Honestly at my school the higher math levels have instructors that aren't the best at teaching. I'm really thankful that you put time into making your videos Patrick because they honestly make learning math much easier. I would probably not pass my math class if it wasn't for you. Your basically my math teacher.
i've just recovered from a horrible accident which kept me away from sittin for my finals for bout a year.Totally, all my memories gone,i had nothing but your videos as a hope to get back on the track.Seriously Patrick,u just cant imagine how much of help u have done.Thank a billion
The integral of tan(x) dx is either -ln|cos(x)| OR ln|sec(x)| The reason for this is that in the first instance, -1 is being multiplied by the ln. By the properties of logs, we can take the -1 and put the inside of the log to the power of -1. cos(x) to the power of -1 is 1/cos(x) which is the same as sec(x). Hope this clears it up for anyone confused like I was. :)
yes the final answer is different only because the constants of integration +c is different for both cases but if you have and a definite integral evaluated between to points. His answer in the video is = to sec^4x/4 - sec^2x+ln(sex(x)). Hope this helps
Hello from 2022! I gotta say, that "Knowledge trascends time" is such a nice phrase, your way of explaining things with such casual words like "Hanging out" just make this fucking mess SO.MUCH.EASIER! Thank you so much from Mexico.
You are such a great teacher! I don't even have to rewind because your pacing and explanations are always perfect. Thank you so much for helping me pass calculus!
i have midterm tomorrow. I was trying to understand from my book with no use ! after looking at ur videos i believe i will do good =D thanks for all your help and time. ( ur the best )
If you split intgeral tanxsec^2x into tanxsecxsecx, cant you use a v=secx substitution, making the answer (sec^4x)/4 - sec^2x + ln|secx| + C ? and btw THANK YOU SOOOOOOOOOOOO MUCH, your videos are the only reason I'm passing Calculus right now!
Hey, I was wondering why in the second example (∫tan^5x), why you didn't just break apart the ∫sec^4xtanx into ∫sec^2x(tanx)(sec^2x) and let done ∫(tan^2x+1)(tanx)(sec^2x) and let u=tanx. from there, du=sec^2x. doing that though would've resulted in an answer of (u+u^3) rather than the answer you got of u^3. What would be the difference between the two and are they both interchangeable?
Patrick you scared me, I thought my new comp had a dead pixel but turns out its the video haha here I am worrying about dead pixels when I have a final tomorrow... more like today.
Right around the end in the second substitution, why didn't you substitute secx again instead of tanx? I did it and got the same answer but when changing back the values of u I got -sec^2x instead of -tan^2x. Just looked it up on a calculator and it appears like it was the correct way to go.
Its the same because tan^2 x +1 = sec^2 x. Both expressions are the same in this case, where they only vary with the +1 or -1, which is a constant. If your antiderivatives vary by only a constant it would still accepted as valid antiderivatives. That is also related to why we add a +c in the end of the antiderivatives.
hi patrickJMT, firstly thanks for these videos! secondly, i have a question about integrating tan^5(x): instead of using tanx for my v, i used secx. I ended up with an answer just like yours, except instead of -tan^2(x) i got -sec^2(x). These differ by the -1. Does this -1 go into the C, or did I do something incorrectly?
Wow! U r the best teacher ever!!.. Great tutorials..!!.. Wish I could learn under you, but i'd havta fly over to attend your lectures.. :) Anyways, my question is in the last part of the solution in the 2nd problem wherein:
I did the problem for tan(x)^5 and got the same answer as you and just like you I tested it on my calculator and it works with different limits of integration. This is bothering me as well I'd love to get some feedback.
Hmmm, patrick, at 7:20, shouldn't there be a 1/secx there? I say this because of when the 'u' was taken to equal secx. In that case, only the tanx would have remained in the integrand. Thus, when taking the derivative of the u value(secx) which equals 'secxtanx', that derivative would have only crossed out the tanx in the integrand. So it would seem to leave you with a 1/secx. Anyway, I might be brutally wrong, please take me in the right direction if so.
