@@Morris-NL Thank you, Morris. Should/when I return to creating videos, I will try to incorporate some clips about these calculations. Unfortunately, I have been drawn away by a host of real life matters that need(ed) addressing. I am still working towards this goal. Kind regards and thank you for your observation and encouragement (and request).

@@JRueTee Wow! Ten years! You are doing well, JRueTee. I am glad that my video helped you. Kind regards, and thank you very much for your feedback.

@@Jaolay Thank you very much, kind colleague. I am glad that you enjoyed the video (and certainly hope that it helps your students). Warm regards from just south of Sydney, Australia.

You have me intrigued, friend. You have been delving into some interesting mathematics. I greatly appreciate your sharing these details with me, but please forgive me if I do not follow through at the moment. There was a time when I would have leapt at the chance before the day was finished ... but I am currently dealing with cancer and a lot of its implications and am not taking time out to explore these things any more ... at least, not until I am significantly better. In the meantime, I hope some enterprising viewers will take up your suggestions and follow in your footsteps. I would be quite curious to hear more. Thank you very much for sharing ... and thank you, in advance, for your understanding of my situation. Kind regards and best wishes to you.

Thank you very much. I am glad that you enjoyed the video and found it helpful (and really appreciate your taking the time to tell me).

Oh wow! That one slipped through! Thank you for your correction, hs. You are quite right. I hope other viewers read your comment. Thank you for picking that up.

I greatly value your contributions to my videos, hs. There are multiple methods for evaluating many integrals and, in each case, I have adopted one of those methods in order to instruct viewers in a particular method or approach. I have wondered about showing multiple methods in the one video (leading to much longer videos), or creating extra videos. I would like to know your thoughts (and those of any other viewers who would care to comment).

Absolutely!

I very much valued your comment on my other video, friend. Would you mind being more explicit about the identity that you would use here, please? Were you suggesting dividing by sin²(θ/2) + cos²(θ/2) and then dividing numerator and denominator by cos²(θ/2)? I agree that that would be a very good way to proceed. Thank you.

@@CrystalClearMaths Yes that is what I mean You have sin(x) = 2sin(x/2)cos(x/2) and we know that 1 = cos^2(x/2)+sin^2(x/2) so let's divide it by this version of one to get 2sin(x/2)cos(x/2)/(cos^2(x/2)+sin^2(x/2)) now we can divide numerator and denominator by cos^2(x/2)

@@holyshit922 Thank you, hs. I appreciate your clarification. It is an excellent suggestion.

Thank you, friend, for your excellent input. Your first suggestion is the other (alternative) method I would use and recommend. I must admit that I am a bit 'old fashioned' and am not familiar with the notation you are using when describing your second option, so I cannot comment on it. I value your input and guidance for those who might view this video. Thank you.

@@CrystalClearMaths Maybe it would be more readable in some Latex editor (and it seems that i missed square in denominator in quotient rule) This second method comes from Миxаил Васильевич Остроградский (Метод Остроградского выделения рациональной части интеграла) In Russian textbooks you can find stuff like this I found this method in Russian textbook translated to my native language

@@holyshit922 Yes. Latex is something that I have not had the time or inclination to invest in (I have had other priorities), so I plead ignorance in that case. I am glad that you are finding value in texts from other cultures/languages. One of my favourite books in my mathematics library is a Russian one. They produce some excellent material.

I am glad that you have been finding my videos useful, friend. I stopped making video due to a number of calamities ... illness and death of my mother and father and a handful of other close relatives, renovating our house (my video room became a storage room), significant ill health (ongoing) and more. I am still planning what to do when our heads get above water again. I used to have a time frame, but am dealing with cancer at the moment and it is dominating a lot of our decisions. Hopefully, I will return to video creating again soon. In the meantime, I greatly appreciate your kind words and encouragement. Thank you.

I am so delighted to learn that you are now doing well in college, vikhyat. That has made my day :-). Thank you. I remember your name and it is so encouraging to learn of your progress. I am dealing with some challenging health issues at the moment, friend. Hopefully, I will get over them and return to making videos. Very best wishes to you (and thank you for remembering to let me know how you are 'travelling').

I am really pleased that this video has clarified things for you, KaasKikker. Please let me know how you go with your test. I hope you do well (better than you might expect). Thank you very much for taking the time to leave your feedback for me.

You are very welcome. I am glad that my video helped you :-).

