Integral of 1/(1+tan(x)) (substitution)

✍🏼 integralsforyo... - Integral of 1/(1+tan(x)) - How to integrate it step by step using integration by substitution!
✅ 𝐃𝐞𝐫𝐢𝐯𝐚𝐭𝐢𝐯𝐞 𝐭𝐨 𝐜𝐡𝐞𝐜𝐤 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Derivative of (1/2)ln(sin(x)+cos(x)) + (1/2)x = • Derivative of (1/2)ln(...
🔍 𝐀𝐫𝐞 𝐲𝐨𝐮 𝐥𝐨𝐨𝐤𝐢𝐧𝐠 𝐟𝐨𝐫 𝐚 𝐩𝐚𝐫𝐭𝐢𝐜𝐮𝐥𝐚𝐫 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥? 𝐅𝐢𝐧𝐝 𝐢𝐭 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝐬𝐞𝐚𝐫𝐜𝐡𝐞𝐫:
► Integral searcher 👉integralsforyo...
🎓 𝐇𝐚𝐯𝐞 𝐲𝐨𝐮 𝐣𝐮𝐬𝐭 𝐥𝐞𝐚𝐫𝐧𝐞𝐝 𝐚𝐧 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝? 𝐅𝐢𝐧𝐝 𝐞𝐚𝐬𝐲, 𝐦𝐞𝐝𝐢𝐮𝐦 𝐚𝐧𝐝 𝐡𝐢𝐠𝐡 𝐥𝐞𝐯𝐞𝐥 𝐞𝐱𝐚𝐦𝐩𝐥𝐞𝐬 𝐡𝐞𝐫𝐞:
► Integration by parts 👉integralsforyo...
► Integration by substitution 👉integralsforyo...
► Integration by trig substitution 👉integralsforyo...
► Integration by Weierstrass substitution 👉integralsforyo...
► Integration by partial fraction decomposition 👉integralsforyo...
👋 𝐅𝐨𝐥𝐥𝐨𝐰 @𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬𝐟𝐨𝐫𝐲𝐨𝐮 𝐟𝐨𝐫 𝐚 𝐝𝐚𝐢𝐥𝐲 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥! 😉
📸 / integralsforyou
𝐅𝐨𝐥𝐥𝐨𝐰 𝐈𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬 𝐅𝐨𝐫𝐘𝐨𝐮
▶️ KZbin: www.youtube.co...
📸 Instagram: / integralsforyou
👍 Facebook: / integralsforyou
𝐃𝐨𝐧𝐚𝐭𝐞
🙋‍♂️ Patreon: / integralsforyou
#integralsforyou #integrals #integrationbysubstitution

[@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/

[@vikasgoogle579]

I think cbse students class12 come to understand example 6 part3 chapter 7 question

[@dfla5472]

Thank you silent teacher.

[@IntegralsForYou]

You're welcome! I like silent teacher :)

[@beproductive-g5h]

😅

[@demonslayerkimetsunoyaiba9139]

Silent teacher 😂

[@215PR]

😂

[@SSBCOMRADE]

Great brother i passed my exam THANK YOU 🔥🔥🔥

[@IntegralsForYou]

Yeah! Congratulations!! 💪💪💪

[@mayukhintesarislam306]

thanx a lot, it was on my test today but i couldn't figure out what to do after the sin-cos transformation

[@IntegralsForYou]

Glad it helped! ❤️

[@ramirobarba4881]

Should we just have this memorized whenever we deal with integrals?

[@IntegralsForYou]

Let's say you learnt another trick to use in case of emergency 😉

[@arnavpandey6986]

yes

[@sourabhaggarwal4509]

Thanks but why did you write so many elementary steps? It was getting on my nerves.

[@IntegralsForYou]

Hi! I show you all the steps to make it easy to understand but you can skip the steps you prefer 😉

[@DanielGiles-hh6p]

I'm so glad with you teacher.

[@IntegralsForYou]

Thank you! ❤

[@reshmakushwaha9718]

Where is your voice ?

[@IntegralsForYou]

Hi! I don't use my voice or any sound for my videos 😉

[@vinayakjoshi5027]

You can substitute 1 as sec^2(x)-tan^2(x) in the numerator the soln would be much shorter.