I get the answer (1/4)*(tan x)^4 -(1/2)*(tanx)^2 + ln|secx| + C. I have varified this answer from other sources which claim it to be right? How do I know which answer is the right one? Yours? or The one above? or both? My exam of calculus will be based on multiple choice questions.
u = something its the same than v= something and its the same than w= something z= something in fact you can use any variable for subtitution, same idea
It's the same thing, you just plug in the trig. identity value of sec^2x into Patrick's final answer and it becomes what you are saying. Also just 2 key things, you are gonna get an additional constant in the form of a fraction, just mesh it up and into "c", 2 constants added together make a constant after all. 2nd, log/secx/ = log/cos^-1 (x)/ , using property of log the power gets multiplied to log, so log/secx/ = -log/cosx/ . In the end, yours and Patrick's answers are the same :) (I know this is like 3 yrs late, but just writing this for anyone else who may be confused and happens to read this thread)
IMPORTANT QUESTION: at the end of the video, why did you make v=something....instead of just sub in u=......?? I have seen all ur other videos, and i fully understand the rules....but, in this case, what do i do? thanks
When you integrated (secx)^2(tanx), I paused and tried it myself. I had v = secx and dv = secxtanx instead. I substituted v back in after I was done, and I ended with (secx)^2 which definetly does not = (tanx)^2. What went wrong?
Hey Patrick couldn't you have just set sec(x) = u in the very beginning and gotten ((sec^2 x)^2)/sec x and after doing a u sub, have (u^4 - 2u^2 +1)/u which ends up making you integrate u^3 - 2u + 1/u thus being (u^4)/4 - u^2 +ln|u| which ends up (sec^4 x/ 4) - sec^2 x + ln|sec x| ?
The user substitution wasn't used for the secxtanx part because the derivative covered it. So only sec^3x was substituted with u and so the secxtanx became the du. I realize I am a year late :/
I know this is kinda late but yes! because of log rules, you can move the negative out the front of -ln |cosx| into an exponent so it would look like +ln|cosx^-1| which is the same thing as +ln|secx|
Thanks for your videos. They are very helpful. But.. The answer for (tan^5 dx ) in secound example is wrong. I checked manually and by computer too. The final answer is (1/4) * tan^4(x) - (1/2) * tan^2(x) - ln( cos(x) ) + C.
it's not that its wrong per se... it's just that due to log rules, you can move the negative out the front of -ln |cosx| into an exponent so it would look like +ln|cosx^-1| which is the same thing as +ln|secx|
+Pauline Tabajonda v = tan(x) and dv = sec^2(x). In the problem there, you have 2(integral) tan(x)sec^2(x) dx, so replace it with dv and v; So you get 2(integral) v dv which is (2v^2)/2. and now you replace v back with tan(x). I know this is 7 months late, but I don't see an answer anywhere. Toodles.
+Zike i think you can split the sec^2xtanxdx into secx(secxtanx) then let u=secx, du=secxtanx for -2integ of udu thats where the -sec^2x came up. Got the same answer ^
when the integral was broken into 3 integrals ,, in the first 2 we had tanx to an odd power an secx to even powers why we took u secx in the first one and u=tanx in the second one,, both cases look the same to me ' odd power on tanx ad even power on sec?!!
Wait a sec.I just need a clarification about the integral of tanx is it the ( ln |secx| + c ) or the ( ln |cosx| + c ) ? I'm confused a bit. Sorry for the question i just want to answer my quiz correctly.
On the last question where part of the integral is 2tanxsec^2x can you make do it like this instead integral of tanxsecxsecx u = secx, du = secxtanx integral of udu sec^2x/2 + c But another way of doing it as shown in the video u = tanx du= sec^2x integral of udu tan^2x/2 + c Are these both equivalent?
This goes back to the last video, where if you have an intigrand in the form of (Tanx)^n * (secx)^m. And m is an even number, you use tanx as your u in the substitution.
