9:38 Yeah, but that worked only because the repeating factor was just "x", how can you prove the formula to be valid if the repeating factor was anything else?
Thanks for the reply! I tried the t=x+2 substitution and I ended up with the same four partial fractions of the video, except in the third I got C-4B in the numerator instead of just C and in the fourth I got 4B-2C+D instead of just D, but since they're all some sums of constant terms that do not involve x, I'm assuming I can easily replace those two numerators with, say, C' and D' respectively.
I am an engineering student, in a course, we should those partial fraction decomposition to solve Laplace questions in transform calculus for advanced math engineering, and I couldn't remember how to decompose them, and you SAVED my time and helped to remember all. Thanks!
Thank you! Currently taking Differential Equations and I forgot everything I learned from Calc 2 about partial fractions. Your videos always save the day
2:01 Thank you for the shot showing the difference between the options for the various forms the rational forms should be in! Also, I really liked how the different segments describe the individual steps/possible setups in order (with a couple of examples regarding a particular step) instead of merely solving different problems like most videos! Awesome explanations & video!
I love how you learn something like partial fractions in highschool or longdivision in elementary school, and then you dont use it for 4+ years and then get into calc 3 or diff eq at college and you need to look it up again because its expected to be memorized. I think it's less the fault of the teacher on either end but the curriculum in the middle not reinforcing some of these things that aren't just esoteric one offs for non math majors.
You have a test already by September 15? My kids just started school today. They'll learn how to do basic fractions this year, not partial fraction decomposition. They're only in 3rd grade.
@@Eltodofull Diff Eq was the hardest class I took. I studied philosophy, not math, but I really wanted to learn about chaos properly to apply it in philosophy. So I struggled through the class. Really interesting stuff, but hard.
I use patrial fractions to calculate following things 1. Integrate rational functions 2. Inverse Laplace transform I like solving recurrence relations with generating functions (ordinary and exponential) After you get generating function you have to express it as a sum of geometric series and their derivatives which looks similar to partial fraction decomposition but it is not quite the same thing In integration in repeated complex root case i prefer to use reduction formula which can be derived with integration by parts and linearity In inverse Laplace transform in repeated complex root case i use convolution theorem
There is a way to work with repeated complex roots, without using convolution. You can construct a linear combination of the transforms of sin(t), cos(t), t*sin(t), and t*cos(t). Properties of even and odd Laplace transform functions can often eliminate two of these choices. You then use the s-derivative theorem to find L{t*sin(t)} and L{t*cos(t)}, and solve for the unknown coefficients.
Perfect format for this kind of overview video - where you are laying out many cases and examples so we can compare side-by-side. Thank you so much! I also love that you gave us a few practice questions. 🙂
My first cal 2 exam I failed but i started watching your videos and doing the practice sheets you post and have became very fluent in my trig identities, trig sub, partial fractions, u sub, and improper integrals✊🏻
@@oralia3340 awesome sauce girl!! I ended up with a A in Cal 2. You got this! Currently in differential equations and its so much easier compared to cal 2.
@@12degreesnowman11 depends on the university. Im a mechanical engineering student so we need dif q for fluids. Didn't use much of cal 3 in dif q though
Hey! Great video! I'm taking Differential Equations and I needed to refresh on PFD! Thank you very much! You included all the details! Have a great week ❤
Sometimes the degree on the numerator may appear to be higher than the degree on the denominator. Which we can reduce by long division. I expect to gonna explain that too 💪💪
Love these examples! They’re similar to what my professor showed us! And then to prove how amazing my professor is, he then proceeded to quiz us on PFD integration of 1/(u^8+u^6+u^2+1) :’’’’)
That can be simplified to 1/(u^6+1) (u^2+1) But still, that's just evil if you expant it and try to do it by partial fraction. Yaa, it's just too long 😂😂
For 2:00, I worked it through on my own before to generalize it, and I prefer thinking of it for n > 1, since you don't really see negative exponents. So, it's just -1/ [ (a*(n-1)) * (ax+b)^n-1].
The technique of partial fraction decomposition (not really the motivation or the final steps of integrating the resulting fractions) is one you can easily do in algebra 2 or precalculus. In some books it is part of the systems of equations chapter.
That it looks interesting is useful data about yourself! That’s awesome. Just keep learning new math skills, and this will eventually become easy for you. 🙂
Wow! Such a wonderful video, Blackpenredpen, you are my favorite youtuber, always try to be the first one to watch your videos, it would be nice if you could respond to this comment.