[@abayontunay5734]

How about Integral of (1/(1+cosx)) dx

[@IntegralsForYou]

Integral of 1/(1+cos(x)) dx = kzbin.info/www/bejne/mWimYYJsoLeLnbc 😉 If you are looking for a particular integral I suggest you to try my integral searcher: integralsforyou.com/integral-searcher 💪

[@futuredr.8513]

Without talking ,your pen have talked a lot.

[@IntegralsForYou]

I agree 😉

[@ashmitadutta-3497]

it was really helpful sir/madam

[@IntegralsForYou]

Thank you! ☺

[@ernestschoenmakers8181]

There's another method to solve this integral: Multiply the integral by sec^2(x) and divide by 1+tan^2(x). I leave this to you to solve this integral this way.

[@IntegralsForYou]

Thannks! I'll add it to my "to do" list 😉

[@sumayeaamin2729]

Thanks ❤️

[@IntegralsForYou]

My pleasure! 😘

[@studywalker1994]

mannnnn, thanks so much

[@IntegralsForYou]

My pleasure! ☺☺

[@bhaskarpandey8586]

That was brilliant and unnecessary !! Just sub t = tan(x/2) ,then dt (2/(1+t*t))=dx , 1/(1+tan(x)) = (1+t*t)/((1+t)^2) Then the expression becomes Integration of 2/(1+t)^2 dt = -2/(1+tan(x/2)) + C

[@ageli3384]

False man because (when you put t=tan(x/2) then dt=1/2 ( 1+tan^2(x/2)) ); Then angle x/2 and x are not the same

[@bhaskarpandey8586]

@@ageli3384 What are you even saying, your expression for dt is meaningless because there is no dx term. You are not understanding this step(dt = (2/(1 + t*t))dx) I think so here you go, dt = (1/2)*(sec^2(x/2))dx = (1/2)*(1 + tan^2(x/2))dx. Tell me if you still don't understand.

[@ageli3384]

@@bhaskarpandey8586 yes you are right man

[@ageli3384]

@@bhaskarpandey8586 i was thinking in other thing

[@bhaskarpandey8586]

@@ageli3384 :)

[@arnavpandey6986]

thank u teacher😊

[@IntegralsForYou]

You're welcome 😊❤

[@bajrangbhakti2028]

Thank ❤️ sir

[@IntegralsForYou]

My pleasure! 💪

[@meerajanagal1669]

Thnxxx alott ☺u helped me for my Test

[@IntegralsForYou]

I'm very happy to know it! Thanks for your comment! :-D

[@dimpalthakur32]

Meera Janagal

[@groundexlight7850]

which pen

[@IntegralsForYou]

Hi! It is a black bic pen 💪

[@Dauselhafiz]

Thank you.

[@IntegralsForYou]

You're welcome! 😊

[@digitaldiaries8064]

love from Bangladesh. really great Brother 😘

[@IntegralsForYou]

Thank you very much!! 💪💪💪

[@daljeetkaur4687]

Thank you sir it's very helpful 👍👍

[@IntegralsForYou]

Thanks! Enjoy the channel! 💪💪

[@musta.moufassih2825]

is it possible to use the méthode of weierstrass subtitution t=tan(x/2) and thank you

[@IntegralsForYou]

