Thanks for your suggestion, sir. I will keep it in mind. My initial plan was to use this fall for calculus - 1, and then work back on calculus - 2 during next Spring. Kind of planning a systematic collection of videos as time permits.

Please check this: kzbin.info/www/bejne/gaHWpaFvepaGm9Usi=iRS3L4m4DTcgllvh

Thank you.

Thanks for the comment. I’m glad you find it helpful for your studies.

👍🏻👍🏻👍🏻👍🏻😊👍🏻👍🏻👍🏻👍🏻

👍🏻👍🏻👍🏻 thank you so much.

Yes, you’ve got a valid point. Ideally to keep all math majors calm, and without getting too complicated to non-math majors, the problem should’ve been given as X>2. The suggested answer in your comment works, and an alternative popular answer for the same is using two indicator functions, I(x<2) and I(x>2), instead of the absolute values. I’ll have another video at some future point on this. Thanks for the comment.

@@gayansamarasekara no problem. Are you assuming that everything is positive if that's what you're talking about?

@@justabunga1well, I think I wasn’t clear enough, sorry about that: For indicator functions, there will be a negative sign with I(x<2) and positive sign with I(x>2), with ( (x-2)^2)/2 multiplied with each, so not limiting x to be greater than 2. However, if it was given x>2 in the problem itself, no indicator functions or absolute sign needed, since x-2 is positive only. I think that’s what I was trying to comment that day. Thanks.

You may integrate those two, one by one. The integral of e^-x is easy: -e^-x. Similarly, use the result integral of a^x, for a>0, which is: (a^x)/(ln a), for the other part with a=1/3. Just a tiny bit more simplifications can be done. Hope this will help. I’ll try to have a video on that next week or so. Thanks for your question.

@@gayansamarasekara actually 3^x + e^-x are together like 1/(3^x + e^-x) dx

Thank you for your nice comment. Those days I was trying to make travel videos, which nobody watched. Then switched to math, hah ha 😁🤗.

Thanks for your nice comment. Yes it’s for sure a very useful theorem. Many substitutions can be eliminated, and many standard results (e.g: integral of cot x, etc.), can be obtained directly without much effort. I’m not too sure what the names of those theorems are, I learned them some 25+ years ago, in a book of worked examples, which had those theorems labeled as theorem 1, and 2, without any specific names (if I remember right). In my calculus classes, I teach them as special theorem 1 and 2, under the section of integrations by substitutions. I haven’t seen those two theorems in any popular textbooks that they use for teaching calculus in the US universities.

My pleasure. Indeed it’s a very useful theorem, it has many applications, plenty of substitutions can be eliminated using the theorem. Unfortunately many US calculus textbooks don’t have this theorem, so they usually don’t teach it. I learned it some 25+ years ago while getting ready for an extremely competitive university entrance exam, I found it those days in a book of target problems, under a set of worked examples. Thanks for your comment.

You are welcome.

Thanks ma’am 👍🏻

Thank you.

Wa Alaikum Salam ✅

Yes, a generalization of the theorem can be found here: CAN YOU EVALUATE THIS DEFINITE INTEGRAL? kzbin.info/www/bejne/gaHWpaFvepaGm9U

Thank you.

Thank you.

I’m good, thank you. How are you?

I'm okay

Wa Alaikum Assalam (this time I used AI to generate the response - still kind of difficult to remember the phrase, and spelling).

Assalamu Alaikum is an Arabic word that means I wish you health and safety. As-Salam is a name of Allah

❤️❤️❤️❤️ thank you

Indeed 👍🏻

Wasalamu Alaikum Mr

Thank you, I’ll always try.

Wasalamu Alaikum

You are welcome 👏

Please check it here: THEOREM 2 OF 2: INTEGRATIONS BY SUBSTITUTIONS kzbin.info/www/bejne/gIKmeJKkrpuUmq8

That is correct, it’s a popularly used name for the method in many textbooks.

Cheers

You are welcome a lot.

❤️❤️❤️❤️

Thank you…!

Wa-Alaikum Salam…! I think I got it right (last time I had a Muslim friend sitting next to me, he helped me figuring out the correct response. Today it’s all by myself 😊).

Nice to hear that. To go beyond college, there are scholarships available at grad schools around the US. Please check their entry requirements. Almost all the PhD programs are available at no cost out of pocket for the graduate students, who are willing to work as GRA or GTA, which will be 20 hours per week (practically less than that, but they pay for 20). You’ll need GRE, recommendation letters (2 or 3), a personal statement, online application and fee, under graduate transcript, technically for any graduate program. I haven’t authored any textbooks yet, I have some research publications only. Thanks for checking.

My pleasure. Thanks for your comment.

Thank you so much. Your comment gives me confidence.

My pleasure, and thanks for the comment. Yes, the power reducing formula can also be used, though it may be better for larger powers, such as the discussion here: HOW TO DERIVE AND USE REDUCTION FORMULA FOR INTEGRATING SINE N-TH POWER - A WORKED EXAMPLE INCLUDED kzbin.info/www/bejne/aJyQaaiMaLqlaqM

Good question. Yes, n can be technically any real number greater than 2, in order to reduce. However, after reducing it to an integral whose power is less than 2, you’ll have to manually evaluate an integral, which may be difficult. For example, if n = 5/2, after applying the formula once, you’ll have to evaluate an integral involving 3/2 power. Since that power is less than 2, further reductions won’t help, and integrating tangent raised to 3/2 power is difficult (could be not so difficult as a definite integral, if the limits are given, at the very least the Simpson’s Rule can be applied, therefore, for any practical engineering/design matters, the solution is easily reachable, and therefore reduction still helps to convert a higher order integral into a lesser order integral).

Thanks for the comment…! Thanks for staying with my channel.

👍🏻👍🏻👍🏻

Kansas.

Nice to meet you too. I go by Sam, I am a professor. I am also interested in mathematics.

Very good Mr ​@@gayansamarasekara

Of course. Thanks for the comment…!

Of course. Glad to hear you already knew the method. Thanks for the comment.

Wonderful to hear that. Thank you so much for the comment…!

Thanks for your question. Right now, I’m working on something, I’ll get back to you tomorrow. Sorry about the delaying. If you paid attention to the changing + / - signs of the integrals (because sine/ cosine repeated integration leads to alternating signs), along with the method specific changes of + / - signs, your answer should be correct, provided you followed differentiation and integration in each step correctly.

Correct. Also called the DI method.

No. I’m a Buddhist.