Logical Proofs I will reply if I want to. You presented a problem, and I presented you a fact about the problem. It is not my problem if you do not like it. How about you present the solution yourself, since you are acting so high and mighty? I would love to see you solve this equation step by step and provide all 3 complex solutions in exact form. If you are not able to do so, then maybe you should stay silent and stop acting all arrogant for no reason.
Thank you for this video! At 10:17, in the 6th example, how to find the values of A,B and C? Whatever we substitute to make the B and the C terms zero would also make the A term zero.
For the case of repeated roots, you can use Heaviside cover-up to find the coefficient on the highest power of the repeated term. I prefer to assign the first letters for terms I can get with Heaviside cover-up, and put those first. Then, I descend the power for all other terms of the repeated root. In his case, that's coefficient B. Coefficient C can also be found with Heaviside cover-up. Once you use x=0 for Heaviside cover-up, that value is spoken-for, and can't directly be used again. You can use other unrelated values of x, and you just need as many choices for x as you have unknowns. There is a general method for Heaviside cover-up, that involves taking derivatives and using the same value again, until you find all coefficients of the repeated power. However, I find it is more trouble than it is worth. There is a trick with repeated roots, where x=infinity can be one of your values. It involves partially clearing the fraction, by multiplying by the repeated root just once. Then take the limit as x goes to infinity. Here's how I'd solve his example: Given: (2*x - 5)/[x^2*(x + 1)] Set up partial fractions: A/(x + 1) + B/x^2 + C/x Use Heaviside coverup for A & B: At x = -1, A = (2*(-1) - 5)/[(-1)^2] = -7 At x = 0, B = (2*0 - 5)/(0 + 1) = -5 You could plug in x=1 as a strategic value to find C. Alternatively, you can partially multiply by one copy of x: (2*x - 5)/[x^2*(x + 1)] = -7/(x + 1) - 5/x^2 + C/x Multiply by 1 copy of x: (2*x - 5)/[x*(x + 1)] = -7*x/(x + 1) - 5/x + C Take the limit as x goes to infinity: 0 = -7 - 0 + C C = +7 Result: -7/(x + 1) - 5/x^2 + 7/x
That's because I think his wrong starting @5:30 case 2: irreducible quadratic factors. (I commented this to him) "I think you're wrong on Quadratic partial fraction form, the first Constant (in this case B) should be multiplied to the derivative of the denominator + the 2nd Constant (in this case C). The partial fraction form @6:18 should be (B(2x)+C)/(X^2+4). And the rest of this video has a wrong Quadratic partial fraction form."
@@janconfusio7178 no, what he’s done is correct for case 3, as I’ve done multiple times since I watched this and it’s always like this, as that’s what your meant to do for a brackets/symbol to a power. Case 2 I’m not so certain on as I’ve always done it a different way, but case 3 he’s done correctly
Will you upload old worksheets back to your website? (e.g. work, hydrostatic force, some other stuff) Thank you for keeping my feed interesting these past years!
You could treat (x - 2)^2 as a quadratic, and go through the same exercise to set it up as if it were an irreducible quadratic, and you'd get an equally valid result. It's generally easier if you use the fact that it is a repeated linear factor, rather than a quadratic, to your advantage, since Heaviside coverup will help you get one of the coefficients on it. Using a repeated linear factor helps get you closer to where you ultimately want to be, for an application of this process, such as integration or Laplace transforms. As an example, consider: 1/((x - 3)*(x - 2)^2) Treating (x - 2)^2 as if it were an irreducible quadratic, we get: 1/((x - 3)*(x - 2)^2) = A/(x - 3) + (B*x + C)/(x - 2)^2 H-cover-up for A, at x=3: A = 1/((3 - 2)^2) = 1 Thus: 1/((x - 3)*(x - 2)^2) = 1/(x - 3) + (B*x + C)/(x - 2)^2 Let x=0, to solve for C: 1/((0 - 3)*(0 - 2)^2) = 1/(0 - 3) + C/(0 - 2)^2 -1/12 = -1/3 + C/4 C = 1 Let x=1 to solve for B: 1/((1 - 3)*(1 - 2)^2) = 1/(1 - 3) + (B*1 + 1)/(1 - 2)^2 -1/2 = -1/2 + B + 1 B = -1 Result: 1/((x - 3)*(x - 2)^2) = 1/(x- 3) + (-x + 1)/(x - 2)^2 And this is equivalent to what you'd get, if you did use the repeated linear factor to your advantage, which is: 1/(x - 3) - 1/(x - 2)^2 - 1/(x - 2)
can u use your old technique of teaching with red and black pens? I loved the old version where you had a ball shaped mic in your one hand and another hand with two pens busy on the board..