They are equivalent, if you take into account the constant you add to each one. sec^2(x) = tan^2(x)+1, and therefore (1/2)sec^2(x)+C = (1/2) tan^2(x)+1+C = (1/2) tan^2(x)+D.
Arg this sucks, these problems can be broken up in a few difference ways and they all yield different but equivalent answers... My professor gives multi choice problems :((((. Going to have to do each problem a few times to be sure it isn't a (None of these) choice.
I actually ended up doing excellent in his class. He does get really tricky with some of the answers but I've been able to figure it all out. Final next week and I'm carrying a 97.25 average on the tests.
Nah, he is tricky with it so bad. I guess one thing that could be lucky is I have a 1/5 to 1/12 chance to guess right instead of almost 0 percent change. (he sometimes has up to 12 options). I don't know how he gets some of his answers, they are simplified all crazy. No partial credit for work either.
Honestly at my school the higher math levels have instructors that aren't the best at teaching. I'm really thankful that you put time into making your videos Patrick because they honestly make learning math much easier. I would probably not pass my math class if it wasn't for you. Your basically my math teacher.
i've just recovered from a horrible accident which kept me away from sittin for my finals for bout a year.Totally, all my memories gone,i had nothing but your videos as a hope to get back on the track.Seriously Patrick,u just cant imagine how much of help u have done.Thank a billion
The integral of tan(x) dx is either -ln|cos(x)| OR ln|sec(x)|
The reason for this is that in the first instance, -1 is being multiplied by the ln. By the properties of logs, we can take the -1 and put the inside of the log to the power of -1. cos(x) to the power of -1 is 1/cos(x) which is the same as sec(x). Hope this clears it up for anyone confused like I was. :)
thank you
Guys, ( ln |secx| ) = ( -ln |cosx| ). its the same thing so calm down
SeabasR LOL. Thank you, I was just about to ask
Can I ask why it's negative??
@@myrepolda ln a^b = b ln a, in this case ln sec x = ln cos^-1 x, thus making it -1 ln cos x
yes the final answer is different only because the constants of integration +c is different for both cases but if you have and a definite integral evaluated between to points. His answer in the video is = to sec^4x/4 - sec^2x+ln(sex(x)). Hope this helps
Hello from 2022! I gotta say, that "Knowledge trascends time" is such a nice phrase, your way of explaining things with such casual words like "Hanging out" just make this fucking mess SO.MUCH.EASIER! Thank you so much from Mexico.
You are such a great teacher! I don't even have to rewind because your pacing and explanations are always perfect. Thank you so much for helping me pass calculus!
Ur the best
Pls don't stop making vdos n helping students :)
2022 and your videos are the only ones that helped me. Thank you so much
SkipiXX (1 week ago)
(secx)^2 = (tanx)^2 + 1. So the results differ only by constant (C) value. Your result is also correct...
Hi patrick from 10 years ago
I don't know what I'd do without you! Thanks, man!
This has really increase my understanding.We want more of this video
You make this so amazingly simple, my cal2 professor drives me nuts with his overly complicated examples.
This is so difficult. I understood it but it is very hard to do such problems by yourself at once attempt.
no problem!
That had me confused for a while but your comment just straightened it out for me
i have midterm tomorrow.
I was trying to understand from my book with no use !
after looking at ur videos i believe i will do good =D
thanks for all your help and time. ( ur the best )
Hey really want to just say again that I really appreciate your videos! Thanks a lot man!
Both are correct. Differentiating both, d/dx [ln (secx)] = (1/secx)(secxtanx) = tanx, while d/dx [-ln (cosx)] = -(1/cosx)(-sinx) = sinx/cosx = tanx. :)
Splendid demonstration of trig integration and substitution. Encore!
yes, by logarithmic property, you can think of it as ln abs( (cos u)^ (- 1) ) +c)
therefore it becomes ln abs(sec u) + c
If you split intgeral tanxsec^2x into tanxsecxsecx, cant you use a v=secx substitution, making the answer (sec^4x)/4 - sec^2x + ln|secx| + C ?
and btw THANK YOU SOOOOOOOOOOOO MUCH, your videos are the only reason I'm passing Calculus right now!
