(Method 1) Integral of 1/sin(x)cos(x) (trigonometric identities + substitution)

[@IntegralsForYou]

🔍 𝐀𝐫𝐞 𝐲𝐨𝐮 𝐥𝐨𝐨𝐤𝐢𝐧𝐠 𝐟𝐨𝐫 𝐚 𝐩𝐚𝐫𝐭𝐢𝐜𝐮𝐥𝐚𝐫 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥? 𝐅𝐢𝐧𝐝 𝐢𝐭 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝐬𝐞𝐚𝐫𝐜𝐡𝐞𝐫: ► Integral searcher 👉integralsforyou.com/integral-searcher 🎓 𝐇𝐚𝐯𝐞 𝐲𝐨𝐮 𝐣𝐮𝐬𝐭 𝐥𝐞𝐚𝐫𝐧𝐞𝐝 𝐚𝐧 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝? 𝐅𝐢𝐧𝐝 𝐞𝐚𝐬𝐲, 𝐦𝐞𝐝𝐢𝐮𝐦 𝐚𝐧𝐝 𝐡𝐢𝐠𝐡 𝐥𝐞𝐯𝐞𝐥 𝐞𝐱𝐚𝐦𝐩𝐥𝐞𝐬 𝐡𝐞𝐫𝐞: ► Integration by parts 👉integralsforyou.com/integration-methods/integration-by-parts ► Integration by substitution 👉integralsforyou.com/integration-methods/integration-by-substitution ► Integration by trig substitution 👉integralsforyou.com/integration-methods/integration-by-trig-substitution ► Integration by Weierstrass substitution 👉integralsforyou.com/integration-methods/integration-by-weierstrass-substitution ► Integration by partial fraction decomposition 👉integralsforyou.com/integration-methods/integration-by-partial-fraction-decomposition 👋 𝐅𝐨𝐥𝐥𝐨𝐰 @𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬𝐟𝐨𝐫𝐲𝐨𝐮 𝐟𝐨𝐫 𝐚 𝐝𝐚𝐢𝐥𝐲 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥! 😉 📸 instagram.com/integralsforyou/

[@pierreangelobacarisas961]

this video literally saved me from having a mental breakdown. thank you very much!

[@IntegralsForYou]

You're welcome, Midome Hiragi! Thanks for telling me it! 😉

[@rahuldsouza1985]

Divide numerator and denominator by (cos x)^2 Numerator becomes (sec x)^2 and denominator becomes tan x We can then sub tan x as t Result will be log t log(tan x)

[@IntegralsForYou]

kzbin.info/www/bejne/sHLYqKygp515jbs 😉

[@sanrai5480]

Ahaha😂😂😂 I spend lot of time for this "1/(sin^2 x * cos^2 x)" luckily I ended up being here. Thank you so much!!!

[@IntegralsForYou]

;-D You're welcome! ;-D

[@theengineeringstudent2678]

You are the king of the tricks. Regards from Angola-Africa

[@IntegralsForYou]

Thank you Florindo! 👍

[@theengineeringstudent2678]

The power of number one (1).

[@lepetitskieurhaut-savoyard4920]

You could simplify the solution in ln(tan(x))+C

[@IntegralsForYou]

Totally agree! 💪

[@rivercuperus1158]

Thanks bro, this video was a lifesaver.

[@IntegralsForYou]

Glad it helped! ☺☺

[@madmax3718]

Or, how about ('{' use this as int sign) In second step, Putting{ (Sinx/cosx)=tanx and { (cosx/sinx=cotx) Then, we get { tanx= ln|secx| and {cotx= ln|sinx| As they have addition in between, it will be, ln|secx| + ln|sinx| As it's logarithm, we can multiply secx and sinx, which will give, ln|secx.sinx| =ln|(1/cosx.sinx|= ln|tanx|

[@IntegralsForYou]

It is ok for me 😉

[@irajnaghash]

-Ln|cosX| + Ln|sinx|+ C= Ln (sinx/Cosx) + C= Ln|TanX|+ C

[@Physics-j6c]

Btw I wanted to ask that:- If I create sin2x in downside then, just by doing reciprocal of it we get cosec2x so, integral of cosec2x is just = log|cosecx-cotx| would be answer ????