tan(2x) = 2·tan(x)/[1 - tan(x)^2] = 2·tan(x)/([1 - tan(x)]·[1 + tan(x)]) = ([1 + tan(x)] - [1 - tan(x)])/([1 - tan(x)]·[1 + tan(x)]) = 1/[1 - tan(x)] - 1/[1 + tan(x)]. This may be useful in a situation where you have an integrand with several trigonometric functions with an input of x but with the exception of the tangent, which has input 2x. Otherwise, I cannot imagine this being particularly useful.
![Megan Thee Stallion - Roc Steady (feat. Flo Milli) [Official Video]](http://i.ytimg.com/vi/HozPPyqW0dY/mqdefault.jpg)
9:38 Yeah, but that worked only because the repeating factor was just "x", how can you prove the formula to be valid if the repeating factor was anything else?
You can do substitution. Like for ex7, you can let t=x+2
And it will work like ex6 : )
Good point tho I will pin it so that others will see it. Thanks
Thanks for the reply! I tried the t=x+2 substitution and I ended up with the same four partial fractions of the video, except in the third I got C-4B in the numerator instead of just C and in the fourth I got 4B-2C+D instead of just D, but since they're all some sums of constant terms that do not involve x, I'm assuming I can easily replace those two numerators with, say, C' and D' respectively.
@@nomoremathhere just subbed to your channel, extra cool content!!
@@nomoremathhere Prove it without taking substitution
As a math tutor, this video explains it perfectly and it is not taught with this level of detail in most calc courses. Thanks BPRP!
Glad to hear. Thank you!
This man saved me in 11th and 12th grade. Now is coming clutch again when Im in college! He's da GOAT
Altough the board is a sacred tradition, I also enjoy this format of videos. Keep it up!
Thanks
Did he try to bait us into the video with a sneak peak of him cancelling x like a vlog youtuber?!
I’m deadddd
I am an engineering student, in a course, we should those partial fraction decomposition to solve Laplace questions in transform calculus for advanced math engineering, and I couldn't remember how to decompose them, and you SAVED my time and helped to remember all. Thanks!
My professor showed us partial fractions today, this helps a lot, thank you!
Thank you! Currently taking Differential Equations and I forgot everything I learned from Calc 2 about partial fractions. Your videos always save the day
2:01 Thank you for the shot showing the difference between the options for the various forms the rational forms should be in! Also, I really liked how the different segments describe the individual steps/possible setups in order (with a couple of examples regarding a particular step) instead of merely solving different problems like most videos! Awesome explanations & video!
I love how you learn something like partial fractions in highschool or longdivision in elementary school, and then you dont use it for 4+ years and then get into calc 3 or diff eq at college and you need to look it up again because its expected to be memorized. I think it's less the fault of the teacher on either end but the curriculum in the middle not reinforcing some of these things that aren't just esoteric one offs for non math majors.
I literally have a test on this tomorrow thanks so much for a good review!
Good Luck :))
Play 2x speed for review
You have a test already by September 15? My kids just started school today.
They'll learn how to do basic fractions this year, not partial fraction decomposition. They're only in 3rd grade.
Mine was yesterday, a differential equation, autonomous, it had 3 factors.
Guess what, the time gave negative :D
It's likely wrong
@@Eltodofull Diff Eq was the hardest class I took. I studied philosophy, not math, but I really wanted to learn about chaos properly to apply it in philosophy. So I struggled through the class. Really interesting stuff, but hard.
Best math channel on RUclips, you know the deal!
LOVED THIS! I was confused and after watching this, it clicked. Thanks!
You’re welcome! I am happy to help!
Your videos help me A LOT. It's all so well-structured and straightforward. Thank you!
I salute to your timing
I required this video today and you uploaded it
It took just the first 7 seconds of the video, folks, for him to answer my biggest lingering question with partial fraction decomposition. 7 SECONDS!
: )))))) thank you!
I found the first 2 minutes super helpful as strategy for what formats to aim for !
