Just a little clarification for any one who needs it. At 1:39, he doesn't explicitly say it, but he's splitting the integral of the sum into the sum of two integrals. The first part of the sum is done being integrated at 2:21. the second half of the sum is done being integrated at 2:27.
It might be easier to use integration by parts where u = sec^2(x) and dv = sec^2(x). This approach yields tan(x) sec^2(x) - 2 tan^3(x) / 3 + C, which is equivalent to your answer.
This is a great explanation. I had a problem almost identical to this on an exam but it was sec(3x) instead. It worked the same way but I had to substitute u for 3x and then apply the method you did. I made the mistake though of not separating the 4th power into 2 separate powers and instead I tried to square the (tan^2(u) + 1). Then I tried expanding and it made more of a mess than I originally had lol.
hi I have a quick question, I have worked out the problem sec^4 (theta) [shouldn't be any different right?] and have come up with the same answer however the answer in the back of the book does not confirm this and instead says it is (for the sake of ease i will use "x" instead of theta: 2/3 tan (x) + 1/3 sec^2 (x) tan (x) + C Can you explain why?
Ray D Probably a different method of integration was used, but for the sake of argument, both functions are the same (at least they are graphically). You can confirm this fact with a sketch of each functions graph. As an old friend used to say "6 and two 3s".
how about if i solve with this =integral((sec^2 x) (sec^2 x)) dx =integral((tan^2 x +1) d(tan x) =integral(tan^2 x) d(tan x) + integral(1) d(tan x) =1/3 tan^3 x + tan x + c why its wrong?
This method was my thinking as well. Just keep the integrand as (tan^2x + 1)sec^2x, then let u = tanx. That way you're just integrating (u^2 + 1). I guess he just thought of his method first and went with it, but this question is one of those situations where there's more than one possible approach.
Just a little clarification for any one who needs it. At 1:39, he doesn't explicitly say it, but he's splitting the integral of the sum into the sum of two integrals. The first part of the sum is done being integrated at 2:21. the second half of the sum is done being integrated at 2:27.
thanks m8
Best explanation of these types of questions ive seen! Helping me with revision for my degree
+Tom Frost thank you!
do you solve this for your degree
@@mujibmahaldar3621 yes!! These are for 11-12th standards.
@@saturn-ke1um eggzaktly.....
buena chino me salvas de varias
Luis Gutiérrez Garcia Jajaja x2
Aguante el chino
I tried by a so complicated way. I did not see this way you did it.. So easy and so fast. Thanks man !
After 5 years i am watching this video since it was uploaded this video is very helpful i wish you will achieve 10million soon love from india🇮🇳🇮🇳
It might be easier to use integration by parts where u = sec^2(x) and dv = sec^2(x). This approach yields tan(x) sec^2(x) - 2 tan^3(x) / 3 + C, which is equivalent to your answer.
Thank you! I was stuck in the middle of an exercise thad had this one!
Beautiful explanation, we love your channel
We can also make it 1/cos^4(x)
And multiply by (sin^2(x)+cos^2(x))^2 wich is technically just 1
And we will have
Int(tan^4(x))+int(tan^2(x))+int(1)
Once again mr Pen doing his thing! Thank you very much.
THANK YOU. also love the microphone haha
Thank you sir for your kind help .
Though I was unable to understand your language but I get your point ..
Love from India 🙏
Thank you I understood every explanation of every step in your calculation.
you, my friend, are a legend
You are saving my degree! Thank you so much!
Check out my new video on integrals of trigonometric ratios raised to the fourth power
ua-cam.com/video/9o4b-Kj29L8/v-deo.html
One step process: integral = integral of (1 + (tanx)^2) d(tanx) = tanx + (1/3) (tanx)3 + C
This is a great explanation. I had a problem almost identical to this on an exam but it was sec(3x) instead. It worked the same way but I had to substitute u for 3x and then apply the method you did. I made the mistake though of not separating the 4th power into 2 separate powers and instead I tried to square the (tan^2(u) + 1). Then I tried expanding and it made more of a mess than I originally had lol.