mmmm... I don't recommend this method but let's see if we can do it: Integral of 1/(1+tan(x)) dx = Weierstrass substitution: t = tan(x/2) sin(x) = 2t/(1+t^2) cos(x) = (1-t^2)/(1+t^2) tan(x) = sin(x)/cos(x) = (2t/(1+t^2))/((1-t^2)/(1+t^2)) = 2t/(1-t^2) dx = 2/(1+t^2) dt = Integral of 1/( 1 + 2t/(1-t^2) ) 2/(1+t^2) dt = = Integral of 1/( (1-t^2)/(1-t^2) + 2t/(1-t^2) ) 2/(1+t^2) dt = = Integral of 1/( (1 - t^2 + 2t)/(1-t^2) ) 2/(1+t^2) dt = = 2*Integral of (1-t^2)/( (1 - t^2 + 2t)(1+t^2) ) dt = = 2*Integral of (t^2 - 1)/( (t^2 - 2t + 1)(1+t^2) ) dt = = 2*Integral of (t^2 - 1)/( (1+t^2)(t - 1)^2 ) dt = Partial fraction decomposition: (t^2 - 1)/( (1+t^2)(t - 1)^2 ) = (A+Bt)/(1+t^2) + C/(t-1) + D/(t-1)^2 = = (A+Bt)(t-1)^2/(1+t^2)(t-1)^2 + C(t-1)(1+t^2)/(1+t^2)(t-1)^2 + D(1+t^2)/(1+t^2)(t-1)^2 = = (A+Bt)(t^2-2t+1)^2/(1+t^2)(t-1)^2 + C(t+t^3-1-t^2)(1+t^2)/(1+t^2)(t-1)^2 + D(1+t^2)/(1+t^2)(t-1)^2 = = (At^2 - 2At + A + Bt^3 - 2Bt^2 + Bt + Ct + Ct^3 - C - Ct^2 + D + Dt^2)/(1+t^2)(t-1)^2 = = [ (B + C)t^3 + (A - 2B - C + D)t^2 + (-2A + B + C)t + (A - C + D) ]/(1+t^2)(t-1)^2 ==> 0 = B + C ==> B = -C 1 = A - 2B - C + D 0 = -2A + B + C -1 = A - C + D ==> B = -C 1 = A - 2(-C) - C + D ==> 1 = A + 2C - C + D ==> 1 = A + C + D 0 = -2A + (-C) + C ==> 0 = -2A - C + C ==> 0 = -2A ==> A = 0 -1 = A - C + D ==> B = -C 1 = C + D ==> 1 - C = D A = 0 -1 = - C + (1-C) ==> -1 = -C + 1 - C ==> -2 = -2C ==> C = 1 ==> B = -C = -1 D = 1 - C = 1 - 1 = 0 A = 0 C = 1 ==> A = 0 B = -1 C = 1 D = 0 ==> (t^2 - 1)/( (1+t^2)(t - 1)^2 ) = (A+Bt)/(1+t^2) + C/(t-1) + D/(t-1)^2 = = -t/(1+t^2) + 1/(t-1) = 2*Integral of (t^2 - 1)/( (1+t^2)(t - 1)^2 ) dt = = 2*Integral of [ -t/(1+t^2) + 1/(t-1) ] dt = = 2*[ Integral of -t/(1+t^2) dt + Integral of 1/(t-1) dt ] = = 2*Integral of -t/(1+t^2) dt + 2*Integral of 1/(t-1) dt = = - Integral of 2t/(1+t^2) dt + 2*Integral of 1/(t-1) dt = = - ln(1+t^2) + 2*ln|t-1| = = - ln|1+tan^2(x/2)| + 2*ln|tan(x/2)-1| + C = = - ln|1/cos^2(x/2)| + 2*ln|tan(x/2)-1| + C = = ln|cos^2(x/2)| + 2*ln|tan(x/2)-1| + C = = 2*ln|cos(x/2)| + 2*ln|tan(x/2)-1| + C = = 2*ln|cos(x/2)*(tan(x/2)-1)| + C = = 2*ln|cos(x/2)*(sin(x/2)/cos(x/2) - 1)| + C = = 2*ln|sin(x/2) - cos(x/2)| + C It is very long to check, but if you find any mistake tell me please! The partial fraction decomposition can be done easier and faster using: (t^2 - 1)/( (1+t^2)(t - 1)^2 ) = = ((t - 1)(t+1))/( (1+t^2)(t - 1)^2 ) = = (t+1)/( (1+t^2)(t - 1) ) = = (A+Bt)/(1+t^2) + C/(t-1)

[@musta.moufassih2825]

thank you

[@IntegralsForYou]

You're welcome!

[@pulkitsinha3736]

I'm dizzy.

[@brandbabu4888]

Thanku sir ji

[@IntegralsForYou]

😊😊

[@Rajuramsirvi]

Super very nice 👍I am very 😁Happy

[@IntegralsForYou]

Nice! Enjoy the channel! 💪😉

[@ashta_bruh3823]

bro saved my life man i had my finals and this question came like thanks a lot my broo

[@IntegralsForYou]

Glad it helped! ❤

[@ernestschoenmakers8181]

It's much easier to write the cosx in the numerator as (1/2)(cosx + sinx + cosx - sinx) then you get the integral of 1/2 + (1/2)d(sinx + cosx)/(sinx + cosx) which gives your answer immediately.

