Hi! It works too! You will end up doing integration by parts with a different expression: Integral of cos(ln(x)) dx = Substitution: t = ln(x) ==> e^t = x dt = 1/x dx = 1/e^t dx ==> (e^t)dt = dx = Integral of cos(t) (e^t)dt = = Integral of (e^t)*cos(t) dt = = th-cam.com/video/w2U98_vQPsY/w-d-xo.html = = (1/2)(e^t)(sin(t) + cos(t)) = = (1/2)*x(sin(ln(x)) + cos(ln(x))) = = (x/2)*(sin(ln(x)) + cos(ln(x))) + C 💪
Ciao! Grazie a te per guardare il video! In questa playlist trovi tutti gli integrali per parti che ho fatto. th-cam.com/play/PLpfQkODxXi4-GdH-W7YvTuKmK_mFNxW_h.html
x^i = e^(i.ln(x)) = cos(ln(x))+i.sin(ln(x)) Do the integral of x^i using the power rule (x^1+i/(1+i)) and extract its real part, you'll get the answer to integral(cos(ln(x))) :)
Hi! Here you have the solution: Integral of (e^x)sin^2(x) dx = = Integral of (e^x)( (1/2)(1-cos(2x)) ) dx = = (1/2)Integral of ( e^x - (e^x)cos(2x) ) dx = = (1/2)( Integral of e^x dx - Integral of (e^x)cos(2x) dx ) = Integral of (e^x)cos(2x) dx = Parts: Integral of u dv = uv - Integral of v du u = e^x ==> du = (e^x)dx dv = cos(2x)dx ==> v = (1/2)sin(2x) = (e^x)(1/2)sin(2x) - Integral of (1/2)sin(2x) (e^x)dx = = (1/2)(e^x)sin(2x) - (1/2)Integral of (e^x)sin(2x) dx = Parts: Integral of u dv = uv - Integral of v du u = e^x ==> du = (e^x)dx dv = sin(2x)dx ==> v = (-1/2)cos(2x) = (1/2)(e^x)sin(2x) - (1/2)( (e^x)(-1/2)cos(2x) - Integral of (-1/2)cos(2x) (e^x)dx ) = = (1/2)(e^x)sin(2x) - (1/2)( (-1/2)(e^x)cos(2x) + (1/2)Integral of (e^x)cos(2x) dx ) = = (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x) - (1/4)Integral of (e^x)cos(2x) dx ==> Integral of (e^x)cos(2x) dx = (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x) - (1/4)Integral of (e^x)cos(2x) dx (1/4)Integral of (e^x)cos(2x) dx + Integral of (e^x)cos(2x) dx = (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x) (5/4)Integral of (e^x)cos(2x) dx = (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x) Integral of (e^x)cos(2x) dx = (4/5)( (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x) ) Integral of (e^x)cos(2x) dx = (2/5)(e^x)sin(2x) + (1/5)(e^x)cos(2x) + C = (1/2)( Integral of e^x dx - Integral of (e^x)cos(2x) dx ) = = (1/2)[ e^x - ( (2/5)(e^x)sin(2x) + (1/5)(e^x)cos(2x) ) ] = = (1/2)[ e^x - (2/5)(e^x)sin(2x) - (1/5)(e^x)cos(2x) ] = = (1/2)(e^x) - (1/5)(e^x)sin(2x) - (1/10)(e^x)cos(2x) + C Hope it helped! ;-D
Hi! I'm sorry, I did the last step very quick... Let me explain it here with more detail: 2*Integral of cos(ln(x)) dx = x*cos(ln(x)) + x*sin(ln(x)) 2*Integral of cos(ln(x)) dx = x*[cos(ln(x)) + sin(ln(x))] Integral of cos(ln(x)) dx = ( x*[cos(ln(x)) + sin(ln(x))] )/2 Integral of cos(ln(x)) dx = (x/2)*[cos(ln(x)) + sin(ln(x))] + C Hope it is clear now! 💪
Hi Yared! Don't worry, I understand... In fact I would really like to answer you, but when I created this channel I decided to not answering personal questions. Sorry again! ;-D
