not really see cosecant and cotangent that often. I would love to see more of them :D

[@blackpenredpen]

Krukow Studios thanks!! In fact chase did it two years ago on my channel already : ) kzbin.info/www/bejne/jYqrdJ2crb56oLM

[@krukowstudios3686]

blackpenredpen Yea, I have seen that :D. I was actually thinking about things like differential equations, usage og their identities, or other problems that involve those two :)

[@dr.rahulgupta7573]

Excellent presentation ! 👌 black pen - red pen -green pen.

[@maliciousmarka]

Great explanation on this video! Thanks for helping me understand :)

[@thomasarch5952]

Actually once you get the integral into the form of ((cos x ) / ( 1 - (sin x)^2 ) ) dx you are done as the answer is artanh (sin x) + C! Just use the hyperbolic tangent function. Finshed!! No fooling around with partial fractions.

[@hmm1778]

Do you noticed something is wrong? In case you didn't Derivative of arctan is 1/1+x^2 not 1/1-x^2

[@ssdd9911]

@@hmm1778 he is saying the hyperbolic version of arctan

[@ssdd9911]

but the hyperbolic function does not have the absolute value so it is less rigid

[@shambosaha9727]

This video really made my day

[@ieqvilamaria619]

Bprp Thanks for yours videos, them are awesome so much, I am your fan!! :)

[@ernestschoenmakers8181]

I did it the other way around, i equated both solutions and here are the steps: lnIsec(x)+tan(x)I = (1/2)*lnI(1+sin(x))/(1-sin(x))I, now i put 1/2 in front of the left-hand term and squared it: (1/2)*ln(sec(x)+tan(x))^2=(1/2)*ln(1+sin(u)/(1-sin(u)) so the 2 terms within ln are equal: (sec(u)+tan(u))^2=(1+sin(x)/(1-sin(x)) now i expand the left-hand term and this should be equal to the right-hand term of the equality sign: sec^2(u)+tan^2(u)+2*sec(u)*tan(u)= (1/cos(x))^2+(sin(x)/cos(x))^2+2*sin(x)/cos^2(x): the denominators are the same so: (1+2*sin^2(x)+2*sin(x))/cos^2(x)= (1+sin(x))^2/(1-sin^2(x))= (1+sin(x))^2/(1+sin(x))(1-sin(x))= (1+sin(x))/(1-sin(x)) so left-hand side and right-hand side are equal.

[@CornishMiner]

I miss the "Black Pen Red Pen YAY!" intros.

[@edu10th47]

I love this video!!! I always hated that trick!

[@Peter_1986]

blackpenredpen is one of the most likeable guys I know of on KZbin.

[@blackpenredpen]

Laurelindo awww thank you!

[@thomasarch551]

So using the inverse hyperbolic arc tan function avoids have to mess with paartial fractions and gives a perfectly correct answer too.

[@allaincumming6313]

Actually, this is a much more didactic/heuristic way to solve ∫ sec(x)dx, and the tricky path to the classic ln|sec(x)+tan(x)| is even more "findable" and logical, by just applying a clever algebra. That trick never satisfied me, but it and other tricks taught me the power of the convenient 1's and 0's.

[@anis786]

you could also start with u=sinx, same process different order

[@adityakumarvishwakarma7282]

Sir make a video on e! e factorial,I have been searching for it since many days but I have not found it

[@marcuslautier5304]

If you use the substitution u=tanx, du=sec^2xdx and you arrive at the integrand du/secx. From there, manipulate the initial substitution u=tanx x to obtain secx=sqrt(1+u^2) and substitute to arrive at the integrand du/(sqrt(1+u^2)) which becomes ln(u+sqrt(1+u^2)) +C by the table of standard integrals. Substituting u for tanx into the indefinite solution arrives at the required result. No tricks needed 👍

[@marcuslautier5304]

Similarly for the integral of cosecx, let u=cotx

[@Sg190th]

I wish my professor did that in Calc 1 but we had a bit of a delay due to the hurricane so he probably didn't have time to explain how to do the integral. Now I can't wait for you to do integral of Sec^2(x) with Weierstrass substitution!

[@stephenbeck7222]

Did you cover integrations with partial fractions in calculus 1? Normally I’ve seen that as a calc 2 topic, and most students don’t even learn how to decompose partial fractions until they get to it in calculus (some algebra 2 or precalculus books have it but it never gets taught).