[@IntegralsForYou]

Hi, Ernest! Thank you very much for your solution! I write it here in case anyone wants to see the details: Integral of 1/(1+tan(x)) dx = = Integral of cos(x)/(sin(x)+cos(x)) dx = = Integral of (1/2)(cos(x) + sin(x) + cos(x) - sin(x))/(sin(x) + cos(x)) dx = = Integral of [ (1/2)(cos(x) + sin(x))/(cos(x) + sin(x)) + (1/2)(cos(x) - sin(x))/(sin(x) + cos(x)) ] dx = = Integral of [ (1/2) + (1/2)(cos(x) - sin(x))/(sin(x) + cos(x)) ] dx = = (1/2)Integral of dx + (1/2)Integral of 1/(sin(x) + cos(x)) (cos(x) - sin(x))dx = Substitution: u = sin(x) + cos(x) du = (cos(x) - sin(x))dx = (1/2)Integral of dx + (1/2)Integral of 1/u du = = (1/2)x + (1/2)ln|u| = = (1/2)x + (1/2)ln|sin(x)+cos(x)| + C ;-D

[@Snow_Leopard_Uncia_uncia]

Хм.

[@Snow_Leopard_Uncia_uncia]

Khe-khe

[@educationalfly3066]

Thank you sooo much brother 😍😍😍😍😍

[@IntegralsForYou]

😊😊😊

[@sachinvishawakarma5525]

Sir thodi tameez se banaya karo pen hee aage ghushede rahte ho😂

[@aluneesan1916]

sometime less is more

[@MadhuMadhu.-kk8rs]

May God gives you voice soon 🥺💗so that you could teach us audibally. praying for you ...great fan of youuu 🌚

[@IntegralsForYou]

Thank you!! I will try my best!! ❤

[@MadhuMadhu.-kk8rs]

@@IntegralsForYou really you don't have voice sir !!??

[@IntegralsForYou]

Yes, I have, but I don't want to use it for my videos: integralsforyou.com/frequently-asked-questions#why-dont-videos-have-sound ☺

[@Md_Nasim_Reza_Rabby]

আপনি হাত দেখায় কি মজা পান?

[@IntegralsForYou]

I am sorry, I don't understand your language and the translator is not helping me...

[@blink_u2126]

Baler moto koracche

[@thaliaquintana378]

Hi! How did you get =1/2int(2cos(x)/sinx+cos))dx? Thank you!

[@IntegralsForYou]

Hi! In this case we multiply by 2 inside and divide by 2 outside the integral. Why? Short answer: Because it works hehe Long answer: First I wanted to do +sin(x)-sin(x) but it splits cos(x)/(sin(x)+cos(x)) into 1 - sin(x)/(sin(x)+cos(x)). In this case, we haven't gone forward, we have the same problem but with sin(x) instead of cos(x) on the numerator. From here we thought that we need to have cos(x)+cos(x) and the only way to do it is by multiplying by 2 the cos(x). Then, fortunately, the original function splits into two "easy-integrable" functions. Hope it helped!

[@thaliaquintana378]

@@IntegralsForYou Yes, it does. Thank you!

[@IntegralsForYou]

@Thalia Quintana Nice! You're welcome! Enjoy this channel 😉

[@wordsunspoken7391]

Sir , Wht is the difference between log and ln

[@IntegralsForYou]

The ln is the log to the base e: en.wikipedia.org/wiki/Natural_logarithm 😉😊

[@batuhanbozkurt7838]

Hello! ∫ (a + 𝑡𝑎𝑛𝑥)/(1 − 𝑎𝑡𝑎𝑛𝑥) dx ? can you make thıs question?

[@IntegralsForYou]

Hi! atan(x) is arctan(x) or a*tan(x)?

[@batuhanbozkurt7838]

@@IntegralsForYou integral of (a+tan×)/(1-atanx)dx?

[@IntegralsForYou]

