I used to struggle so much with these kind of integrals, I never knew when to use u=secx and when to do u=tanx. I have a really good understanding of this now! Thank you, and God bless you
on prob 1 if we take u = sec x, and rewrite tanxset^2x as tanxsecx.secx then du/dx = secxtanx which will workout seamlessly afterwards. (just optional)
At 7:32, during the U sub operation why do we transform only the sec^2(x) into U? If we made U == sec(x) then wouldn't the other "sec(x)" in the integral also become a u? Shouldn't the equation be "(U^2) * (U) * (tan(x)) " divided by du, sec(x)tan(x). I am probably over thinking this, but I've had this question since I started doing trig integrals.
When doing u sub you can choose what you want to set as u, as long as it helps you integrate it. In this case he only chose sec^2x because if you do u sub du=secxtanx which cancels out the other part and leaves you with something you can integrate easily (u^2)
I have a problem for question 1. integral tanx sec²x dx you wrote it as (tan²x)/2 but it can also be (sec²x)/2 But if both are the same then tan²x = sec²x tan x = sec x which means x is (1/2) (4nπ + π) Why are we getting the values of x??
U can't really be helping someone and then go like "now the answer is obvious, figure it out on your own" this is why I find it hard to follow your videos sometimes.
Little late, but for anyone needing clarification on this in the future: the integral of tanx is indeed -lncos x, but the in the case of this video in problem 1, we arent finding the integral of tan x per se and using u-sub instead. What i mean is we let u = secx, du = secx tanx dx, thus dx = 1/secx tan x....but dont forget to replace the secx in the dx we calculated with u. So dx=1/u tanx. If you then replace dx in our original setup and replace secx with u: (u^2-1)tanx * (1/u tanx). Tanx cancels out and we get (u^2-1)/u. Rewrite as (u^2/u)-(1/u) and simplify to get u - (1/u) and thus the integral of this would be (u^2/2)-ln(|u|), replace the u's for sec x and we get the final answer (sec^2x/2) - ln|secx|. Hope that helps.
nope he's right. the tan(x) is sin(x)/cos(x). the derivative of cos(x) is MINUS sin(x). so the integral of tan(x) is -ln(cosx). Then using log rules, we can bring the minus up, so that it is ln(1/cosx) which is ln(secx).
@moonlight_studies hey. You're right it was a typo and I fixed it. The final form of the answer was correct tho with the negative being present. I don't really remember any of this tbh lmao but point of my comment was to further explain the steps used to show that a mistake wasn't made by the video (as per the original comment). Granted I'm just another student so could wrong but the work made sense at the time and I asked on chegg and symbolab at a later point. Gl!
I used to struggle so much with these kind of integrals, I never knew when to use u=secx and when to do u=tanx. I have a really good understanding of this now! Thank you, and God bless you
So, I used TH-cam search to look for the integral of secx dx and I saw your video from six years ago
It was really nice 🙌
Really good thanks for you explanation
Thank you so much it's exactly good and also pleas do the integration of √tsint
This is so easy to understand. Thank you!
on prob 1 if we take u = sec x, and rewrite tanxset^2x as tanxsecx.secx then du/dx = secxtanx which will workout seamlessly afterwards. (just optional)
great videos thanks
Thanks for your help
Suppose secant has an odd power, how do we integrate such?
In the first can't we just let u as tanx and then integrate u^3 and tanx and multiply them
At 7:32, during the U sub operation why do we transform only the sec^2(x) into U? If we made U == sec(x) then wouldn't the other "sec(x)" in the integral also become a u? Shouldn't the equation be "(U^2) * (U) * (tan(x)) " divided by du, sec(x)tan(x). I am probably over thinking this, but I've had this question since I started doing trig integrals.
When doing u sub you can choose what you want to set as u, as long as it helps you integrate it. In this case he only chose sec^2x because if you do u sub du=secxtanx which cancels out the other part and leaves you with something you can integrate easily (u^2)
i love you havent even watched the video yet
Nice solution
I have a problem for question 1.
integral tanx sec²x dx
you wrote it as (tan²x)/2
but it can also be (sec²x)/2
But if both are the same then
tan²x = sec²x
tan x = sec x
which means
x is (1/2) (4nπ + π)
Why are we getting the values of x??
Yea. That’s called “off by a constant”. I have an old video explaining it in details.
Mmmmm no I don't agree to this.
Giraffe has learned all that now.
@@aashsyed1277 yeah but less than before.
@@aashsyed1277 Rising Star Math TH-camres Challenge
U can't really be helping someone and then go like "now the answer is obvious, figure it out on your own"
this is why I find it hard to follow your videos sometimes.
1:17 min Mistake spotted 🥸, integrale of tan x isn t ln sec x actually it is ln cos x . since secx equal to 1/cosx
Little late, but for anyone needing clarification on this in the future:
the integral of tanx is indeed -lncos x, but the in the case of this video in problem 1, we arent finding the integral of tan x per se and using u-sub instead. What i mean is we let u = secx, du = secx tanx dx, thus dx = 1/secx tan x....but dont forget to replace the secx in the dx we calculated with u. So dx=1/u tanx. If you then replace dx in our original setup and replace secx with u: (u^2-1)tanx * (1/u tanx). Tanx cancels out and we get (u^2-1)/u. Rewrite as (u^2/u)-(1/u) and simplify to get u - (1/u) and thus the integral of this would be (u^2/2)-ln(|u|), replace the u's for sec x and we get the final answer (sec^2x/2) - ln|secx|. Hope that helps.
nope he's right. the tan(x) is sin(x)/cos(x). the derivative of cos(x) is MINUS sin(x). so the integral of tan(x) is -ln(cosx). Then using log rules, we can bring the minus up, so that it is ln(1/cosx) which is ln(secx).
@@ital3nt3d nah bro u said the integral of tanx is ln cosx. its not. it is -ln(cosx). which is the same as ln(secx) using log rules :)
@moonlight_studies hey. You're right it was a typo and I fixed it. The final form of the answer was correct tho with the negative being present. I don't really remember any of this tbh lmao but point of my comment was to further explain the steps used to show that a mistake wasn't made by the video (as per the original comment). Granted I'm just another student so could wrong but the work made sense at the time and I asked on chegg and symbolab at a later point. Gl!
@@ital3nt3d haha no worries, Gl! :D