[@Sg190th]

Oh yeah partial fractions... yeah you're right I had learned that in calc 2.

[@logiciananimal]

Guess: the traditional method with the sec + tan multiplication does not require partial fractions, so is easier. (Stewart's book at least from years ago doesn't do PF until later.)

[@c_ruizl]

I don't speak English very well, but I still see your videos, greetings from Mexico.

[@liltop5930]

I’m only in 9th grade and we’re doing solving for x on Both sides only( which is easy) but this is so complicated omg

[@matthewbradley4644]

You should watch 3blue1brown's essence of calculus, i really like him. Did you even learn about logarithms yet?

[@gurvishwassingh5542]

Sir what about 1/cosx=sinxtanx+cosx and Integrating sinxtanx by parts? I know the answer it would be same but still fun way

[@gurvishwassingh5542]

I have some interesting question with good solutions if you want i can share them.

[@carultch]

For integrating sin(x)*tan(x) by parts: Let tan(x) be differentiated and sin(x) be integrated. S _ _ _ _ D _ _ _ _ I + _ _ _ tan(x) _ _ sin(x) - _ _ _ sec(x)^2 _ _ -cos(x) Connect signs with D-column entries, and then multiply by 1 term down in the I-column. For the final row, integrate across the row. tan(x)*-cos(x) + integral sec(x)^2 * cos(x) dx Simplify: -sin(x) + integral sec(x) dx We've already solved integral of secant of x, so we can construct the solution: ln(|sec(x) + tan(x)|) - sin(x) + C

[@aviator_2401]

I am from India . Lol I had those formulas written in my book . Search the book ML Aggarwal . All formulas are there . Integration of Sec x dx = log| sec x + tan x | + C or log | tan (π/4 + x/2)|+ C Integration of tan x dx = - log|cos x| + C Integration of cot x dx = log| sin x| + C Integration of cosec x dx = log| cosec x - cot x| + C or log | tan x/2| + C Nice video btw dude !

[@joananathan6571]

Thank you for the videos! It's helping me a lot :)

[@KamleshSharma-gc8jv]

at 8:24 cant we use the identity sin^2 x/2 + sin^2 x/2 + 2(sin x/2 cos x/2) in the numerator and sin^2 x/2 + sin^2 x/2 - 2(sin x/2 cos x/2) in the denominator?

[@pichass9337]

Lol, just went over this today in my class

[@JSSTyger]

Yes but did your teacher use different colored markers?

[@pichass9337]

@@JSSTyger no. Just a blue one...

[@blackpenredpen]

Nice!!!

[@holyshit922]

People who dont use secant also have trick Int(1/cosx,x)=Int(1/sin(Pi/2+x),x)= Int(1/(2sin(Pi/4+x/2)cos(Pi/4+x/2)),x)=Int(1/(2tan(Pi/4+x/2)cos^2(Pi/4+x/2)),x)

[@thomasarch551]

the integral of 1/(1-x^2) is artanh (x) + C

[@Jhev1000]

Another nice way: a Weierstrass substitution!

[@blackpenredpen]

Jhevon Smith oh yea! For some reason that isn’t a popular method but I think it’s very cool!

[@Jhev1000]

@@blackpenredpen what's cool is you still reply to comments like this!

[@blackpenredpen]

Thanks! I try to reply to as many comments as possible!

[@vaibhavbhardwaj1886]

Love from India 🇮🇳

[@dychaneng]

Hello Sir, Please help to to figure out of this sum of sequence: S_n=1/2+1/8+1/24+....+1/(n*2^n). Thank you

[@ericzhang634]

this guy is a god

[@AhmedAli-co7ob]

Hello, you really help me learn calculus and I really love your videos! do you have one of those 6 hour videos for Calculus 2 series? Would really love to see that.

[@blackpenredpen]

If you are talking about convergence/divergence of series, then it's here kzbin.info/www/bejne/oIXYhXiZrNuehpY 100 series!

[@AhmedAli-co7ob]

blackpenredpen you’re actually amazing and I wish I can sub to you 10000000 times! You don’t know how much this will help me. This is exactly what i am looking for :) do you also have something to find what a series converges to and not just if it converges or diverges?

[@blackpenredpen]

Ahmed Ali yea they are in there as well. Check out the pdf. Link in description

[@AhmedAli-co7ob]

blackpenredpen perfect! Thank you so much man. Keep up the good work :)

[@anzarrabbani3766]

4:05 is it possible to make this -arctanh(u)?

