We had integral of sin^2(x)+tan^2(x)+cos^2(x) dx on the test. After half an hour of pain I finally got the answer tan(x)+C. First then I saw the quick way to do it (sin^2(x)+cos^2(x)=1).
From the third step integral sec x is ln|secx + tan x| and for integral of tan^2x sec x you can put tan x = t which will lead the answer as tan^3x/3 . So the final ans would be tan^3x/3 + ln|secx +tanx| + *C*
One step prior to the final step, we know that we are getting back to the same thing as in question but still write it expecting some miracle to happen.
The key for these types of integrals is to use integration by parts with u = sec x and dv = sec^2. You can then use algebra to add to other side and then divide by two.
U did it correct till 2nd step,solve integral of secx differently and secx.tan^2x differently. In case of solving secx.tan^2x consider secx=t and then dt=secxtanxdx so in terms of t the integral would be √(t^2 - 1) dt and that's how u solve it
Sunteți extraordinar. Așa profesor îndrăgostit de matematică nu am mai văzut. La noi în România notațiile funcțiilor trigonometrice sunt puțin schimbate, însă urmărindu-vă pe D-tră, le-am învățat și cum le utilizați D-tră. Învăț de la D-tră niște subtililtăți și artificii de calcul matematic, cum nu am mai făcut. Succes în continuare.
Well it's lil poopy because idk. Once you get integral of secxtan²x + secx which is his 3rd step, use algebraic property of calculus which gives u integral of secx + integral of tan²xsecx equal to integral of sec³x. Integral of secx is Ln|tanpi/4 + x/2| (here |▪︎▪︎▪︎▪︎| represents modulus) and integral of tanx.tanx.secx.dx, we know secx derivative is secxtanx so take secxtanx as alpha, then d(alpha) comes out to be secxtanxdx which is noice for us. So you will get integral tanxdx, tanx we will get as root over alpha²-1 by trignometery. So it will become integral root over alpha²-1. That is aplha/2×root aplha²-1 + 1/2×sin inverse alpha. Now just put in substituted values and don't forget da C!!
*True Emotional Fact That Checks Out The Background Music:* That "c³" there is the only barrier that is keeping the eternal lovers "se" and "x" apart from eachother 😔..
we can write that sec³x as secx sec²x and then take √(1+tan²x) in place of secx then take tanx = t and sec²x dx = dt so we will be left with integral of √(1+t²) which can be easily solved
You did wrong. If you do this ∫ sec^3 (x) dx = ∫ {sec(x) sec^2 (x)} dx = ∫ [sec(x) {1 + tan^2(x) }] dx = ∫ [sec(x) + sec(x) tan^2(x) ] dx = ∫ [sec(x) + sec(x) {sec^2(x) - 1}] dx = ∫ [sec(x) + sec^3(x) - sec(x) ] dx = ∫ [sec^3(x) ] dx = ∫ sec^3(x) dx Here +C is not possible. You just went back to your original equation by reversing the method.
I just remembered all the pain from the countless times this has happened to me.. And not just after 3 lines of working... Sometimes after 2 pages of solving
Bhaiya I think you can solve this question :- In third step , 1)You can put secx=t. ( 1st integral: secxtanx ) secxtanx dx= dt 2) For the second integral you can easily solve as we know the formula for integral of secx.