@K Bee I become prouder of this channel when I read comments like yours 😊This channel was created to help people to solve integrals and I think it is doing it successfully! 💪💪
Integral of cos(2ln(x)) dx = Substitution: t = ln(x) ==> e^t = x dt = 1/x dx = 1/e^t dx ==> (e^t)dt = dx = Integral of cos(2t) (e^t)dt = Integral of (e^t)cos(2t) Parts: Integral of u dv = uv - Integral of v du u = e^t ==> du = (e^t)dt dv = cos(2t) dt ==> v = (1/2)sin(2t) dt = (e^t)(1/2)sin(2t) - Integral of (e^t)(1/2)sin(2t) dt = = (1/2)(e^t)sin(2t) - (1/2)Integral of (e^t)sin(2t) dt = Parts: Integral of u dv = uv - Integral of v du u = e^t ==> du = (e^t)dt dv = sin(2t) dt ==> v = (-1/2)cos(2t) dt = (1/2)(e^t)sin(2t) - (1/2)[ (e^t)(-1/2)cos(2t) - Integral of (e^t)(-1/2)cos(2t) dt ] = = (1/2)(e^t)sin(2t) - (1/2)[ (-1/2)(e^t)cos(2t) + (1/2)Integral of (e^t)cos(2t) dt ] = = (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t) - (1/4)Integral of (e^t)cos(2t) dt ==> Integral of (e^t)cos(2t) dt = (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t) - (1/4)Integral of (e^t)cos(2t) dt Integral of (e^t)cos(2t) dt + (1/4)Integral of (e^t)cos(2t) dt = (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t) (5/4)Integral of (e^t)cos(2t) dt = (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t) Integral of (e^t)cos(2t) dt = (4/5)[ (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t) ] Integral of (e^t)cos(2t) dt = (2/5)(e^t)sin(2t) + (1/5)(e^t)cos(2t) Integral of cos(2ln(x)) dx = Substitution: t = ln(x) ==> e^t = x dt = 1/x dx = 1/e^t dx ==> (e^t)dt = dx = Integral of cos(2t) (e^t)dt = Integral of (e^t)cos(2t) = = (2/5)(e^t)sin(2t) + (1/5)(e^t)cos(2t) = = (1/5)(e^t)[ 2sin(2t) + cos(2t) ] = = (1/5)*x*[ 2sin(2ln(x)) + cos(2ln(x)) ] = = (x/5)*[ 2sin(2ln(x)) + cos(2ln(x)) ] + C ;-D
Hi! The last part consist of solving the next equation: Integral of cos(ln(x)) dx = x*cos(ln(x)) + x*sin(ln(x)) - Integral of cos(ln(x)) dx Let's say L = Integral of cos(ln(x)) dx and we have to find the value of L: L = x*cos(ln(x) + x*sin(ln(x)) - L L + L = x*cos(ln(x) + x*sin(ln(x)) 2L = x*cos(ln(x) + x*sin(ln(x)) L = (1/2)*(x*cos(ln(x) + x*sin(ln(x)) L = (x/2)*(cos(ln(x)) + sin(ln(x))) Integral of cos(ln(x)) dx = (x/2)*(cos(ln(x)) + sin(ln(x))) + C Hope it helped! 💪
Hahaha Dave, that's not a question about the integral, isn't it? :-D Five, six if we consider mathematics as a language.... Thanks for your comment, you made my day :D
последний шаг такой очевидный (когда получившийся оригинальный интеграл надо перенести в левую часть) а я сам так и не додумался. стыдно :( спасибо за решение!)
Hi Axel! I don't think you need it! Here is what I was listening while doing this video th-cam.com/video/8iMMoGuUxks/w-d-xo.html , and it helped! Now seriously, maybe one day I will add the sound, but by the moment I don't think it is necessary. Thanks for asking and enjoy the music!