[@IntegralsForYou]

Hi! Be careful with the constants 2 from the formula: sin(2x) = 2sin(x)cos(x) ==> sin(2x)/2 = sin(x)cos(x) Then: Integral of 1/sin(x)cos(x) dx = = Integral of 1/(sin(2x)/2) dx = = Integral of 2/sin(2x) dx = = 2*Integral of 1/sin(2x) dx = = 2*Integral of csc(2x) dx = Substitution: u = 2x du = 2 dx ==> du/2 = dx = 2*Integral of csc(u) du/2 = = (2/2)*Integral of csc(u) du = = Integral of csc(u) du = = -ln|csc(u) + cot(u)| = = -ln|csc(2x) + cot(2x)| + C 💪

[@Physics-j6c]

@@IntegralsForYou Thank you so much for your efforts... ❤️❤️ Yeah 2 was the constant which I haven't type in the comment as it was already been divided by another 2 of csc x I'm bit a confused in my answer which I got !

[@O_Tobi-e6h]

Sinxcosx= sin(2x)/2 1/sinxcosx=2/sin(2x)

[@IntegralsForYou]

Agree! After a u=2x substitution you will only have to solve the integral of 1/sin(u) 💪

[@IITD2005]

Thank you sir 😊😊

[@IntegralsForYou]

Most welcome! ❤❤

[@minadragonrider8695]

Thank you so much 😍 You saved me

[@IntegralsForYou]

My pleasure! 😊 Have you watched both methods? Integral of 1/sin(x)cos(x) ( Method 1: 1=sin^2(x)+cos^2(x) ): kzbin.info/www/bejne/annTnoevlNuFgs0​ Integral of 1/sin(x)cos(x) ( Method 2: u=tan(x) ): kzbin.info/www/bejne/sHLYqKygp515jbs

[@danielahernandezaleman8028]

Thank you so much, it really helped me

[@IntegralsForYou]

My pleasure! 😉

[@abhishekdhepe3249]

btw, integration of tanx dx is log l secxl

[@IntegralsForYou]

Totally agree! Integration of tan(x) is ln|sec(x)| and integration of cot(x) is -ln(csc(x)) but I try to not use sec(x) and csc(x) in my videos. It is not wrong, it is just that I don't like them hehe

[@rajatverma3692]

I don't know why am I studying all this in BCA 😭😭...save me.....this ...fourth sem is gonna break my ass but hey ty for this awesome video.... it really helped me

[@IntegralsForYou]

Thank you! And good luck with your studies! 😉

[@luluuu9096]

Well at least you are in college.. I am still in school😃😂.. Regardless best of luck🤞

[@ENG-Z313]

Thanks ALOT Buddy 😊

[@IntegralsForYou]

My pleasure! 😉

[@theengineeringstudent2678]

I respect u man

[@IntegralsForYou]

Thanks! 😀

[@muazkhan9089]

Can i do it by like sinx/cosx = tanx and then integrate tanx??

[@IntegralsForYou]

Hi! Yes, I did it here 👉 kzbin.info/www/bejne/sHLYqKygp515jbs 😉

[@shailpatel0330]

*Just multiply and divide by two then do integeration of 2cosec2x*

[@IntegralsForYou]

That works too! 😉

[@ayoubab2120]

is it correct like this ? : 1/sinxcosx = 2/sin2x with : t=2x => dx = 1/2 dt and we will have 1/sin(t) => sin(t)/1-cos(t)^2 =( -argth(cos(t)))' =(-argth(cos(2x)))' argth is the inverse of the fonction th(x)