IntSec^3x = sec^2x.secx =(1+tan^2x).secx =secx+tan^2xsecx =log(secx+tanx)+u + c u= put tanx=t,then dt=sec^2xdx Then, it will be t^2dt/secx Now putting secx=(1+t^2)^1/2 Then it will become ( t^2)/(1+t^2)^1/2 Now adding 1 and subtracting 1 in numerator We get =(t^2+1)^1/2dt -1/(1+t^2)^1/2dt Using formula We get t/2(1+t^2)^1/2 + 1/2.log(t+(t^2+1)^1/2) -log(t+(t^2+1)^1/2) Hence this is the answer if you want to get full then put t=tanx
The standard method for antidifferentiating sec(x)^3 is to antidifferentiate by parts, differentiating sec(x) to sec(x)·tan(x) and antidifferentiating sec(x)^2 to tan(x). This gives us sec(x)·tan(x) minus the antiderivatives of sec(x)·tan(x)^2 = sec(x)^3 - sec(x). Antidifferentiating sec(x) is fairly easy, so what remains is doing some algebraic manupulations. You can also do this without relying on antidifferentiating by parts, though. sec(x)^3 = 1/cos(x)^3 = cos(x)/cos(x)^4 = cos(x)/[1 - sin(x)^2]^2. Let y = sin(x), hence dy = cos(x)·dx, so we antidifferentiate 1/(1 - y^2)^2 with respect to y by parts. This is can be done using partial fraction decomposition. The reason I prefer this method over the standard method is precisely because it avoids antidifferentiation by parts, and it avoids the risk of running into loops if a wrong but intuitive choice is made. In this regard, the logic of this method is simpler, even if it does have slightly more steps. I am not a fan of antidifferentiating by parts, though.
Ans tanxsecx/2 + 1/2log |tanx + secx| + c Um in my case i didnt use intergation by parts i simply instead of changing sec²x changed secx as √1+tan²x substituted tanx and arrived at this answer
That's the same question that I got wrong in this term exam 😂, but now i know how to solve it 😅, its ans should be 0.5secxtanx + 0.5 log(secx+tanx) + c
A half mark for remembering +C
technically the integral implies +C already so the explicit one is redundant
Miracle, i always forgot it...
@@ParadoxV5 so technically he got 0
@@TechnoSan09 Vanakkam bro :D
@@saipranavm1112 வணக்கம் 😄
That +C at the end is the icing on the cake.
But what about taste of cake.
Lol 😂😆😆😂
@@codedaily365 OPEN UP!
I keep writing dx, as I mostly do definite integrals.
i dont get it
ah yes, the integral of the function is of course the integral of the function + C
@S. Fahim Nabeel any integral is like that
Yes
Proof that the value of c is 0 and it is just a made up conspiracy by teachers to cut marks.
@@rohithninan8785 I think you're bad at math because the prove isn't rigorous
#Only valid for e^x 😂
How can something be *funny* and *sad* at the same time?
_Quantum Mechanics Intensifies_
Funny for viewers
Sad for the calculus studentz
@@princerajan3235 so how about a calculus student viewing the vid? 😈
@@ByakuyaKuchiki006 Like me :|
@@ByakuyaKuchiki006 who get to solve this question in reality !!!
R u solving along with him ??
BTW I am also calculus student.
*I'm writing this with a heavy heart*
_You are now stuck in the never-ending loop called the _*_loop of Integration_*
_Which nobody can escape_
Aa...a...are you still stuck?
@@_mobasshir_ i guess i wl be stuck until someone differentiates me
@C P true friendship
B... but the C keeps getting added up, I'm at C222
@@thebatsbury8053 hundreds of thousands of C's ....one derivative to kill them all..
I feel the pain man i've literally been in this situation n number of times :(
Where n approaches _______.
@@Shreyas_Jaiswal infinity 😭😭😭😭😭😭
@@Shreyas_Jaiswal idk ask Sir RD sharma
XD clever
Bhaiya 4th attempt dia jee main ka
One must imagine a calculus student happy
😢🎉
I tried to imagine it and ended up with a huge i(iota) dancing around in my brain
Im happy but thats only because im self studying. No pressure and i dont have to expose how bad i am at math in a class.
Hmm, yes
1+1 is indeed equal to 1+1
+c
😂@@sohamacharya171
To lighten up the mood of this #sad tragedy, one can say that it is technically true by the reflexive property of equality.
I feel your pain Steve.
Is his name Steve? I've never known his name, but I've been watching for years.
No way his name is Steve. I've been watching him since 10 grade and now I am college
His real name is Chen Lu. He’s Chinese
Someone said Shortcut leads to deep cuts :)
Ohhhh
😂😂😂
"If the shortcut was a shortcut, it would be called a route."