I am glad that this video helped you, friend. Hopefully, my logic/explanation made the solution more obvious, too. Kind regards to you ...

@@something3476 I am delighted to learn that this video has been of value to you. Thank you very much for taking the time to leave your feedback. It is encouraging and greatly appreciated. Beat wishes for your continuing studies.

You are very welcome, Micah. Best wishes from 'Down Under.'

I am glad that you found my video useful, friend, and wish you well for your November exam(s). Fortunately, you have sufficient time to think deeply about past (careless) errors and to develop and practise new habits that should help you avoid such serious losses in the future. These methods certainly worked for me. Warm regards and best wishes for your studies! Please let me know how you 'go.'

Hello friend. That is an EXCELLENT question ... and I think you are the first student that I have had ask it! Congratulations. I cannot think immediately of a formal geometric proof, but one could place the triangle on a Cartesian plane with the base vertices at A(a,0) and B(-a,0), so that the perpendicular bisector is the y-axis. The apex could be located at C(x,y) where y > 0. Set ∠ABC = α and let ∠BCA = 2θ. The angle between CA and the positive x-axis will be α + 2θ. You can then determine the value of θ in terms of a, b and c by setting up equations for gradients and, thereby determine the equation of the bisector of the angle at C in terms of a, b and c. This will allow you to show that the y-intercept of this line is negative (blow the origin). This method lacks finesse, but would certainly prove that the bisector of the angle at C and the perpendicular bisector of AB will meet outside the triangle. I do not have time to pursue a more elegant proof at this time, but would welcome any suggestions or insights from people. Thank you for such an interesting question.

You are welcome, Terrence. I am glad that you found it to be useful!

@@Dysoma Excellent question, Dysoma. In that case, you may still find all the factors and know whether the leading coefficient is positive or negative (from the behaviour of the graph for large values of x) ... but you willne unable to determine the SIZE of the leading coefficient, as you no longer have a further reference point that is not on the x-axis. Thank you for asking such a searching question.

@@srinathortho Thank you very much. Your comment is encouraging.

In the units column, we cannot subtract 7 from 0. We add ten to the units column in the top number (forming a 10 in the units column), and add ten to the lower number by placing a 1 in the tens column. In the units column, we now subtract 7 from 10 to obtain the answer 3. In the tens column, we now have 0 minus 1. We perform the same manoeuvre to obtain 10 minus 1, resulting in 9 for the tens column. The hundreds column now contains the subtraction 9 minus 1 which results in an 8. The final answer is 893.

You are very welcome, friend. Thank you for leaving your message. Kind regards from Australia.

That is definitely one way to perform the calculation, Antonio! Another would be to calculate 10% of £8 (which equals 160 shillings) ... i.e. 16 shillings. Then one could calculate 10% of 5s 4d (which equals 64 pence) ... i.e. 6.4 pence. The total would be 16s 4d, and one would proceed in the same way that you described. Which method would be more appropriate would depend largely on the figures involved (and how one felt at the time). Thank you very much for your helpful contribution!

@@CrystalClearMaths Your method sounds a lot better, in separating pounds from shillings and pence, and I will try to memorise it. I didn't realise how straightforward it was, and thought that's why the £sd system was abandoned. I can calculate percentages mentally using your method. I bought an old pre 1971 arithmetic book, via ebay, to practise the maths. I've improved my arithmetic skills since doing this: I've memorised the 12 times table, and all the times-tables as we did when children at school. The half-crown was 2s/6d, (12.5 new pence) and there were 8 to a pound. A guinea is 21 shillings. There are four farthings to a penny.

@@atonio2476 I am glad that you have been discovering all this, Atonio. In some ways, it is quite an enjoyable exercise (changing bases between measurements). Thank you for your reflections.

@@CrystalClearMaths We weren't told about the older system at school, and would have benefited from one lesson, to familiarise us. The 'shilling' is a feature in older literature such as Charles Dickens, and this cultural aspect has been lost with decimalisation. I find that with decimal addition, the loss of the shilling creates a gap in the psychological processes.

@@atonio2476 I agree. It is mathematically (arithmetically) beneficial to learn other base systems and culturally and historically important to be aware of these measurements, too. I enjoyvreading historical books and it helps hugely when we understand references like 'shillings.' Thank you for your important observation.

Hehehehe ... thank you :-).