Screenshotted the board at 2:10 for review!
same bro, neither my teacher could explain it to me in 5 minutes jeje
I use patrial fractions to calculate following things
1. Integrate rational functions
2. Inverse Laplace transform
I like solving recurrence relations with generating functions (ordinary and exponential)
After you get generating function you have to express it as a sum of geometric series and their derivatives
which looks similar to partial fraction decomposition but it is not quite the same thing
In integration in repeated complex root case i prefer to use reduction formula which can be derived with integration by parts and linearity
In inverse Laplace transform in repeated complex root case i use convolution theorem
There is a way to work with repeated complex roots, without using convolution. You can construct a linear combination of the transforms of sin(t), cos(t), t*sin(t), and t*cos(t). Properties of even and odd Laplace transform functions can often eliminate two of these choices. You then use the s-derivative theorem to find L{t*sin(t)} and L{t*cos(t)}, and solve for the unknown coefficients.
One of the only videos I could find that explained how to break up the fraction when it's like the last example, very useful thank you
The new format of the video is amazing, thanks!
Perfect format for this kind of overview video - where you are laying out many cases and examples so we can compare side-by-side. Thank you so much! I also love that you gave us a few practice questions. 🙂
This is my favorite integration techinque and you managed to perfectly explain it in less then 10 minutes i hope you do other techinques as well
Hlo all might midoriya here 😂
🙌
Thanks for simplifying this. This is the first time to open it and I may have an exam on it tomorrow.
Wow this channel just gets better and better
I just wanted to say thank you for everything! I just took my second cal 2 exam today and know I did awesome thanks to you!.
My first cal 2 exam I failed but i started watching your videos and doing the practice sheets you post and have became very fluent in my trig identities, trig sub, partial fractions, u sub, and improper integrals✊🏻
@@JesusMartinez-zu3xl I know this was 7 months ago but great job!! I'm tryna be like you haha
@@oralia3340 awesome sauce girl!! I ended up with a A in Cal 2. You got this! Currently in differential equations and its so much easier compared to cal 2.
@@JesusMartinez-zu3xldo you need calc 3 for differential equations or is calculus 2 enough ?
@@12degreesnowman11 depends on the university. Im a mechanical engineering student so we need dif q for fluids. Didn't use much of cal 3 in dif q though
Was doing complex calculus and needed a quick refresher on partial fractions when this popped up on my feed. Great video!
I mostly used in fractional integration, Thnx for enhancing in advanced level. 👍👍👍👍
Hey! Great video! I'm taking Differential Equations and I needed to refresh on PFD! Thank you very much! You included all the details! Have a great week ❤
Thank you! we just finished integrations in school and this helped a lot!!
PFD also stands for Personal Floatation Device aka a life jacket. A good analogy.
Sometimes the degree on the numerator may appear to be higher than the degree on the denominator.
Which we can reduce by long division. I expect to gonna explain that too 💪💪
Thank you so much Sir, I have a test coming up tomorrow.
Great video and a very clear explanation.
Thanks Dave!
Love these examples! They’re similar to what my professor showed us! And then to prove how amazing my professor is, he then proceeded to quiz us on PFD integration of 1/(u^8+u^6+u^2+1) :’’’’)
That can be simplified to 1/(u^6+1) (u^2+1)
But still, that's just evil if you expant it and try to do it by partial fraction. Yaa, it's just too long 😂😂
For 2:00, I worked it through on my own before to generalize it, and I prefer thinking of it for n > 1, since you don't really see negative exponents. So, it's just -1/ [ (a*(n-1)) * (ax+b)^n-1].
I got a completely different math test Friday. Linear Algebra! Already completed Calc I and II
this format was probably more work for you but it was just as good.
Love from india man❤❤. The way of teaching was just fab😍
I swear!! you're the best ever
We truly appreciate your hard work sir 👏👏 👏👏 you have taught us to do maths with fun....😃
Honourable sir, You are genius 💙👼🙏👌🏻👌🏻😍😊
He's back to regular uploads, I hope. :)
So you have changes the format of your lesson?
Well done 💪💪
this is like a pep talk:
you know in life you won't always get what you want
It has really met my expectations
Me: Doesn't understand -everything- anything
whatever it looks interesting
Michael Wu first let me wooooosh myself, then I’ll say that this is an integration technique which you will learn if you learn calculus
@@ericw2391 also useful in telescoping series sometimes
王Eric You also use it in Laplace transform.