@@batuhanbozkurt7838 Integral of (a+tan(x))/(1-a*tan(x)) dx = Substitution: u = tan(x) du = 1/cos^2(x) dx = (1 + tan^2(x)) dx = (1+u^2) dx ==> du/(1+u^2) = dx = Integral of (a+u)/(1-au) du/(1+u^2) = = Integral of (a+u)/(1-au)(1+u^2) du = Partial fraction decomposition for (a+u)/(1-au)(1+u^2) : (a+u)/(1-au)(1+u^2) = A/(1-au) + (Bu+C)/(1+u^2) = = A(1+u^2)/(1-au)(1+u^2) + (Bu+C)(1-au)/(1-au)(1+u^2) = = (A + Au^2)/(1-au)(1+u^2) + (Bu + C - Bau^2 - Cau)/(1-au)(1+u^2) = = (A + Au^2 + Bu + C - Bau^2 - Cau)/(1-au)(1+u^2) = = ( (A-Ba)u^2 + (B - Ca)u + (A+C) )/(1-au)(1+u^2) ==> 0 = A - Ba ==> A = Ba ==> B = A/a 1 = B - Ca ==> 1 = A/a - Ca ==> 1/a = A/a^2 - C ==> C = A/a^2 - 1/a = (A-a)/a^2 a = A + C ==> a = A + (A-a)/a^2 ==> a^3 = Aa^2 + A - a ==> a^3 + a = A(a^2+1) ==> a(a^2+1)/(a^2+1) = A ==> a=A ==> B = A/a = a/a = 1 A = a C = A/a^2 - 1/a = a/a^2 - 1/a = 1/a - 1/a = 0 ==> A = a B = 1 C = 0 ==> (a+u)/(1-au)(1+u^2) = = a/(1-au) + u/(1+u^2) ==> = Integral of (a+u)/(1-au)(1+u^2) du = = Integral of [ a/(1-au) + u/(1+u^2) ] du = = Integral of a/(1-au) du + Integral of u/(1+u^2) du = = Integral of -a/(1-au) du + (1/2)Integral of 2u/(1+u^2) du = = ln|1-au| + (1/2)ln(1+u^2) = = ln|1-a*tan(x)| + (1/2)ln(1+tan^2(x)) + C

[@hdanandkumar8189]

Where did plus one go before you start with substitution method!!

[@IntegralsForYou]

Hi! It is in the (1/2)Integral of dx. Since the 1 is multiplying dx, we can avoid write it. It is like wirting 1*A=A (1/2)Integral of dx = (1/2)Integral of 1 dx

[@nehavinod7421]

Thank you sir...the way u write....everything seems so simple.... It was really helpful... Good handwriting....🤗

[@IntegralsForYou]

Thank you! 😀😊😊

[@catherinegilbert2428]

How do you get 1/2 integral dx 10th line

[@IntegralsForYou]

I'm sorry, in which minute did I write the 1/2 of the question?

[@ashutoshsoni9975]

*op*

[@user-pp8lk8nl9u]

Could you tell me from which book you found this problem ?

[@IntegralsForYou]

Hi! I didn't find it on a book, sometimes I think an example by myself but the most part comes from the people who ask for a concrete example 😉

[@user-pp8lk8nl9u]

@@IntegralsForYou

[@IntegralsForYou]

@xx My pleasure! 😊

[@salar159]

Thank you so much!

[@IntegralsForYou]

My pleasure! 😉

[@Manash_Pratim_sharma]

mee boltaa hoo third class because aapnee yesee hii liktee jaa rohee hoo sound dee deta tuu atchaa hootaa

[@IntegralsForYou]

I'm sorry, I don't understand hindi and Google doesn't help... 😞

[@Manash_Pratim_sharma]

@@IntegralsForYou I don't know English..very well...so sorry for this..but I said that please give the voice when you are teaching...beacuse I don't understand easily the answer of this question...after I was trying to solve...then I was successfully doing this...where are you living?why you don't know hindi??But Hindi is the national language of india...

[@IntegralsForYou]

​ @Manash Pratim sarma Hi! I am from Spain, that's why I don't understand hindi 😉 . I know that it would be easier for you if I talked but I prefer to let you think why am I doing each step and if you don't understand it you can ask me in a comment. In my opinion, it is the best way to learn to integrate 😊

[@Manash_Pratim_sharma]

@@IntegralsForYou sorry..i thought that you are from india...