We had integral of sin^2(x)+tan^2(x)+cos^2(x) dx on the test. After half an hour of pain I finally got the answer tan(x)+C. First then I saw the quick way to do it (sin^2(x)+cos^2(x)=1).
Ouch
omg...lol...u would always remember that moment for ur entire life....XD
Prof gotchu
Tanx-x +c
Calculus growing pains.
Lol 😂😂
He was even thought not to forget +C at the ending..
And he did that..
As a math student, I cannot relate to this.
Edit: after reconsidering my life choices, I can now relate to this
😅😂😂
Keep strong bro, keep it strong. #prayforTrrig
the level of frustration xD
This integral is quite simple:
Integration by parts and then the classic move the integral to the other side then divide by 2 thingy
Are secx ko convert kar de √1+tan^2x mai
@@decodedunia6486 yeah same method bro
@@decodedunia6486 cengage op
Yup
Wait seriously?! I did this integral and it was ridiculously long! Man I could have gotten it easier, shoot!
From the third step integral sec x is ln|secx + tan x| and for integral of tan^2x sec x you can put tan x = t which will lead the answer as tan^3x/3 .
So the final ans would be tan^3x/3 + ln|secx +tanx| + *C*
If we put tanx=t , dt=dxsec^2(x) not secx as you seemed to have assumed so this answer is incorrect
@@pratyushshrivastava8791 ya u r right
We could put secx as t then the integral would become root t^2-1 ..which is a standard one ...otherwise we could use by parts
Tanxsecx-tan^3x/3
@@harshitarora2005 do integration by parts after sec²x × sec x that is easiest way
Put u=secx v=sec²x
The pleasure you put in people's minds is what makes your community grow faster 🤣 💪
Look at the third letter of each reciprocal, ull get coSecant for sinus, seCant for cosinus and coTangent for tangent, easy.
You should do shorts like these where you can teach how to do a problem in a short amount of time.. would be a gift for me man.
That's most relatable video I have ever scene😭 I can understand the pAiN😭
Ok I'll remember this in my JEE advanced paper
My recurrent nightmare.
one must imagine sisyphus happy
One step prior to the final step, we know that we are getting back to the same thing as in question but still write it expecting some miracle to happen.
That smile in the intro
The key for these types of integrals is to use integration by parts with u = sec x and dv = sec^2. You can then use algebra to add to other side and then divide by two.
U did it correct till 2nd step,solve integral of secx differently and secx.tan^2x differently. In case of solving secx.tan^2x consider secx=t and then dt=secxtanxdx so in terms of t the integral would be √(t^2 - 1) dt and that's how u solve it
He knows how to solve it, this was just a joke
Ahh the class reversereverse trigonometric identity substitution
Long ago I tried hard to solve this exact integral and I finally got somewhere when I chose to substitute u = sin x
Your facial expression after completing what you have written is Awesome!!
Use chain rule if the function is a composite function like this one.
Let u= sec x
Sunteți extraordinar. Așa profesor îndrăgostit de matematică nu am mai văzut. La noi în România notațiile funcțiilor trigonometrice sunt puțin schimbate, însă urmărindu-vă pe D-tră, le-am învățat și cum le utilizați D-tră. Învăț de la D-tră niște subtililtăți și artificii de calcul matematic, cum nu am mai făcut. Succes în continuare.
I face palmed right when you wrote tan^2•x+1 because I knew exactly what would happen
Well its quite easy, you can solve it by parts. Answer would be 1/2×[secx•tanx + log(secx + tanx)] +c
This had me rolling, ive been in this situation a couple of times 🤣🤣
I love the integral of secant cubed because for some reason it’s equal to the integral of secant minus the derivative of secant divided by two
Taking sec^2 x as t
Secx as t^0.5
Dx as dt/2t√t^2-1 and solving it
Well it's lil poopy because idk. Once you get integral of secxtan²x + secx which is his 3rd step, use algebraic property of calculus which gives u integral of secx + integral of tan²xsecx equal to integral of sec³x. Integral of secx is Ln|tanpi/4 + x/2| (here |▪︎▪︎▪︎▪︎| represents modulus) and integral of tanx.tanx.secx.dx, we know secx derivative is secxtanx so take secxtanx as alpha, then d(alpha) comes out to be secxtanxdx which is noice for us. So you will get integral tanxdx, tanx we will get as root over alpha²-1 by trignometery. So it will become integral root over alpha²-1. That is aplha/2×root aplha²-1 + 1/2×sin inverse alpha. Now just put in substituted values and don't forget da C!!