[@akshatahuja2523]

RED

[@عشتارللرياضيات]

lived your hand

[@henrylin65]

x^(x+1)+1=(x+1)^x Solve for x. I know the answer is 0,1,2, how do you solve that?

[@ernestschoenmakers8181]

I tried a lot of ways to solve this one like factoring out x^x and squaring both sides but it didn't lead to a solution, maybe anyone else have a suggestion?

[@fantiscious]

I don't know how to solve algebraically, but it seems very related to Catalan's conjecture, which is already proved. Maybe you could show that there's no other solutions than those?

[@adedejiyemi8268]

U are wow sir,good, perfect.. And get minus wen solving into partial fraction ?

[@rarebeeph1783]

i stumbled upon this method in reverse while trying to integrate 1/(4-x^2) dx. my mind went to trig sub first, stuck me at sec(x) dx, and then i realized i could use partial fractions, which then implied that i could have used partial fractions to integrate sec(x) by reverse trig sub to 1/(1-x^2).

[@Rajuramsirvi]

lntan|π/4+x/2|+c explains sir

[@ivankaticos]

Pls do integral battles!!!

[@shiveshkumar1965]

Lim n tends to infinity nx^n=? for x

[@lordcezar4657]

I did this on a test in school 😂. Didnt know about the trick

[@moshebr-c9q]

I have seen this integral before no secret there....in the book

[@danielsimiyu774]

Hey fam greetings from Kenya......... am asking how do i know that i have to consider one to be D and the other one to be I

[@wryanihad]

There is another one Easy but wolfram doesnt agree with it!!! May be you know the reason. The techniq is Put x=arcsec(t) dx=dt/(t)sqr(t²-1) Intg(secx)=intg(dt/sqr(t²-1)=arc(cosh)(secx) Can you make an vedio on this? I checked many app that doesnt agree with my solution Where im getting wrong?

[@mubarakali657]

It's long method but easy

[@jakub_pham]

great stuff

[@AnthonyRobledo-ce8jl]

Could've used this during my test earlier 😢

[@theelectro15]

FINALLY, THANKS

[@TouhidulIslam]

Hello Blackpenredpen! I'm one of your students from Bangladesh. Can you please inform me that how can I get your hoodie?

[@blackpenredpen]

Touhidul Islam hi there! You can see the link in the description thanks

[@feelthereal5742]

Hello. I'm also from Bangladesh. I love black pen red pen's videos.

[@TheAnhard]

Amazing, thank you!

[@aannurwahidi5497]

Infinite and definite series pls

[@TheInterestingInformer]

Can someone explain the 1/2 over 1+u thing he did?

[@YorangeJuice]

That was magical

[@einsteingonzalez4336]

I remember that there was a suspense tone. The video was edited, wasn't it?

[@gustavorc25]

Very original, nice 😎👌

[@Sir_Isaac_Newton_]

the organic chemistry tutor is seething

[@abdulazizmemesh2791]

Can you make a video about descartes rule of signs?

[@bahaloicperrial8964]

It is the same as 2tanh inverse (tan(x/2)) ur method is just to long but it is good

[@layer1087]

My question is why can't we say that the integral is just not ln|cosx| +C

[@Nightmare-ps9sk]

I wanted to buy you a derivate pullover but there is only the integrale one :(

[@GayAnnabeth]

kinda interested in stuff about cosecant and cotangent...

[@DengGum]

😮😮That's real I have been looking 4

[@shlomozerbib388]

@ 10:00 how can you put out the square which is within the absolute values?

[@blackpenredpen]

shlomo zerbib Abs(x^2)=abs(x*x)=abs(x)abs(x)=(abs(x))^2

[@shlomozerbib388]

@@blackpenredpen ok

[@TheStrafendestroy]

What was the method to get the constants when you separated the 1-u^2

[@FrostDirt]

It's called Partial Fraction Decomposition

[@carultch]