🔍 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
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Even without words, it is better than others.
;-D
Thank u. This question came in jee Mains exam of India
You're welcome! ;-D
Thanks a lot... I just got confused before watching this video... This clarified me... :-D
I am very happy to know it! Enjoy the channel! 😉
What about t = Inx
Hi! It works too! You will end up doing integration by parts with a different expression:
Integral of cos(ln(x)) dx =
Substitution:
t = ln(x) ==> e^t = x
dt = 1/x dx = 1/e^t dx ==> (e^t)dt = dx
= Integral of cos(t) (e^t)dt =
= Integral of (e^t)*cos(t) dt =
= th-cam.com/video/w2U98_vQPsY/w-d-xo.html =
= (1/2)(e^t)(sin(t) + cos(t)) =
= (1/2)*x(sin(ln(x)) + cos(ln(x))) =
= (x/2)*(sin(ln(x)) + cos(ln(x))) + C
💪
I LOVE YOUUU, MASSIVE THANK YOUU!!
Massive "de nada"'s!! :-D Se agradece mucho comentarios como el tuyo!
i go to uw and this video helped me!
Nice! 💪😊
a thank you to you from Brazil, helped me a lot
My pleasure! 😊
You have help me a lot for my calculus exam. Thank you very much
My pleasure!! Good luck for the exam! 💪
OH MY GOD!!! I can't belive that I understand it. Man you're THE MAN!
hehe thank you for you comment :D :D
GRAZIE FRATELLO !
ho finalmente capito l'integrazione per parti
Ciao! Grazie a te per guardare il video! In questa playlist trovi tutti gli integrali per parti che ho fatto.
th-cam.com/play/PLpfQkODxXi4-GdH-W7YvTuKmK_mFNxW_h.html
Keep on doing these vedios bro...so helpful!!!
I will! Thank you for your comment! 😊
x^i = e^(i.ln(x)) = cos(ln(x))+i.sin(ln(x))
Do the integral of x^i using the power rule (x^1+i/(1+i)) and extract its real part, you'll get the answer to integral(cos(ln(x))) :)
Hi Quentin! I agree with you, but I try not to use complex numbers. Thank you, I didn't know we can do it as you say! :-D
You're welcome, I just wanted to share this strategy as I find it impressively beautiful :)
But the other one you used it just as nice too!! :')
;-D
best channel i have ever seen
Woah! Thank you very much!! 😀😀
Thank you so much !!!
You're welcome! ☺
Que increible te encontre por google poniendo el ejercicio jeje, muchas gracias
De nada! Me alegro que te sirviera! ;-D
lord bless this channel
😊
Please answer the integral of e^x sin^2x dx
Hi! Here you have the solution:
Integral of (e^x)sin^2(x) dx =
= Integral of (e^x)( (1/2)(1-cos(2x)) ) dx =
= (1/2)Integral of ( e^x - (e^x)cos(2x) ) dx =
= (1/2)( Integral of e^x dx - Integral of (e^x)cos(2x) dx ) =
Integral of (e^x)cos(2x) dx =
Parts: Integral of u dv = uv - Integral of v du
u = e^x ==> du = (e^x)dx
dv = cos(2x)dx ==> v = (1/2)sin(2x)
= (e^x)(1/2)sin(2x) - Integral of (1/2)sin(2x) (e^x)dx =
= (1/2)(e^x)sin(2x) - (1/2)Integral of (e^x)sin(2x) dx =
Parts: Integral of u dv = uv - Integral of v du
u = e^x ==> du = (e^x)dx
dv = sin(2x)dx ==> v = (-1/2)cos(2x)
= (1/2)(e^x)sin(2x) - (1/2)( (e^x)(-1/2)cos(2x) - Integral of (-1/2)cos(2x) (e^x)dx ) =