I was bored meow and no method is poopy mate u is a g
Me on the Question: 😮
1st line: 😊(hehe now he will substitute tanx = t and its solved)
2nd line: 😮
Haha that's actually hilarious and quite relatable😂
I like how he added the constant like a true homie
1/2(tanxsecx+log|secx+tanx|)+c
*True Emotional Fact That Checks Out The Background Music:* That "c³" there is the only barrier that is keeping the eternal lovers "se" and "x" apart from eachother 😔..
Just tried to spice up the mood of depressed guys and gals here to make them enjoy :)
Legends say the C³ stands for Condom³
I just spent half the day screaming about trig, needed this laugh 😂
we can write that sec³x as secx sec²x and then take √(1+tan²x) in place of secx then take tanx = t and sec²x dx = dt so we will be left with integral of √(1+t²) which can be easily solved
Wait till u see the word without the intergral sign, c^3 and dx
One must imagine sysisphus happy.
Thank you for reminding me to take a look at integral of sec³x.
It's done by forcing by-parts integrals.
Thanks a lot!
The way he say eeeeee😂😂
hehe exactly like 18 minus 9 problem
Ah, I see you're a man of meme as well
You did wrong. If you do this
∫ sec^3 (x) dx = ∫ {sec(x) sec^2 (x)} dx
= ∫ [sec(x) {1 + tan^2(x) }] dx
= ∫ [sec(x) + sec(x) tan^2(x) ] dx
= ∫ [sec(x) + sec(x) {sec^2(x) - 1}] dx
= ∫ [sec(x) + sec^3(x) - sec(x) ] dx
= ∫ [sec^3(x) ] dx
= ∫ sec^3(x) dx
Here +C is not possible. You just went back to your original equation by reversing the method.
Me during calc exam with only 20 mins left on question 1
sec cubed is easier to integrate by splitting it and then integrating by parts
I just remembered all the pain from the countless times this has happened to me.. And not just after 3 lines of working... Sometimes after 2 pages of solving
My man couldn't integrate it with that much grief.
u^4/4 . (1+tan square)
Me who barely understand geometry: Ah yes, very sad. I definitely feel you pain. Mhm.
You don't need geometry to integrate trigonometry function
Its calculus bro!
The exact thing had happened to me 😅 but i recognised it before going further.
1/2secxtanx + 1/2 intg secx dx
1/2secxtanx + 1/2 ln | secx + tanx | + C
You can use trig. Identities early but don't use same one twice(which will lead you to question)
Oh man🥺 I failed my exams because of this solution and my feelings now is just like the background song 😭😭😭
Let the whole integral be I. Then split sec³x = secx•sec²x and apply by parts integration. Substitute I wherever needed.
the frustration when writing +C 😂
one must imagine sisyphus happy
Math is the most beautiful language in the world. I wish I was better at it. I’m struggling in my Calculus class right now.
🤣 I actually did this when doing that problem in your 50 integrals video.
Bhaiya I think you can solve this question :-
In third step ,
1)You can put secx=t. ( 1st integral: secxtanx )
secxtanx dx= dt
2) For the second integral you can easily solve as we know the formula for integral of secx.
Honestly this hurts my feelings.
My internal dialogue:
"No.... No... No ... Don't do that ..."