Using the log properties you can bring the negative sign as the exponent of the cosx. Which makes it into 1/cosx = secx.
awesome, i just finished th 6 parts, will go to the exam in one hour and this part (trig integral subs) drove me crazy, but not anymore ;)
@christinajl you can use the ln properties, -ln(cosx) = ln(cos^(-1)x) = ln(secx)
Only patrickJMT's terms can hang out.
Hey, I was wondering why in the second example (∫tan^5x), why you didn't just break apart the ∫sec^4xtanx into ∫sec^2x(tanx)(sec^2x) and let done
∫(tan^2x+1)(tanx)(sec^2x) and let u=tanx. from there, du=sec^2x. doing that though would've resulted in an answer of (u+u^3) rather than the answer you got of u^3. What would be the difference between the two and are they both interchangeable?
No,,,your ans is also right
Patrick you scared me, I thought my new comp had a dead pixel but turns out its the video haha here I am worrying about dead pixels when I have a final tomorrow... more like today.
How'd it go?
Obrigado senhor para estes videos excelentes
thanx, my calc teacher explained it just the other day.
Right around the end in the second substitution, why didn't you substitute secx again instead of tanx? I did it and got the same answer but when changing back the values of u I got -sec^2x instead of -tan^2x. Just looked it up on a calculator and it appears like it was the correct way to go.
Hi Patrick in the last example the middle term should be sec^2 x not tan^2 x
i am also confused about that... which one is right? because i got sec^2 x but he got tan^2 x so ... idk
Its the same because tan^2 x +1 = sec^2 x. Both expressions are the same in this case, where they only vary with the +1 or -1, which is a constant. If your antiderivatives vary by only a constant it would still accepted as valid antiderivatives. That is also related to why we add a +c in the end of the antiderivatives.
no Hamed. u substitution results in tanx being u
term is tan^2x
You rock dude, saved my life
hi patrickJMT, firstly thanks for these videos! secondly, i have a question about integrating tan^5(x):
instead of using tanx for my v, i used secx. I ended up with an answer just like yours, except instead of -tan^2(x) i got -sec^2(x). These differ by the -1. Does this -1 go into the C, or did I do something incorrectly?
dore mon
U rocked once again....................Thanks
Yeah, I did this same problem with a reduction formula and got a different (but hopefully equivalent) answer.
Wow! U r the best teacher ever!!..
Great tutorials..!!..
Wish I could learn under you, but i'd havta fly over to attend your lectures.. :)
Anyways, my question is in the last part of the solution in the 2nd problem wherein:
you are an amazing person
great video. Helped me so much!
there is no error, which is why you have been flagged as spam.
thanks!!
this helped me!!
@jaleed well, after you change it, how do you integrate it?
final tomorrow and this makes it so much easier!!!
I did the problem for tan(x)^5 and got the same answer as you and just like you I tested it on my calculator and it works with different limits of integration. This is bothering me as well I'd love to get some feedback.
@da1booger13 you are probably right, i am too lazy to rewatch
you made calculus soo much easier for me
Hmmm, patrick, at 7:20, shouldn't there be a 1/secx there? I say this because of when the 'u' was taken to equal secx. In that case, only the tanx would have remained in the integrand. Thus, when taking the derivative of the u value(secx) which equals 'secxtanx', that derivative would have only crossed out the tanx in the integrand. So it would seem to leave you with a 1/secx. Anyway, I might be brutally wrong, please take me in the right direction if so.
I get the answer (1/4)*(tan x)^4 -(1/2)*(tanx)^2 + ln|secx| + C. I have varified this answer from other sources which claim it to be right?
How do I know which answer is the right one? Yours? or The one above? or both? My exam of calculus will be based on multiple choice questions.
