Why the integral of csc x has 'different' solutions

In this video, I showed how to integrate cscx and then use trig identities to obtain other forms of the antiderivative.

[@utuberaj60]

Nice one Sir. Even your illustrious name sake (I.Newton) would give you a 👍. That said, I am more familiar with the standard answer = ln |sec x+tan x| that we learnt by another smart algebraic trick i.e., multiply and divide the expression by (sec x + tan x), that makes the integration as simple as can be. Using this as the starting point, you can get the answers in terms of sin x and cos x and also tan (x/2) , which I derived and is quite fun doing. Why don't you make a similar video for int sec x? The answers are all the more interesting.

[@patstanton1885]

Wow! This guy is really good. I was able to follow him all the way through without getting hung up. Learned some new tricks too. Thanks.

[@patstanton1885]

Wow! This guy is really good. I was sable to

[@bsam4860]

Woow! This is very very self explanatory! God job sir ! ....... but sir why did yu decided to let u=cosx why not the whole 1-cos^2x is it because if yu dy 1 its becomes 0??? Thanks sir

[@PrimeNewtons]

I needed the derivative to be a scalar multiple of sinx. If u was 1-cos^2x, then dy/ dx would be 2cosxsinx which is not what I need.

[@user-ti9xi1nt1i]

Dear sir I wish to be like you in teaching mathematics

[@Nutshell_Mathematica]

These are cool But can u do bit advanced calculus problems

[@richardbraakman7469]

I don’t really follow how the ln keeps growing and then losing abs bars

[@nicolascamargo8339]

Genial

[@shehabsarder]

how does the modulas came form?

[@utuberaj60]

As you are aware ln (x) is valid only if x is positive. So we typically place the argument inside modulus.

[@rachitchauhan8164]

Apparently if you take 1/sinx =(sin²(x)+cos²(x))/sinx you will get -cosx-cosecsx+C as an answer which does not contain logarithmic function🫤

[@PamComPollo]

Me tenia loco esta integral la mayoría solo fue por el camino de "facil" multiplicando arriba y abajo por cscx + ctgx si bien da una respuesta, no explica muy bien el porq. Tendre que estudiar más trigonometría.