Heaviside coverup. The idea is that you cover up the denominator of the term you're interested in, and plug in the value of the variable that makes the covered up term equal to zero, in to the rest of the expression. Evaluating that, will tell you the coefficient that belongs on top of that term. Here's why it works. Consider a general case of: (c*x + d)/[(x - p)*(x - q)], which we'd like to equate to A/(x - p) + B/(x - q) (c*x + d)/[(x - p)*(x - q)] = A/(x - p) + B/(x - q) To find A, multiply through by (x - p), to partially clear fractions: (c*x + d)/(x - q) = A*(x - p)/(x - p) + B*(x - p)/(x - q) Take the limit as x approaches p. This makes the B-term disappear, since (x - p) = 0, and (x - q) will equal something else, so we aren't dividing by zero there. This is a huge advantage, because now we can solve for A independently of solving for B. However, there appears to be a problem with the A-term, until you realize that it's just a removable discontinuity, since the top and bottom are identical terms that cancel. Once we remove the removable discontinuity, we now have a more direct way to solve for A: A = (c*p + d)/(p - q) Likewise, a similar formula, solves for B: B = (c*q + d)/(q - p)

[@kingbeauregard]

Two questions: 1) Why does anyone even teach the traditional crappy proof when this much better proof is available? 2) Why didn't my calculus education include teaching the importance of factoring fractions?

[@stephenbeck7222]

The fraction decomposition is an ‘intermediate’ integration technique typically reserved for calc 2 (along with integration by parts and trig subs). The typical method is kind of a trick but can be taught using ‘beginner’ integration methods (simple u-subs and recognizable anti-derivatives). It’s nice to have a catalog of integrals of all simple functions with known antiderivatives as early as possible, so this trick method is shown in calc 1. By the way, the trick is not much more advanced than the ‘trick’ needed to prove the product rule for derivatives, which is quite important.

[@kingbeauregard]

@@stephenbeck7222 Fraction decomposition seems like something they should have taught back in algebra, but didn't. I'm pretty sure I was paying attention, but maybe I forgot. Even so, I never saw partial fractions used in any of my calculus classes (both high school and college). My first calculus teacher emphasized the point over and over that, in any problem, there's typically only one line of calculus and the rest is algebra, and this seems like it would have been a really good thing to utilize in the algebra. Shrug.

[@zaidhamdaan1905]

Sir , can you help me with integrating a problem I want to see how u approach to the problem . int (0 to 1/2 ) 1/{(1+x^2)√(1-x^2)} dx. Note : int = INTEGRAL SIGN AND 0 IS LOWER BOUND AND 1/2 IS UPPER IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED

[@puremaths2444]

Kindly sir From this ln(1+u)+ln(1-u) ,I don't know why ur finding derivative of 1-u and changing the sign to negative ie. ln(1+u)-(1-u)

[@tonymunene1592]

Its because of the chain rule, the derivative of ln[1+u] is done as such: d/du ln A where A=1+u = 1/A * derivative of 1+ u which is just 1 because u is positive, and the derivative of 1 is 0 so 0+1 =1...I guess you can take it from there

[@AeyZeiNS]

Why do you do conjugate?

[@nimmira]

:) someone was peeking thru the door?

[@coniemich]

Hi! I still don’t understand why you put 1/2 in the numerator

[@carultch]

You can use the Heaviside coverup method to more directly get the answer, but I'll show the first principals for it. Given: 1/(1 - u^2) Factor bottom: (1 - u) * (1 + u) Set up partial fractions, with a constant on top of each linear denominator: A/(1 - u) + B/(1 + u) Equate to the original: 1/[(1- u)*(1 + u)] = A/(1 - u) + B/(1 + u) Multiply to clear fractions and expand: 1 = A*(1 + u) + B*(1 - u) 1 = (A - B)*u + A + B There are no u-terms on the left, so this means the coefficient on the u-term on the right must be zero. The remaining A+B must add up to 1. A - B = 0 A + B = 1 Add up equations to get: 2*A = 1, which implies A = 1/2 Plug back in to solve for B, and get B = 1/2 Thus: 1/(1 - u^2) = 1/2/(1 - u) + 1/2/(1 + u) Alternatively, the Heaviside coverup method allows you to plug in u = 1, and cover-up (1 - u), as a way to directly find A. Likewise, covering up (1 + u), and letting u = -1, will allow you to directly find B

[@vena7429]

4:34 Can anyone tell me in detail how 1/2 was coming out?

[@domanicmarcus2176]

Can you do sech(x) in the same way? I tried that and got stuck because the denominator ended up being 1+u^2 and I did not know how to factor it. Can you do a video on how to integrate sech(x) without the trick, please?

