... Newton, what I forgot to say is, keep doing your presentations on the blackboard with just a simple piece of chalk, your handwriting is excellent for this! Jan-W
You can actually solve it much more simply by multiplying the numerator and denominator by the conjugate of the denominator, to get ∫ (1-sin(x))/cos^2(x) dx.
Yeah but this method is very very useful for integrals that are impossible to figure out using other methods without already knowing the answer and working back. Its hard to explain but the "t method" is a very useful tool that some rusty person could use to solve integrals he is unprepared for or cant see the shortcut. Put simply its inductive not deductive. This is an easy example of the t rule.
Nice way to get the answer. But Mr Newton, I did it WITHOUT any substitution. In fact all I did was multiply and divide by the conjugate of the DENOMINATOR ( 1 - Sin X) and there onwards ,it is quite simple really for a good Calc 1 or Calc 2 level student. Your method, though a nice U-sub, seems quite lengthy, if I may say, Mr Newton. To make this a little harder, why not try to solve the same Integral as a DEFINITE integral from 0 to pi/2? It is quite an interesting one indeed, trust me❤❤
Well for your information "SIR", this method (sir newton's one) really saved me since i had to do the integration USING this method in my calculus paper. Although your method is right, i guess your knowledge has made you arrogant, and you are trying hard to prove your supremacy to everyone by spreading hatred on the internet, and i guess that's why you're not teaching here with thousands of subscribers. have a great day "❤❤❤❤"
Professor Prime Newtons, thank you for the video. Calculus Two playlist on RUclips does not cover this topic. This topic is called special substitution, which is part of Techniques of Integration in Calculus Two. This is an error free video/lecture on RUclips TV with Professor Prime Newtons.
Great video professor👏🏽 I have two questions: 1 - can I use this t substituion whenever I have a integral of cosine and sine? And if I have another trigonometric identity can I rewrite this identity in terms of sine or cossine or both to apply this substituion? 2 - how can I apply this substituion in those cases I have in the answer a angle in radians summing the variable on the argument of some trigonometric identity. Like for exemplo how to apply t substituion on the integral of dx/ sen(x) + cos(x). The answer of this integral is 1/sqrt 2 that multiply ln( csc( x + pi/4) - cot( x + pi/ 4) How can I reach the same result with t substituion.
I multiplied the numerator and denominator by the conjugate, 1 - sin X, got 1 - sin x/1-sin²X, substituted Cos²X for 1-sin²X, split the fraction, took the integral and ended up with tan x - sec X + c
Newton you can also solve that integral using the conjugate of 1+sinx , that is multiplying up and the bottom by 1- sinx, and the final result is tanx-secx, just another way to do it. greetings
... A good day to you Newton, Right out of one of my " old and trusted " little math notebooks regarding integrals the following solution path in short: Given INT(1/(1 + sin(x))dx [ multiply top and bottom of the integrand by (1 - sin(x)) ] --> INT((1 - sin(x))/(1 - sin^2(x)))dx = INT((1 - sin(x))/cos^2(x))dx = INT(1/cos^2(x))dx + INT(- sin(x)/cos^2(x))dx [ u = cos(x) --> du = - sin(x)dx ] = INT(sec^2(x))dx + INT(1/u^2)du = tan(x) + INT(u^-2)du [ applying the good old REVERSE power rule, remember Newton? (lol) ] = tan(x) - 1/u [ u = cos(x) ] = tan(x) - 1/cos(x) + C = tan(x) - sec(x) + C = (sin(x) - 1)/cos(x) + C ... etc etc ... I leave the outcome to everyone's preference ... Thank you too Newton for your great performance; I really mean this, an eye opener for me; isn't it called the Weierstrass method? A pleasant weekend to you, Jan-W
Yes! This is an alternative. I call it 'rationalization'. I just get scared with the integral of secx or sec²x or sec³x. They scare me. But certainly, in recent times I have used that until I found my old Engineering Math book by K. A. Stroud. Then it all came back to me. We are dealing with the cold and rains here. It's never been like this before. Now I appreciate sunshine 🌞. And thank you for helping with the name. Truly it's called the Weierstrass Substitution
Can we go backward? We now that sin^2(x)+cos^2(x) =1. Can we sub sin^2(x)+cos^2(x) for 1 and then separate our fractions like this: sin^2(x)/(1+sin(x)) +cos^2(x)/(1+sin(x)) and then keep manipulating our algebra to solve the problem in the video? Please let me know if it is possible? Thank You
... Newton, what I forgot to say is, keep doing your presentations on the blackboard with just a simple piece of chalk, your handwriting is excellent for this! Jan-W
Thanks for your feedback. It means a lot.
This channel has really come at the right time, thank you Newton🙏🙏
Which level
I am a teacher, but whenever I watch your video, everything is just fine to teach
You can actually solve it much more simply by multiplying the numerator and denominator by the conjugate of the denominator, to get ∫ (1-sin(x))/cos^2(x) dx.
Yeah but this method is very very useful for integrals that are impossible to figure out using other methods without already knowing the answer and working back. Its hard to explain but the "t method" is a very useful tool that some rusty person could use to solve integrals he is unprepared for or cant see the shortcut. Put simply its inductive not deductive. This is an easy example of the t rule.
Big man you're doing it more than anyone. I like the way you Tutor us. Keep up the good work manh.
Thank you. Your comment sounds African 😀. Where are you from?
Thanks Newton, that was so good. I really enjoyed that.
Keep up the great videos.