The technique of partial fraction decomposition (not really the motivation or the final steps of integrating the resulting fractions) is one you can easily do in algebra 2 or precalculus. In some books it is part of the systems of equations chapter.
That it looks interesting is useful data about yourself! That’s awesome. Just keep learning new math skills, and this will eventually become easy for you. 🙂
Wow! Such a wonderful video, Blackpenredpen, you are my favorite youtuber, always try to be the first one to watch your videos, it would be nice if you could respond to this comment.
see if he responds
Better luck next time :)
You save me men ......
I was looking for that like that 😩😩😩,... when I get your Vedio notification my face :😲😲
pov:11:30 * BlackPenRedPen you won't ask that question in the exam but my calc professor will!!! we need more professors like u
9:42 earns you a like.
Could you show how to solve integrals of the type:
sqrt(x^2+a^2)
sqrt(x^2-a^2)
1/sqrt(x^2+a^2)
1/sqrt(x^2-a^2)
Thanks and keep up the good work!!
The first two can be solved using trigonometric substitution, the latter two require you to know the derivatives of inverse trigonometric functions
Thanks for the video,it's very helpful for integrations
I learnt a lot, thank you so much BPRP!!
I see you from Honduras! 🇭🇳
Thank you for explaining this so well!
thanks I am now able to do DEs with laplace transform easily
This video make it easy.. thanks a lote.
Thank you so much for the video, it was well explained 👏👏👏👏
Thank u sir from india
Awesome video !!!!! like the new format
Wow i literally needed it,Thank you sir...
Let's modify the Tejas question
√(5-x)= 5-x³
Solve for x.
From what I understand, it is not possible to express the solution to this using analytic methods.
@@angelmendez-rivera351 if you are not able to do this question then don't reply.
Logical Proofs I will reply if I want to. You presented a problem, and I presented you a fact about the problem. It is not my problem if you do not like it. How about you present the solution yourself, since you are acting so high and mighty? I would love to see you solve this equation step by step and provide all 3 complex solutions in exact form. If you are not able to do so, then maybe you should stay silent and stop acting all arrogant for no reason.
@@angelmendez-rivera351 He is not arrogant.
thanks it was so helpful ❤
How did you know? We were just started doing solving integration by partial fractions in our class
U r back!!!!!!!!!
Love from India
@ BPRP can u plz derive each and everything about integration factor of differential equations, how it came , how to use it etc
Thank you for this video!
At 10:17, in the 6th example, how to find the values of A,B and C? Whatever we substitute to make the B and the C terms zero would also make the A term zero.
For the case of repeated roots, you can use Heaviside cover-up to find the coefficient on the highest power of the repeated term. I prefer to assign the first letters for terms I can get with Heaviside cover-up, and put those first. Then, I descend the power for all other terms of the repeated root.
In his case, that's coefficient B. Coefficient C can also be found with Heaviside cover-up. Once you use x=0 for Heaviside cover-up, that value is spoken-for, and can't directly be used again. You can use other unrelated values of x, and you just need as many choices for x as you have unknowns. There is a general method for Heaviside cover-up, that involves taking derivatives and using the same value again, until you find all coefficients of the repeated power. However, I find it is more trouble than it is worth.
There is a trick with repeated roots, where x=infinity can be one of your values. It involves partially clearing the fraction, by multiplying by the repeated root just once. Then take the limit as x goes to infinity. Here's how I'd solve his example:
Given: (2*x - 5)/[x^2*(x + 1)]
Set up partial fractions:
A/(x + 1) + B/x^2 + C/x
Use Heaviside coverup for A & B:
At x = -1, A = (2*(-1) - 5)/[(-1)^2] = -7
At x = 0, B = (2*0 - 5)/(0 + 1) = -5
You could plug in x=1 as a strategic value to find C. Alternatively, you can partially multiply by one copy of x:
(2*x - 5)/[x^2*(x + 1)] = -7/(x + 1) - 5/x^2 + C/x
Multiply by 1 copy of x:
(2*x - 5)/[x*(x + 1)] = -7*x/(x + 1) - 5/x + C
Take the limit as x goes to infinity:
0 = -7 - 0 + C
C = +7
Result:
-7/(x + 1) - 5/x^2 + 7/x
Thank you sir,it was really helpful.
super helpful, thank you so much!
Glad to help 😃
thank you,it's really helpful.
