... 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.
Professor Prime Newtons, thank you for the video. Calculus Two playlist on UA-cam 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 UA-cam TV with Professor Prime Newtons.
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 "❤❤❤❤"
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
... 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
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
i had the integral 1/sin(x) but nothing worked I tried for like an hour and then I figured out with some googling that apparently it only works with tan
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
Newton.. sincerely speaking you have helped me alot..🙏
Glad to hear that
@@PrimeNewtons thank u..I'm a Kenyan student
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?
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.
you are a perfect teacher, moreover a perfect human
I am a teacher, but whenever I watch your video, everything is just fine to teach
Thanks for this lesson . You are a good teacher .
Thanks Newton, that was so good. I really enjoyed that.
Keep up the great videos.
👍
Same here. 👍
You are a good teacher, sir
Excellent refresher. Thank you.
Wow, this is a great video. You have such an excitement inducing voice. You're really getting the beauty of maths across.
Professor Prime Newtons, thank you for the video. Calculus Two playlist on UA-cam 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 UA-cam TV with Professor Prime Newtons.
Brilliant way to solve sir
best teacher ever
Thank you for the nice example and exposition. Added another tool to the toolkit.
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 "❤❤❤❤"
That is great, but I have question could we use t= tanx
Or we have to make angle x/2
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.
life saver! great explanation, thank you!
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
... 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
Is there a relationship between t-substitution and the half-angle identities?
Thank you so much :D
It’s such a great explanation
Well explained sir❤
Clearly explained 👏👏 Thanks
thank you
Fantastic
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
Excellent !
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
Thanks sir
Thank you! 😊
amazing
Very helpful! ❤
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
What if there are
non linear function
Amazing video. You are so damn smart.
Loved it
Thanks a lot man
Try by multiplying the denominator and numerator with conugate.
Why did you choose x over 2 and not just Theta?
That's the substitution that works
How come the day after ive had my calculus 2 exam this video gets recomended 😒
i had the integral 1/sin(x) but nothing worked I tried for like an hour and then I figured out with some googling that apparently it only works with tan
Sir you also do it by substituting 1+sinx=
l got you
I like u boss
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