[@rajaroychoudhuri7451]

Sir great tq very much for the efforts😃

[@IntegralsForYou]

You're welcome! Happy to help! 😊

[@aditisharma6728]

Thank-you 😍

[@IntegralsForYou]

You're welcome! 😊

[@projectwildrift1705]

1 = អាំងតេក្រាល dx មែនឫ ? ខ្ងុំអត់យល់កន្លែងនឹង

[@IntegralsForYou]

Integral of dx = 1 + C = C' (where C'=1+C is also a constant) 😉

[@projectwildrift1705]

@@IntegralsForYou Thank you very much 🏆🏆🏆

[@IntegralsForYou]

@PROJECT : Wild Rift My pleasure! 💪

[@therealbaklava2340]

Thank you much! SAVED MY EXAMINATION 🎉

[@IntegralsForYou]

My pleasure! Welcome and enjoy the channel! 😊😊

[@洵蒼]

Thank you the method is useful and marvelous

[@IntegralsForYou]

Thanks, 姚梃畯 ! 😉

[@11.arpitamukherjee4]

Thank u...sir..it is very helpful..😊

[@IntegralsForYou]

Thanks! 😊

[@Phiendienthusinh]

Thank you for help me with homework {1/(1+X2)^2 i am from VietNam. [☆]

[@IntegralsForYou]

You're welcome!! 😀😀

[@HeyKevinYT]

kính chào đồng nghiệp VN!! 🇻🇳👋

[@Phiendienthusinh]

@@HeyKevinYT 👋

[@tutucute]

Thank you

[@IntegralsForYou]

My pleasure! 😊

[@dailyfrenchlanguage855]

Thanks sir

[@IntegralsForYou]

You're welcome! 😊

[@Animix4uu]

how to know when to add/subtract +1-1 or divide/multiply by 2???please help i always sucks in this type of cases

[@IntegralsForYou]

Hi! Let's say that this kind of tricks are to be used when anything more is being useful... The +1-1 trick we use it to avoid doing a long polynomial division, for example x/(x+1) = (x+1-1)/(x+1) = 1 - 1/(x+1) The 1/2 and 2 trick is mostly used to have the derivative of the denominator on the numerator. In this case we need it because in the near future we will be able to cancel numerator and denominator and "magically" the derivative of the denominator appears on the numerator. Hope it helped...

[@randomguyfromsouthasia4515]

when an integration tutorial is longer than 5 minutes, you know you are in trouble.

[@IntegralsForYou]

Totally agree! 😉

[@randomguyfromsouthasia4515]

@@IntegralsForYou interesting that you are still giving out hearts after 5 years. Take love❤️

[@IntegralsForYou]

@random guy from south asia I try to answer all comments hehe Thanks for your answer! 😎

[@carlosjunior7936]

Muito bom!!!!

[@IntegralsForYou]

Obrigado! ☺

[@ashta_bruh3823]

bro saved my life man i had my finals and this question came like thanks a lot my broo

[@IntegralsForYou]

Glad it helped! ❤

[@sachinkumar-du5di]

Rest of all is ok,but if you speak and write then it is easy to understand

[@IntegralsForYou]

Hi, sachin! In my channel I want the viewer think about what I am doing instead of explaining it, and if you still don't understand any step, you can ask me in a comment and I will answer! ;-D

[@fatimabouzid6065]

Thank you so much you helped me a lot

[@IntegralsForYou]

Thanks for telling me it! I am very happy seeing how my channel is helping you! 😊😊

[@monikasharma-hf5nf]

thankuuu

[@IntegralsForYou]

You're welcome, monika sharma! 😉

[@akshanshdixit6900]

Thanks !!

[@IntegralsForYou]

You're welcome! ;-D

[@tmt6994]

Thanks,it very good for me 👍👍

[@IntegralsForYou]

😊

[@abhayrevankar6573]

thx bro

[@IntegralsForYou]

You're welcome, abhay revankar :-D

[@shamlikatoch9184]

Tnq...well done sir...👍😊

[@IntegralsForYou]

Thank you! 😊😊

[@rerreinge1884]

Request: 1) Integral (sin x)^x dx= ....? 2) Integral 1 per "(1- x^2)^0.25" dx= ....? 3) Integral Ln cos x dx= ....? 4) Integral (Ln x)^0.5= ....? Thanks.

[@IntegralsForYou]

Hi, Rerre Inge! I'm sorry but the most part cannot be solved in terms of elementary functions: 1) I don't think it can be expressed in terms of standard functions 2) kzbin.info/www/bejne/eJupeaRpfs2UbLs 3) I don't think it can be expressed in terms of standard functions. But I have cos(ln(x)): kzbin.info/www/bejne/oXm3oY18o8igi5o 4) I don't think it can be expressed in terms of standard functions

[@kelebek2105]

@@IntegralsForYou 4) can we not just: Integral (ln x)^½ = integral (1/2) ln x = 1/2 integral ln x = 1/2 xlnx - x +C (integration by parts for ln x)

[@IntegralsForYou]

@@kelebek2105 Hi! I think the question is the integral of (ln(x))^(1/2) instead of the integral of ln(x^(1/2)). The log property is ln(x^n) = n*ln(x)... There is not a property for (ln(x))^n...