= (1/2)(e^x)sin(2x) - (1/2)( (-1/2)(e^x)cos(2x) + (1/2)Integral of (e^x)cos(2x) dx ) =
= (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x) - (1/4)Integral of (e^x)cos(2x) dx
==>
Integral of (e^x)cos(2x) dx = (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x) - (1/4)Integral of (e^x)cos(2x) dx
(1/4)Integral of (e^x)cos(2x) dx + Integral of (e^x)cos(2x) dx = (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x)
(5/4)Integral of (e^x)cos(2x) dx = (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x)
Integral of (e^x)cos(2x) dx = (4/5)( (1/2)(e^x)sin(2x) + (1/4)(e^x)cos(2x) )
Integral of (e^x)cos(2x) dx = (2/5)(e^x)sin(2x) + (1/5)(e^x)cos(2x) + C
= (1/2)( Integral of e^x dx - Integral of (e^x)cos(2x) dx ) =
= (1/2)[ e^x - ( (2/5)(e^x)sin(2x) + (1/5)(e^x)cos(2x) ) ] =
= (1/2)[ e^x - (2/5)(e^x)sin(2x) - (1/5)(e^x)cos(2x) ] =
= (1/2)(e^x) - (1/5)(e^x)sin(2x) - (1/10)(e^x)cos(2x) + C
Hope it helped! ;-D
why is it x/2?
Hi! I'm sorry, I did the last step very quick... Let me explain it here with more detail:
2*Integral of cos(ln(x)) dx = x*cos(ln(x)) + x*sin(ln(x))
2*Integral of cos(ln(x)) dx = x*[cos(ln(x)) + sin(ln(x))]
Integral of cos(ln(x)) dx = ( x*[cos(ln(x)) + sin(ln(x))] )/2
Integral of cos(ln(x)) dx = (x/2)*[cos(ln(x)) + sin(ln(x))] + C
Hope it is clear now! 💪
ありがとうございました！
You're welcome! 😊
Thank You, very much. From Brazil
:-D
thank uh so much your methods are very helpful :)
Your welcome! ;-D
muchas gracias, gringo!!
De nada!
great video very clear. whats your area?
Thanks Yared Reinarz! Sorry, I only answer questions related to this channel... Anyway, thanks for your comment!
ok sorry. just wanted to clarify that with area i meant occupation. (math teacher, physic, engineer, etc.)
Hi Yared! Don't worry, I understand... In fact I would really like to answer you, but when I created this channel I decided to not answering personal questions. Sorry again! ;-D
thank you so much.
My pleasure! 😎
Gracias! Ahora entiendo cómo hacerlo
De nada Melissa! Es un verdadero placer :D
God send.
Thank u, Sir
You're welcome! 😊
W‼️😩
😎
ohhhh thank you so much mate!!
You're welcome! :D
thank u
My pleasure! 😊
You helped me solve it easily
@K Bee I become prouder of this channel when I read comments like yours 😊This channel was created to help people to solve integrals and I think it is doing it successfully! 💪💪
Keep doing broooo👍👍👍
hello, please answei of integral cos(2Lnx) dx
Integral of cos(2ln(x)) dx =
Substitution:
t = ln(x) ==> e^t = x
dt = 1/x dx = 1/e^t dx ==> (e^t)dt = dx
= Integral of cos(2t) (e^t)dt
= Integral of (e^t)cos(2t)
Parts: Integral of u dv = uv - Integral of v du
u = e^t ==> du = (e^t)dt
dv = cos(2t) dt ==> v = (1/2)sin(2t) dt
= (e^t)(1/2)sin(2t) - Integral of (e^t)(1/2)sin(2t) dt =
= (1/2)(e^t)sin(2t) - (1/2)Integral of (e^t)sin(2t) dt =
Parts: Integral of u dv = uv - Integral of v du
u = e^t ==> du = (e^t)dt
dv = sin(2t) dt ==> v = (-1/2)cos(2t) dt
= (1/2)(e^t)sin(2t) - (1/2)[ (e^t)(-1/2)cos(2t) - Integral of (e^t)(-1/2)cos(2t) dt ] =