[@IntegralsForYou]

Hi Ayoub Ab! Yes it is correct! Integral of 1/sin(x)cos(x) dx = = Integral of 2/sin(2x) dx = = Integral of 1/sin(2x) 2dx = Substitution: t = 2x dt = 2dx = Integral of 1/sin(t) dt = = Integral of sin(t)/sin^2(t) dt = = Integral of sin(t)/(1-cos^2(t)) dt = = Integral of 1/(1-cos^2(t)) sin(t)dt = u = cos(t) du = -sin(t)dt ==> -du = sin(t)dt = Integral of 1/(1-u^2) (-du) = = - Integral of 1/(1-u^2) du = = - arctanh(u) = = - arctanh(cos(t)) = = - arctanh(cos(2x)) + C ;-D

[@ayoubab2120]

@@IntegralsForYou thank you but i have a question : if -arctanh(cos2x) and ln(tanx) have the same derivate so they are equal ??

[@IntegralsForYou]

Not always. The truth is that they have the same derivative, I think in this case they are equal, but sometimes two solutions can be different. Let me write you a simple example: Integral of 1/(x+1)^2 dx = = Integral of (x+1)^-2 dx = = (x+1)^(-1)/(-1) = = -1/(x+1) + C So we have "-1/(x+1) + C" is a solution. However, "x/(x+1) + C" is also a solution. Derivative of -1/(x+1) + C = = Derivative of -(x+1)^(-1) + C = = -(-1)(x+1)^(-2) = = 1/(x+1)^2 Derivative of x/(x+1) + C = = [1*(x+1) - 1*x]/(x+1)^2 = = [ x + 1 - x]/(x+1)^2 = = 1/(x+1)^2 = Why is it possible? Because they are the same except for the constant, but it doesn't matter when derivating, because the derivative of any constant is zero. x/(x+1) + C = = (x+1-1)/(x+1) + C = = (x+1)/(x+1) - 1/(x+1) + C = = 1 - 1/(x+1) + C = = -1/(x+1) + C+1 = = -1/(x+1) + C' (where C'=C+1) Hope I answered your question ;-D

[@ayoubab2120]

@@IntegralsForYou thankyou very very much

[@ernestschoenmakers8181]

@@IntegralsForYou I was searching for a method to get this solution and found that if you take the step u=(x+1)/x then indeed you'll get: I = x/(x+1) + C.

[@dictor32]

eventually equal ln tan x + c

[@IntegralsForYou]

Thank you, a note have been added at the end of the video! Thanks!

[@yopajjapoy8680]

Im late but why is it equal to ln tanx + c?

[@ernestschoenmakers8181]

@@yopajjapoy8680Use property of logarithms.

[@mahmoudmoussa6623]

Thank you so much!

[@IntegralsForYou]

You're welcome! 😉

[@sanatani_rashtr]

1k th like by mee😮

[@IntegralsForYou]

Thank you so much 😀 You can like my videos on Derivatives ForYou, it would help a lot to grow my new channel! 👉www.youtube.com/@DerivativesForYou Thank you! ❤

[@Hanzel1583]

What i don’t get is if you derive ln(tan(x)) you will get 1/cosx.sinx so which is right???

[@IntegralsForYou]

Yes, it is correct! ;-D

[@fisicauned-xy9zw]

another solution is dividing by cos^2(x). so: 1/cos^2(x)/tg(x) and so on...

[@IntegralsForYou]

Hi! I did it here => kzbin.info/www/bejne/sHLYqKygp515jbs Thanks for your comment!

[@gairick9]

very cool... where you're from?

[@IntegralsForYou]

Hi Gairick, I'm from Spain, and you?