Thanks so much for the great videos!
u = something its the same than
v= something and its the same than
w= something
z= something
in fact you can use any variable for subtitution, same idea
they are one and the same.
Thanks a lot.
excellent : )
I'm confused , we didn't solve like that!! the result is''' tan^4x\4 - tan^2x\2 - ln /cocx/+c'' :'(
You are right
It's the same thing, you just plug in the trig. identity value of sec^2x into Patrick's final answer and it becomes what you are saying. Also just 2 key things, you are gonna get an additional constant in the form of a fraction, just mesh it up and into "c", 2 constants added together make a constant after all.
2nd, log/secx/ = log/cos^-1 (x)/ , using property of log the power gets multiplied to log, so log/secx/ = -log/cosx/ . In the end, yours and Patrick's answers are the same :)
(I know this is like 3 yrs late, but just writing this for anyone else who may be confused and happens to read this thread)
great again
youre awesome dude
math is beautiful.
IMPORTANT QUESTION: at the end of the video, why did you make v=something....instead of just sub in u=......?? I have seen all ur other videos, and i fully understand the rules....but, in this case, what do i do? thanks
for the last question the middle term you subbed v = tanx but you could have also done u = secx right? although the final answers are different...
When you integrated (secx)^2(tanx), I paused and tried it myself. I had v = secx and dv = secxtanx instead. I substituted v back in after I was done, and I ended with (secx)^2 which definetly does not = (tanx)^2. What went wrong?
So if I wont use the natural logarithm of tanx, it can also be equal to ln(cosx) right?
2nd example can be easily done just by substituting the tan^5xdx with (sec^2x-1)^2 *tanxdx.
what answer did you get. i dont get the same as he did. all my answers r in terms of tan
sir what to do if we have given only secx or its power in the integral??
i think its just another variable..you can use any letter u want
for ur second question with tan^5x, at step 4:52 cant we just take U= sec^2 and thus du = tanx dx ? after that its a simple integral
Thanks man..........
Ln(1/cosx) is the same as Ln(1)-Ln(cosx). since Ln(1)=0, so 0-Ln(cosx).
Hey Patrick couldn't you have just set sec(x) = u in the very beginning and gotten ((sec^2 x)^2)/sec x and after doing a u sub, have (u^4 - 2u^2 +1)/u which ends up making you integrate u^3 - 2u + 1/u thus being (u^4)/4 - u^2 +ln|u| which ends up (sec^4 x/ 4) - sec^2 x + ln|sec x| ?
Amazing, it's as if my textbook started talking and writing.
just factor out 1 sin(x) on top and change sin^4(x) to (1-cos^2(x))^2, then let u=cosx so du=-sinxdx
Would this work?
you're sal's doppleganger. white background, left handed...and badass.
This was an awesome vid. I have one thing to say though. You for to put dx on the du part that was equal to ∫secx*tanx
But It was still a great vid!!!
I want to be just like you when I grow up!
hard to tell because it went to the next line but du= -sinxdx (negative sinxdx)
The user substitution wasn't used for the secxtanx part because the derivative covered it. So only sec^3x was substituted with u and so the secxtanx became the du.
I realize I am a year late :/
Gotta ask: is it okay to use -ln |cosx| + C for tangent?
I know this is kinda late but yes! because of log rules, you can move the negative out the front of -ln |cosx| into an exponent so it would look like +ln|cosx^-1| which is the same thing as +ln|secx|
Thank you! Jonodude0
My prof did this in class today and didn't explain why it was the same. Thanks for clarifying that for me
In the second example, can you simple make it: integral of (tanx)^5 and use u-sub from there?
yeah but you are gonna need sec^2 x where are are you gonna get that from?
Thanks for your videos. They are very helpful. But..