I also have a vid on the integral of sec(x) ua-cam.com/video/CChsIOlNAB8/v-deo.html
This integral is on the 2024 UEC adv. math paper 2. I saw the paper and was curious how to solve. So here I am.
you can also try using the t=tg(x) sub
thanks from india iam not able to understand your languae yeah but i got understand the solution pretty well
Thanks man for the help. This Q confused me for more than 30 minutes.
my pleasure to help!
Nice explanation very helpful
Why wouldn't you use u-substitution right after subbing for one of the sec^2....? With u=tan?
Thanks man.. ...I tried by a harder way.
si lo habia hecho bien ayuda q felicidad
Thank you, I needed to check this integral, because Wolfram Alpha gave me a strange result.
hi I have a quick question, I have worked out the problem
sec^4 (theta) [shouldn't be any different right?]
and have come up with the same answer
however the answer in the back of the book does not confirm this and instead says it is (for the sake of ease i will use "x" instead of theta:
2/3 tan (x) + 1/3 sec^2 (x) tan (x) + C
Can you explain why?
Ray D Replace sec^2(x) with tan^2(x)+1.
Ray D Probably a different method of integration was used, but for the sake of argument, both functions are the same (at least they are graphically). You can confirm this fact with a sketch of each functions graph. As an old friend used to say "6 and two 3s".
Can we use reduction formula for this one?
Hey thanks......🙏
I am from India.. 🇮🇳🇮🇳🇮🇳
Thank you for this helpful video!
Thanks.😃😃
It helped me a lott☺
Tq sir u helped me a lot
Why did u not cancel sec^2x the last one also? Can u not do du/sec^2x in the last one too?
Thank you very much for your help
In the step where you wrote the integral with u substitution, i think u missed one sec^2 (x) i am confused
excelente, muy bueno.
EXCELENTE PRECISO COMO DEBE SER
Question: Did you split tan^2(x)sec^2(x) and sec^2(x) so that you can get (u^2)(sec^2)(du/sec^2(x))? I'm a bit confused
THANKS EVERYONE AND THANKS
Thank you sooo much 🌹
Thankyou ❤
what 's the name of the program which I can write math symbols? Plz waiting for your reply!
Mathtype
Steve thanks a lot man, what time is it?
Explanation was good... And yr mike is looking like Pokemon ball 😅
so proud of you
so proud of U
Could it have been done without distribution. Let sec^2 be your du
Thanks sir
great job thanks a lot.
thanks!
Greetings from India
Why you didn't use the reduction formula??
Pedro Casimiro too easy for him
how about if i solve with this
=integral((sec^2 x) (sec^2 x)) dx
=integral((tan^2 x +1) d(tan x)
=integral(tan^2 x) d(tan x) + integral(1) d(tan x)
=1/3 tan^3 x + tan x + c
why its wrong?
This method was my thinking as well. Just keep the integrand as (tan^2x + 1)sec^2x, then let u = tanx. That way you're just integrating (u^2 + 1).
I guess he just thought of his method first and went with it, but this question is one of those situations where there's more than one possible approach.
I really like the fact that you like Kobe
Thanks Bro.
Thanks
Tnk uh♥️
quick n helpful
my pleasure
thank u bro
thank you
u r godgift
Gracias!!
thank you!!!
How to do integral sec^3x?
He showed it in another video using IBP.
Thanks!!!
: )
Excelent!!
You rock
i love you
gracias chino, espero nunca encontrarte en el lol.
can u make sec^7
+EDWIN GIOVANNI GARCIA GALVEZ it would be very similar to sec^5x, which is here: ua-cam.com/video/37VLsf5lxXM/v-deo.html
eEXCELENTE:)
jimena trujillo thank u
Op solution
I luv u
привет бауманцы!!!!!!!!!
Yoy are so cute😁
wrong ans real ans is tanx + 1/3tan3x +c
Thanks