[@pettermollel2531]

Thnx very much sir,, could you use the t_formula to do it....??

[@IntegralsForYou]

Hi petter Mollel! Yes, we can do it with the t-formula, but I suggest to do a substitution before do it: x=u/2 ==> dx=(1/2)du Then you will have tan(u/2) inside the integral and it will be easier to use the t-formula.

[@NhanNguyen-tw1gn]

Thanks. Very good

[@IntegralsForYou]

Thank you! 😉

[@andresmpa]

THAK YOU!!!!! You save my ss

[@IntegralsForYou]

You're welcome! I am very happy to help you! 😀

[@bblue775]

I still confuse..why suddenly -1+1...where did it get? Pls reply me sir...

[@IntegralsForYou]

Hi! Well... let's say that we get that -1+1 because we have tried everything before... At the begining we only have cos(x) on the numerator and we need some sin(x) in order to have the derivative of the denominator on the numerator so we have to try something like -1+1 or +sin(x)-sin(x) in order to modify the expression but not the solution.

[@bblue775]

Integrals ForYou sir I have Integral dx/tanx + 1 = 1/2 ln |sinx - cosx| + 1/2x + c . Is it true ? #sorryifmyenglishsobad

[@IntegralsForYou]

Hi! You should have (1/2)ln|sin(x)+cos(x)| + (1/2)x + C so you should check this minus, it should be "+" instead of "-": Integral of 1/(tan(x)+1) dx = = Integral of 1/(sin(x)/cos(x) + cos(x)/cos(x)) dx = = Integral of 1/( (sin(x) + cos(x))/cos(x) ) dx = = Integral of cos(x)/( sin(x) + cos(x) ) dx = = (1/2)Integral of 2cos(x)/( sin(x) + cos(x) ) dx = = (1/2)Integral of ( cos(x)+cos(x) )/( sin(x) + cos(x) ) dx = = (1/2)Integral of ( cos(x)+cos(x)+sin(x)-sin(x) )/( sin(x) + cos(x) ) dx = = (1/2)Integral of [ ( cos(x)-sin(x) )/( sin(x) + cos(x) ) + ( sin(x) + cos(x) )/( sin(x) + cos(x) ) ]dx = = (1/2)Integral of [ ( cos(x)-sin(x) )/( sin(x) + cos(x) ) + 1 ] dx = = (1/2)[ Integral of ( cos(x)-sin(x) )/( sin(x) + cos(x) ) dx + Integral of 1 dx ] = Substitution: u = sin(x)+cos(x) du = (cos(x)-sin(x)) dx = (1/2)[ Integral of 1/u du + Integral of 1 dx ] = = (1/2)[ ln|u| + x ] = = (1/2)ln|sin(x)+cos(x)| + (1/2)x + C ;-D #yourenglishisnotbad

[@bblue775]

Integrals ForYou ah! Oke...I little confused in it oke

[@IntegralsForYou]

It's okay ;-D

[@alexxpaul]

what will its value be when we give a limit 0-π/2

[@IntegralsForYou]

Integral of 1/(1+tan(x)) dx = (1/2)ln|sin(x)+cos(x)| + (1/2)x + C Integral of 1/(1+tan(x)) dx from 0 to pi/2 = = [ (1/2)ln|sin(pi/2)+cos(pi/2)| + (1/2)(pi/2) ] - [ (1/2)ln|sin(0)+cos(0)| + (1/2)0 ] = = [ (1/2)ln|1+0| + pi/4 ] - [ (1/2)ln|0+1| + 0 ] = = [ (1/2)ln(1) + pi/4 ] - (1/2)ln(1) = = 0 + pi/4 - 0 = = pi/4

[@mathphysicssupersolution15]

super solution sit

[@houssamzrhalla2773]

Thnknuuu

[@IntegralsForYou]

You're welcome! ;-D

[@grownext8828]

(dx/1+tan x )dx rang 0 to π\2 what is the answer

[@IntegralsForYou]