[@carultch]

Given integral sech(x) dx Rewrite it per its original definition: integral 2/(e^x + e^(-x)) dx Multiply top and bottom by e^x: integral 2*e^x/(e^(2*x) + 1) dx Let u = e^x, thus du = e^x dx. Rewrite in the u-world. integral 2/(u^2 + 1) du In the u-world, this integrates to: 2*arctan(u) + C Translate back to the x-world by replacing u with e^x: integral sech(x) dx = 2*arctan(e^x) + C

[@martinarturoarellanoreyes7467]

Someone can explain me the minus betwen ln on min 6:00

[@circlonianmapper]

If you find the derivative of that expression, I think the answer will become clear. When you take the derivative of ln|1-u| you get -1/1-u. Note that in the integral, it is +1/1-u, not -1/1-u. The negative sign that he put cancels out the negative sign that you get when you take the derivative of ln|1-u|

[@ekueh]

Reverse chenlu? Or lu chen

[@Pspspsss]

But i thought that we cant do this trick bc it changes the value of x

[@ratulbanerjee8456]

So another trick instead the old one

[@ultimatedude5686]

why not tanh^-1(sin x)

[@hhht7672]

Sorry I’m late but @3:55 couldn’t you just say that’s arctanh(u) ?

[@carultch]

We still have to recall the value of u, since we want the solution in the x-world, and not the u-world.

[@Patapom3]

Amazing!

[@brunobautista6316]

1/(1-x²) isn't it arc-coth(x)?

[@brunobautista6316]

I had checled and yes, it could be done by hiperbolic substitution. When you hit that result, you actually are able to say "Answer is Arc-coth(u) + C" and then take that expresion to the "x world" being it "Arc-coth(sin(x)) + C" If you derive that result, yo actually end up with cos(x)/(1-sin²(x)) but that may be simplified into sec(x), wich closes this as a correct interpretation. Edit: put a "minus" were it not belong to be unintentionally

[@sagnik_pal]

why cant you take 1/cos x. please do itttttttttttttttttttttttttttt

[@Pspspsss]

Bc u cant integrate 1/cos x

[@fernandogaray1681]

Whats the idea behind partial fractions? I never figured out how it works and for example, when u have a squared termn in the denominator u had to put a linear expresion on top

[@angelmendez-rivera351]

Fernando Garay The idea of partial fractions is to write any rational expression on x as a polynomial on x plus a sum of rational expressions on x, where these rational expressions on x have numerators of degree at least 1 less than the denominators. This is desired because any expression of this form can be "trivially" integrated. For example, (Ax + B)/(ax^2 + bx + c) can always be integrated provided a, b, c are not all 0, regardless of whether the denominator can be factored or not. Alternatively, you can make partial fraction decomposition simpler if you allow complex numbers. In this case, any rational expression on x can be written as a sum of expressions of the form A/(x + b), where A and b are complex numbers. This is desirable, because it means the integral is a sum of logarithms, which you can then recombine into logarithms of products where you can get the logarithm of a rational expression as the result. It's systematic, which is why it is optimal.

[@carultch]

The idea is that you can rewrite a big complicated ratio of polynomials, in a form that is more calculus-friendly, as individual fractions of each denominator. Just as you add 1/4 + 1/3 to get 7/12, partial fractions does this process in reverse, so you can get the original sum of fractions that adds up to your given expression. Constant-over-linear terms, can integrate with natural log, while linear-over-quadratic terms, can be integrated with integration by completing the square, and using inverse tangent. In a general sense, you always have a degree (n-1) polynomial on top, given a denominator polynomial of degree n. So this is how linear factors of the original denominator, become constant-over-linear terms of the expansion. Also how irreducible quadratics become linear-over-quadratic. Higher degree denominators aren't very useful, so it is uncommon to do this with cubic denominators or anything beyond. Usually, you'd reduce higher degree denominators to combinations of linear and quadratic factors first.

[@mini-cafetos9410]

Wow that's so cool I always thought that was the only way to solve that integral so I assume something similar can be done to cscx?

[@Kanbei11]

Try it for yourself, that's the beauty of maths

[@angelmendez-rivera351]

Mini Cafetos Yes, except with sine.

[@blackpenredpen]

Mini Cafetos yes!!

[@khushalreger]

Hlooooo blackpenredpen i am from INDIA

[@blackpenredpen]

Hello!!

[@deovigildo11]

Good

[@散华-l9m]

it's totally amazing😂