[@yeganehyeganeh5927]

Thank you veryyyyy much. It was great

[@IntegralsForYou]

You're welcome Yeganeh Yeganeh! Thanks for your comment! ;-D

[@thomasarch5952]

An easier was is to note that sin x cos x = (1/2) sin 2x. Then the integral becomes 1/((1/2)sin2x) or 2/sin 2x or 2csc 2x dx which is ln /csc 2x - cot 2x/ + c. much simpler and straightforward than the video though both are correct.

[@IntegralsForYou]

Hi Thomas Arch! You can also multiply by cos^2(x)/cos^2(x) and let 1/sin(x)cos(x) become 1/tan(x) 1/cos^2(x): kzbin.info/www/bejne/sHLYqKygp515jbs. I like when there are a lot of methods to solve the same integral :-D

[@rudrangshu5151]

Wonderful

[@jhonatanninomartinez5482]

Sen/cos = tan, integral of tangente = Ln|cosx|

[@IntegralsForYou]

Hola Jhonatan, tu respuesta es casi correcta. La respuesta es -ln|cos(x)| + C (faltaba el menos delante y la constante de integración). Te dejo el enlace del video para que lo compruebes: Integral de tan(x) dx = kzbin.info/www/bejne/d5PFi4FvpZ2fj7s

[@jhonatanninomartinez5482]

Amigo una pregunta, porque no se puede hacer la integral de tangente directa (?) O sea en las tablas de integración esta la integral de tangente y pues es ln |cosx| + c , tengo esa duda gracias

[@IntegralsForYou]

Hola, supongo que esta respuesta te la tiene que dar tu profesor, ya que me extraña mucho que en las tablas tengas que la integral de tangente es ln|cos(x)| + C. Una razón es porque la respuesta es incorrecta, debería ser -ln|cos(x)| + C, y la segunda razón es porque no me parece una integral directa, ya que hay que transformar tan(x) en sin(x)/cos(x) y luego tener en cuenta el signo menos, es decir, no es directa porque hay que manipularla antes de resolverla. Pero ya te digo que esto es mejor que lo hables con tu profesor y le preguntes cómo lo quiere él, porque es una respuesta un poco subjetiva. En lo que estaremos de acuerdo todos los seres humanos del planeta es que la respuesta lleva un menos delante, ya que la derivada de cos(x) es -sin(x) ;-D

[@ernestschoenmakers8181]

No, it's lnIsec(x)I + C.

[@Khushi-do2vs]

Acha eve c k thk gye krde krde ...tnx lot g😂

[@arielfuxman8868]

what about integrating 2*csc2x

[@IntegralsForYou]

Sorry, are you asking for the integral of 2*csc(2x) or the integral of 2*csc^2(x)?

[@mrkiketo22]

thx good for me

[@IntegralsForYou]

You're welcome!! :-D

[@veerrajput6007]

Thanks

[@IntegralsForYou]

You're welcome!

[@veerrajput6007]

Correct answer

[@IntegralsForYou]

;-D

[@alexandretaranoff714]

eventually equal to ln sin 2x ?

[@IntegralsForYou]

I'm sorry but it is not right. Eventually: -ln(cos(x)) + ln(sin(x)) = ln(sin(x)) - ln(cos(x)) = ln(sin(x)/cos(x)) = ln(tan(x)) Thanks for watching!

[@jaskaur8983]

good

[@theengineeringstudent2678]

The way you think It surprises me lol

[@IntegralsForYou]

;-D

[@rishutechnical0.248]

Wrong answer

[@IntegralsForYou]

Why?

[@abinashmallik5272]

Bot

[@guillermovillalba8552]

Thanks

[@IntegralsForYou]

You're welcome!

[@jayshankarsingh3541]

Wrong answer

[@IntegralsForYou]

Hi Jayshankar singh, I don't agree with you but maybe you're right. Why do you think so?

[@ernestschoenmakers8181]

@@IntegralsForYou Don't worry your answer is correct, maybe he thought that the answer should be lnItan(x)I +C but that's the same.

[@IntegralsForYou]

@@ernestschoenmakers8181 Thanks!