The answer for (tan^5 dx ) in secound example is wrong. I checked manually and by computer too. The final answer is (1/4) * tan^4(x) - (1/2) * tan^2(x) - ln( cos(x) ) + C.
it's not that its wrong per se... it's just that due to log rules, you can move the negative out the front of -ln |cosx| into an exponent so it would look like +ln|cosx^-1| which is the same thing as +ln|secx|
Thanks. I got it. It was my mistake.
'cookie cutter?? Nigga I find this difficult but still an amazing tutorial video. Keep up the good, I will but ur app to show my appreciation
i think the answer at ex. 2 is sec^4(x)/4 - sec^2(x) + ln |sec(x)| + C ..because the 2tanxsec^2x are not satisfied with v^2dv
+Pauline Tabajonda v = tan(x) and dv = sec^2(x). In the problem there, you have 2(integral) tan(x)sec^2(x) dx, so replace it with dv and v; So you get 2(integral) v dv which is (2v^2)/2. and now you replace v back with tan(x). I know this is 7 months late, but I don't see an answer anywhere. Toodles.
+Zike i think you can split the sec^2xtanxdx into secx(secxtanx) then let u=secx, du=secxtanx for -2integ of udu thats where the -sec^2x came up. Got the same answer ^
sec^2(x) + c = tan^2 + c so calm down about the middle term too
what do you mean by one and the same?
tanx = - ln abs(cos u) = - ln abs(sec u) ?
thats not what i meant....watch the end of the video....sub in u or v is not the same thing!..u will get a different answer.
you made a mistake you middle answer should have been sec squared not tan squared
where have you been all my life
Solve sqrt(tanx)
when the integral was broken into 3 integrals ,, in the first 2 we had tanx to an odd power an secx to even powers why we took u secx in the first one and u=tanx in the second one,, both cases look the same to me ' odd power on tanx ad even power on sec?!!
The integral of tan(x) is: -ln(cosx)+C , isn't it? I didn't quite catch why is the ln(secx)+C...
Thanks a looot
Wait a sec.I just need a clarification about the integral of tanx is it the ( ln |secx| + c ) or the ( ln |cosx| + c ) ? I'm confused a bit. Sorry for the question i just want to answer my quiz correctly.
what if tanx is even power and secx is odd power?
I think it is the integral of tanx is ln cos x....
@Miguel11adjr Thank you!
U^4/4 does the +C not tag along? why not?
you saved my future
On the last question where part of the integral is 2tanxsec^2x
can you make do it like this instead
integral of tanxsecxsecx u = secx, du = secxtanx
integral of udu
sec^2x/2 + c
But another way of doing it as shown in the video
u = tanx du= sec^2x
integral of udu
tan^2x/2 + c
Are these both equivalent?
+Sajeed Bakht that's exactly how I did it.
This goes back to the last video, where if you have an intigrand in the form of (Tanx)^n * (secx)^m. And m is an even number, you use tanx as your u in the substitution.
They are equivalent, if you take into account the constant you add to each one. sec^2(x) = tan^2(x)+1, and therefore (1/2)sec^2(x)+C = (1/2) tan^2(x)+1+C = (1/2) tan^2(x)+D.
What happens if I encounter an odd m and an even n
can i use tan^5 = sin^5/cos^5 ??
Can't we then use u substitution, & make u=sinx?
Arg this sucks, these problems can be broken up in a few difference ways and they all yield different but equivalent answers... My professor gives multi choice problems :((((. Going to have to do each problem a few times to be sure it isn't a (None of these) choice.
maybe you should raise that point to the professor. seems valid to me at least.
If your prof gives you multiple choices as the answer, just do their derivatives.
I actually ended up doing excellent in his class. He does get really tricky with some of the answers but I've been able to figure it all out. Final next week and I'm carrying a 97.25 average on the tests.
CrushOfSiel
Multiple choice..........Your lucky.
Nah, he is tricky with it so bad. I guess one thing that could be lucky is I have a 1/5 to 1/12 chance to guess right instead of almost 0 percent change. (he sometimes has up to 12 options).
I don't know how he gets some of his answers, they are simplified all crazy. No partial credit for work either.