Secx (secxsquare ) then make it 1+tansquare x substitute tanx=t then you would be able to solve
His face at the end 😂😂😂
Integration of secx = log |secx + tanx| + c
IntSec^3x = sec^2x.secx
=(1+tan^2x).secx
=secx+tan^2xsecx
=log(secx+tanx)+u + c
u= put tanx=t,then dt=sec^2xdx
Then, it will be t^2dt/secx
Now putting secx=(1+t^2)^1/2
Then it will become ( t^2)/(1+t^2)^1/2
Now adding 1 and subtracting 1 in numerator
We get =(t^2+1)^1/2dt -1/(1+t^2)^1/2dt
Using formula
We get t/2(1+t^2)^1/2 + 1/2.log(t+(t^2+1)^1/2) -log(t+(t^2+1)^1/2)
Hence this is the answer if you want to get full then put t=tanx
Omfg i did this exactly yesterday! So fucking relatable
The standard method for antidifferentiating sec(x)^3 is to antidifferentiate by parts, differentiating sec(x) to sec(x)·tan(x) and antidifferentiating sec(x)^2 to tan(x). This gives us sec(x)·tan(x) minus the antiderivatives of sec(x)·tan(x)^2 = sec(x)^3 - sec(x). Antidifferentiating sec(x) is fairly easy, so what remains is doing some algebraic manupulations.
You can also do this without relying on antidifferentiating by parts, though. sec(x)^3 = 1/cos(x)^3 = cos(x)/cos(x)^4 = cos(x)/[1 - sin(x)^2]^2. Let y = sin(x), hence dy = cos(x)·dx, so we antidifferentiate 1/(1 - y^2)^2 with respect to y by parts. This is can be done using partial fraction decomposition. The reason I prefer this method over the standard method is precisely because it avoids antidifferentiation by parts, and it avoids the risk of running into loops if a wrong but intuitive choice is made. In this regard, the logic of this method is simpler, even if it does have slightly more steps. I am not a fan of antidifferentiating by parts, though.
I was looking for This! Thanks God bless you
Answer:- 1/2(secxtanx + ln|secx+tanx|) + int. constant
the answer : f sec^3 (x) dx = tan^3/3 + In | sec (x) + tan (x) | + C
cheer up, u have disproved the Halting Problem.
Ans tanxsecx/2 + 1/2log |tanx + secx| + c
Um in my case i didnt use intergation by parts i simply instead of changing sec²x changed secx as √1+tan²x substituted tanx and arrived at this answer
Integ secpower3x= secx*sec power2x
Then by applying integration by parts formula
Ans will be -
I = 1/2secxtanx-1/2ln(secx+tanx)+C😊❤
1/2tan(x)sec(x)+log|tanx+secx|+c
Ans :)
I'm willing to learn calc even though I can comprehend math, as a stem student I'm happy to watch this videoed hoping to understand it one day
sec2x÷2 + ln(tanx+secx)
The +C at the end is where I lost it 😂😂
I love the part where he puts integration constant at the end
The +C & That smile at the beginning 😂
Int √1+tan^2 dtanx😊
At short with integrations by parts with integral equation, so, 1/(cosx)^3=1/cosx • 1/(cosx)^2, where u=1/cosx and v'=1/(cosx)^2 ...
*My heart just said "Already broken😮💨"*
Just remember: "Not to use identity(or a given equation ATQ) in circular order. "
It saves you some frustration.
Right answer is [ 1/2 sex.tanx + 1/2 log(secx+tanx) ] +C
he is obsessed with Pythagoras identities😂😂
Bro know the rules but don't know how to apply
The +c at the last is 🔥
One must imagine the calculus student happy
One must assume Sisyphus happy
That's the same question that I got wrong in this term exam 😂, but now i know how to solve it 😅, its ans should be 0.5secxtanx + 0.5 log(secx+tanx) + c
Thanks to the heavens, it was an indefinite integral, +C came in clutch
I feel personally attacked yet understood at the same time.
I've done a sin^2+cos^2 identity and immediately reversed it lmao
the answer is 1/2 sec(x) tan(x) + 1/2 ln | tan(x) + sec(x) | + C
In third step we should apply f(x)f(x)^n formula