9:38 is exactly what I what I was getting confused on
That's because I think his wrong starting @5:30 case 2: irreducible quadratic factors. (I commented this to him) "I think you're wrong on Quadratic partial fraction form, the first Constant (in this case B) should be multiplied to the derivative of the denominator + the 2nd Constant (in this case C). The partial fraction form @6:18 should be (B(2x)+C)/(X^2+4). And the rest of this video has a wrong Quadratic partial fraction form."
@@janconfusio7178 no, what he’s done is correct for case 3, as I’ve done multiple times since I watched this and it’s always like this, as that’s what your meant to do for a brackets/symbol to a power. Case 2 I’m not so certain on as I’ve always done it a different way, but case 3 he’s done correctly
Thank you! I loved the video
Hey, what does the Chinese Math curriculum look like? How are you guys so good at Mathematics?
0:51-0:54 I get a image of you smiling when I first heard this lol
Please can you do a video where you will demonstrate thet we effectively can do this kind decomposition !
Interesting format!
thank you so much for this
Thank you very much senpai☺️
Great work
thank you.
Will you upload old worksheets back to your website? (e.g. work, hydrostatic force, some other stuff)
Thank you for keeping my feed interesting these past years!
Thanks
Very helpful!
You're awesome.
Thank you bro
Ty
Fantastic!
thank you soo much🥰🥰
Amazing ❤️
First Like, then watch!!!
Ingrid Sammie place will be there soon😊😊😊
thanks a lot!
I WANT TO SEE THE BEARD MAN 😭😭😀
Your work is very good.
Is (x - 2)^2 not quadratic?
Why did you use a constant for its numerator instead of a linear function?
Thanks
when you can factorised it, its become linear (x=2). if can't factorised, then remain it as quadratic
You could treat (x - 2)^2 as a quadratic, and go through the same exercise to set it up as if it were an irreducible quadratic, and you'd get an equally valid result. It's generally easier if you use the fact that it is a repeated linear factor, rather than a quadratic, to your advantage, since Heaviside coverup will help you get one of the coefficients on it. Using a repeated linear factor helps get you closer to where you ultimately want to be, for an application of this process, such as integration or Laplace transforms.
As an example, consider:
1/((x - 3)*(x - 2)^2)
Treating (x - 2)^2 as if it were an irreducible quadratic, we get:
1/((x - 3)*(x - 2)^2) = A/(x - 3) + (B*x + C)/(x - 2)^2
H-cover-up for A, at x=3:
A = 1/((3 - 2)^2) = 1
Thus:
1/((x - 3)*(x - 2)^2) = 1/(x - 3) + (B*x + C)/(x - 2)^2
Let x=0, to solve for C:
1/((0 - 3)*(0 - 2)^2) = 1/(0 - 3) + C/(0 - 2)^2
-1/12 = -1/3 + C/4
C = 1
Let x=1 to solve for B:
1/((1 - 3)*(1 - 2)^2) = 1/(1 - 3) + (B*1 + 1)/(1 - 2)^2
-1/2 = -1/2 + B + 1
B = -1
Result:
1/((x - 3)*(x - 2)^2) = 1/(x- 3) + (-x + 1)/(x - 2)^2
And this is equivalent to what you'd get, if you did use the repeated linear factor to your advantage, which is:
1/(x - 3) - 1/(x - 2)^2 - 1/(x - 2)
When will you do Maths in a LIVE? :)
You may give a try
What animation are you using? This is a good way to project math
can u use your old technique of teaching with red and black pens?
I loved the old version where you had a ball shaped mic in your one hand
and another hand with two pens busy on the board..
Number got it pow tks mabrat😅😅😅same methode
My legend.
I haven’t done this in years, forgot how disgusting partial fractions are
great!
We can do partial fraction also in tan2x =a/(1+tanx)+b/(1-tanx) is it useful in the integration or not ?
In some rare circumstances, it may be, but not generally speaking.
tan(2x) = 2·tan(x)/[1 - tan(x)^2] = 2·tan(x)/([1 - tan(x)]·[1 + tan(x)]) = ([1 + tan(x)] - [1 - tan(x)])/([1 - tan(x)]·[1 + tan(x)]) = 1/[1 - tan(x)] - 1/[1 + tan(x)]. This may be useful in a situation where you have an integrand with several trigonometric functions with an input of x but with the exception of the tangent, which has input 2x. Otherwise, I cannot imagine this being particularly useful.