👍
Same here. 👍
you are a perfect teacher, moreover a perfect human
Newton.. sincerely speaking you have helped me alot..🙏
Glad to hear that
@@PrimeNewtons thank u..I'm a Kenyan student
Wow, this is a great video. You have such an excitement inducing voice. You're really getting the beauty of maths across.
Thanks for this lesson . You are a good teacher .
Nice way to get the answer.
But Mr Newton, I did it WITHOUT any substitution.
In fact all I did was multiply and divide by the conjugate of the DENOMINATOR ( 1 - Sin X) and there onwards ,it is quite simple really for a good Calc 1 or Calc 2 level student.
Your method, though a nice U-sub, seems quite lengthy, if I may say, Mr Newton.
To make this a little harder, why not try to solve the same Integral as a DEFINITE integral from 0 to pi/2? It is quite an interesting one indeed, trust me❤❤
I'll give it a shot soon
Well for your information "SIR", this method (sir newton's one) really saved me since i had to do the integration USING this method in my calculus paper. Although your method is right, i guess your knowledge has made you arrogant, and you are trying hard to prove your supremacy to everyone by spreading hatred on the internet, and i guess that's why you're not teaching here with thousands of subscribers. have a great day "❤❤❤❤"
Professor Prime Newtons, thank you for the video. Calculus Two playlist on RUclips does not cover this topic. This topic is called special substitution, which is part of Techniques of Integration in Calculus Two. This is an error free video/lecture on RUclips TV with Professor Prime Newtons.
Excellent refresher. Thank you.
You are a good teacher, sir
best teacher ever
Thank you for the nice example and exposition. Added another tool to the toolkit.
If you remember all your derivatives, the integral could be solved as
(1 - sin x)/(cos^2 x) dx
(sec^2 x - sec x tan x) dx
tan x - sec x + C
life saver! great explanation, thank you!
Great video professor👏🏽
I have two questions:
1 - can I use this t substituion whenever I have a integral of cosine and sine? And if I have another trigonometric identity can I rewrite this identity in terms of sine or cossine or both to apply this substituion?
2 - how can I apply this substituion in those cases I have in the answer a angle in radians summing the variable on the argument of some trigonometric identity. Like for exemplo how to apply t substituion on the integral of dx/ sen(x) + cos(x).
The answer of this integral is 1/sqrt 2 that multiply ln( csc( x + pi/4) - cot( x + pi/ 4)
How can I reach the same result with t substituion.
Very good.
I multiplied the numerator and denominator by the conjugate, 1 - sin X, got 1 - sin x/1-sin²X, substituted Cos²X for 1-sin²X, split the fraction, took the integral and ended up with tan x - sec X + c
That's really smart. Good job
i did the exact same process
Newton you can also solve that integral using the conjugate of 1+sinx , that is multiplying up and the bottom by 1- sinx, and the final result is tanx-secx, just another way to do it. greetings
... A good day to you Newton, Right out of one of my " old and trusted " little math notebooks regarding integrals the following solution path in short: Given INT(1/(1 + sin(x))dx [ multiply top and bottom of the integrand by (1 - sin(x)) ] --> INT((1 - sin(x))/(1 - sin^2(x)))dx = INT((1 - sin(x))/cos^2(x))dx = INT(1/cos^2(x))dx + INT(- sin(x)/cos^2(x))dx [ u = cos(x) --> du = - sin(x)dx ] = INT(sec^2(x))dx + INT(1/u^2)du = tan(x) + INT(u^-2)du [ applying the good old REVERSE power rule, remember Newton? (lol) ] = tan(x) - 1/u [ u = cos(x) ] = tan(x) - 1/cos(x) + C = tan(x) - sec(x) + C = (sin(x) - 1)/cos(x) + C ... etc etc ... I leave the outcome to everyone's preference ... Thank you too Newton for your great performance; I really mean this, an eye opener for me; isn't it called the Weierstrass method? A pleasant weekend to you, Jan-W
Yes! This is an alternative. I call it 'rationalization'. I just get scared with the integral of secx or sec²x or sec³x. They scare me. But certainly, in recent times I have used that until I found my old Engineering Math book by K. A. Stroud. Then it all came back to me. We are dealing with the cold and rains here. It's never been like this before. Now I appreciate sunshine 🌞. And thank you for helping with the name. Truly it's called the Weierstrass Substitution
Excellent !
Brilliant way to solve sir
Just use sinx=(2tan(x/2)) /(1+tan^2(x/2) ) =(2tan(x/2)) /(sec^2(x/2)) it becomes simple by half and double angles relations
Clearly explained 👏👏 Thanks
Fantastic
Very helpful! ❤
That is great, but I have question could we use t= tanx
Or we have to make angle x/2
amazing
Amazing video. You are so damn smart.
Well explained sir❤
Thank you! 😊
Thanks sir
Thanks a lot man
Loved it
How come the day after ive had my calculus 2 exam this video gets recomended 😒
Is there a relationship between t-substitution and the half-angle identities?
Try by multiplying the denominator and numerator with conugate.
What if there are
non linear function
Sir you also do it by substituting sinx with 2tanx/2 upon 1+tan square x/2
I like u boss
Why did you choose x over 2 and not just Theta?
That's the substitution that works
Sir you also do it by substituting 1+sinx=
Can we go backward? We now that sin^2(x)+cos^2(x) =1. Can we sub sin^2(x)+cos^2(x) for 1 and then separate our fractions like this: sin^2(x)/(1+sin(x)) +cos^2(x)/(1+sin(x)) and then keep manipulating our algebra to solve the problem in the video? Please let me know if it is possible? Thank You
this is a very bad method, you should trig identities
love your content bro
Which level sir