= (1/2)(e^t)sin(2t) - (1/2)[ (-1/2)(e^t)cos(2t) + (1/2)Integral of (e^t)cos(2t) dt ] =
= (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t) - (1/4)Integral of (e^t)cos(2t) dt
==>
Integral of (e^t)cos(2t) dt = (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t) - (1/4)Integral of (e^t)cos(2t) dt
Integral of (e^t)cos(2t) dt + (1/4)Integral of (e^t)cos(2t) dt = (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t)
(5/4)Integral of (e^t)cos(2t) dt = (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t)
Integral of (e^t)cos(2t) dt = (4/5)[ (1/2)(e^t)sin(2t) + (1/4)(e^t)cos(2t) ]
Integral of (e^t)cos(2t) dt = (2/5)(e^t)sin(2t) + (1/5)(e^t)cos(2t)
Integral of cos(2ln(x)) dx =
Substitution:
t = ln(x) ==> e^t = x
dt = 1/x dx = 1/e^t dx ==> (e^t)dt = dx
= Integral of cos(2t) (e^t)dt
= Integral of (e^t)cos(2t) =
= (2/5)(e^t)sin(2t) + (1/5)(e^t)cos(2t) =
= (1/5)(e^t)[ 2sin(2t) + cos(2t) ] =
= (1/5)*x*[ 2sin(2ln(x)) + cos(2ln(x)) ] =
= (x/5)*[ 2sin(2ln(x)) + cos(2ln(x)) ] + C
;-D
hello , i just didnt understand the last part
Hi! The last part consist of solving the next equation:
Integral of cos(ln(x)) dx = x*cos(ln(x)) + x*sin(ln(x)) - Integral of cos(ln(x)) dx
Let's say L = Integral of cos(ln(x)) dx and we have to find the value of L:
L = x*cos(ln(x) + x*sin(ln(x)) - L
L + L = x*cos(ln(x) + x*sin(ln(x))
2L = x*cos(ln(x) + x*sin(ln(x))
L = (1/2)*(x*cos(ln(x) + x*sin(ln(x))
L = (x/2)*(cos(ln(x)) + sin(ln(x)))
Integral of cos(ln(x)) dx = (x/2)*(cos(ln(x)) + sin(ln(x))) + C
Hope it helped! 💪
@@IntegralsForYou I figured it out 😉 , thank you so much
@@TiagoNogueirachannel Good to know! 😊
Thank you so much.
You're welcome! ;-D
thank you so much
You're welcome! ;-D
L
J
thanks 🙌🙌
You're welcome!! :D
How many languages do you know?
Hahaha Dave, that's not a question about the integral, isn't it? :-D Five, six if we consider mathematics as a language.... Thanks for your comment, you made my day :D
Integrals ForYou thanks for replying, I saw your replies to other comments and got curious lol. great job on this integral btw.
kitni cute choti si writing hai
;-D
Thank you so much, you made my day! Hahaha
;-D
последний шаг такой очевидный (когда получившийся оригинальный интеграл надо перенести в левую часть) а я сам так и не додумался. стыдно :( спасибо за решение!)
Heheh last step is obvious when you already know it :-D when I fisrt learnt it I was also surprised... Thank you for your comment! :D
ДАЙ БОГ ТЕБЕ ЗДОРОВЬЯ
cпасибо :-D
sound? :(
Hi Axel! I don't think you need it! Here is what I was listening while doing this video th-cam.com/video/8iMMoGuUxks/w-d-xo.html , and it helped! Now seriously, maybe one day I will add the sound, but by the moment I don't think it is necessary. Thanks for asking and enjoy the music!
You are very strong
Thanks! ;-D
Give me your email??
Why will he give his e mail man....
Amuzing!
:-D
Don't u think u have to write it bugger
Hi
Hi! ;-D
Wtf