Hi Rohit only! You have to take the solution and apply the fundamental theorem of calculus as follows: Integral of 1/(1+tan(x)) dx = (1/2)ln|sin(x)+cos(x)| + (1/2)x + C Integral of 1/(1+tan(x)) dx from 0 to π/2 = = [ (1/2)ln|sin(π/2)+cos(π/2)| + (1/2)(π/2) ] - [ (1/2)ln|sin(0)+cos(0)| + (1/2)*0 ] = = [ (1/2)ln|1+0| + π/4 ] - [ (1/2)ln|0+1| + 0 ] = = [ (1/2)*0 + π/4 ] - [ (1/2)*0 + 0 ] = = [ 0 + π/4 ] - [ 0 + 0 ] = = π/4

[@grownext8828]

Integrals ForYou tq bro

[@IntegralsForYou]

;-D

[@roboboyaarushkhanna8199]

Integration of e^x^2 Please tell

[@aresitlol]

ma si strunz?

[@devanshi864]

x*3/3

[@justngouassi7988]

Thank you for all you're the best😊😊

[@IntegralsForYou]

;-D Thanks! ;-D

[@jeffwalston7179]

black pen tried to solve this same problem and the comments were thanks a lot thanks a lot. I watched video twice and I am thinking THIS DOESN'T MAKE SENCE. fo

[@availability7652]

nice😀😀

[@IntegralsForYou]

;-D

[@shubham2747]

😍😍😍

[@IntegralsForYou]

😜

[@shubham2747]

@@IntegralsForYou i m from india bro

[@IntegralsForYou]

Technical Gyan From Spain me 😉

[@shubham2747]

@@IntegralsForYou wana we are friend

[@hrutikabadgeri6505]

thankyou Soo much!

[@IntegralsForYou]

You're welcome! ;-D

[@seagullkilla8252]

the way you drew that 'c' pissed me off

[@kagisodipheko3098]

thank you so much sir

[@IntegralsForYou]

You're welcome Kagiso Dipheko!

[@ernestschoenmakers8181]

What about the integral of x*tan(x)*dx and ln(tan(x))*dx?

[@IntegralsForYou]

Hi Ernest schoenmakers, I think we cannot express the integral of x*tan(x) dx or the integral of ln(tan(x)) dx in terms of standard mathematical functions.

[@ernestschoenmakers8181]

Yeah i thought so too cause i tried it but i couldn't find a solution. Now i'm busy with the integral of sqrt(1+x^3).

[@ernestschoenmakers8181]

I think i've found the solution for this one: x^3= (x^(3/2))^2 so if the solution for the integral of sqrt(1+x^2)=(1/2)*x*sqrt(1+x^2)+(1/2)*ln[sqrt(1+x^2)+x]+c then for the integral of sqrt(1+x^3)=(1/2)*x^(3/2)*sqrt(1+x^3)+(1/2)*ln[sqrt(1+x^3)+x^(3/2)]+c. Note that x^2=(x^1)^2

[@IntegralsForYou]

Hi Ernest! The derivative of (1/2)*x^(3/2)*sqrt(1+x^3)+(1/2)*ln[sqrt(1+x^3)+x^(3/2)] + C is (3/2)sqrt(x)sqrt(x^3+1) so your answer is not correct... If you want to use the integral of sqrt(1+x^2) you should do a substitution in order to have the same integral. Here if we try u=x^(3/2) we have: u=x^(3/2) du =(3/2)x^(1/2) dx = (3/2)(x^(3/2))^(1/3) dx = (3/2)u^(1/3) dx ==> (2/3)u^(-1/3) du = dx Then: Integral of sqrt(1+x^3) dx = = Integral of sqrt(1+u^2) (2/3)u^(-1/3) du = = (2/3)Integral of sqrt(1+u^2)/u^(1/3) du = ... and I don't think we will find the answer this way....

[@ernestschoenmakers8181]

You're right, maybe this integral has no elementary solution.

[@inseanity911]

Jesus Christ when you do it, it just seems so simple... Props to you man! Keep up the good work.

[@IntegralsForYou]

Thank you very much! Enjoy the channel! 😊😊

[@giuseppeforte8687]

Sorry, do you really need the sostitution? Isn't it like the integer of f'(x)/f(x)=log | f(x) |

[@IntegralsForYou]

Hi! You don't really need to do it, but by doing it you are explaining it with more details. At this point, it depends on how your teacher wants the solution.

[@hungomanh1026]

Tks you very much!!!

[@IntegralsForYou]

You're welcome! ;-D