Thank you, kindly, John. I greatly appreciate your encouragement, and am glad that you are finding worthwhile things on this channel. Kind regards, Graeme

You are welcome, Komitet Gosudarstvennoy Bezopasnosti, and thank you for your observation. Kind regards to you from Australia.

@@CrystalClearMaths sir yuh still active here,how’s life going for ya

@@komitetgosudarstvennoybezo5216 A few ongoing challenges, K. Thank you for asking. I am still responding to comments, but have some significant matters in real life to sort out before I can see my way clear to resume posting videos. I am sorry. Kind regards to you, and thank you very much for your support and concern. Both are appreciated!

The second graph certainly LOOKS like a negative quartic graph, but it is not necessarily one. I randomly sketched it and no formula was provided. The object of the video was to explain and teach a principle or method for sketching gradient functions when the formula for the original function is unknown. It is quite possible for a graph of that shape (not a negative quartic) to have a gradient function as I depicted. You are, however, quite correct in this ... IF it was a negative quartic graph, the derivative function would most certainly be a cubic function.

Because the equation for the x-axis is y = 0. When we set y = 0, we are solving two equations simultaneously ... that of the polynomial (or any function for that matter), and the line y = 0. The results are all the points where the polynomial crosses or touches the x-axis (i.e. the points of intersection). Algebraically, we refer to these values as the 'roots' or 'zeros' of the polynomial. Graphically, they are simply referred to as x-intercepts. Thank you for asking, Tran.

Thank you very much, Jim. I have been unwell and have not been monitoring my website for some time. I appreciate your drawing my attention to it and will try to rectify the matter asap. Kind regards to you. Graeme

Greetings, Arshan. Thank you very much for your questions. Unfortunately, I have been dealing with a long list of family matters (currently I am dealing with cancer). This has kept me from making videos for years. I still hope to resume working on them, but need to get my health re-established first. I am sorry. I have seen some quite good KZbin videos in the past about Richard Feynman's integral technique. It is quite ingenious and, if you search KZbin for them, you will find good explanations in that way. Sadly, I do not have the resources at the moment to help you with this matter. I greatly appreciate your concern and interest and hope to be able to produce videos again 'soon.' Kind regards to you!

I would love to hear how you go, Bubblepop, thank you. I am glad that my suggestions have made sense to you and very much appreciate your making contact and telling me about it! Warm regards to you. Graeme

@@CrystalClearMaths Dear sir, these suggestions magically worked for me I got 29/30 in my latest math test... I hope to continue this marks for the upcoming tests also. Thank you again 😊

@@Jess.is.a.mess_08 That is brilliant and exciting news, Bubblepop! Congratulations! I very much appreciate your letting me know of your results. I am glad that my suggestions helped you, and wish you every success for the future.

@@CrystalClearMaths thank you sir, I'll work very hard from now on!!

Thank you, carvelbell181. I am glad that my videos help you 'see and learn concepts clearly.' That is a great encouragement to me. May you always continue to enjoy learning. Warm wishes to you. Graeme

Occasionally, that happens. It is simply an estimate. In that case, choose 9 and see if it 'works.' This is not an exact predictive method. It is an iterative process using approximations. Thank you for seeking clarity, WoC. Sometimes, the first approximation needs to be adjusted.

You are welcome, CMP.

It very likely would be ... a good iterative technique. Thank you for your astute observation, Silly Me Silly.

I agree, Ivan. When I started producing this series, I was intent on demonstrating and teaching particular substitution techniques and realise that, for a number of the integrals, I therefore adopted a less efficient method of arriving at a satisfactory answer. I left the videos because the primary objective was still achieved ... that of showing how to use certain trigonometric substitutions. Students facing this question during an examination would be well advised to adopt the method that you describe so well! Thank you.

Greetings, Vidhya. You have asked a very interesting question. Thank you. Not all children will respond to the "uncomfortable feeling of solving challenging problems." I think of it like asking children to run or hike up a tall hill or mountain, or to engage in a multi-day hike. Some children respond positively. Some do not. I have taken reluctant children on a multi-day hike or had them camp alone in the 'bush' for a night and had some of them respond extremely positively. I believe it is because they realise that they have succeeded at something big and momentous ... something that they did not believe that they could do (or wanted to do). In all cases, I believe it was MY PARTICIPATION and ENCOURAGEMENT at all steps that made the difference. Their sense of achievement is what may motivate them to engage in future challenges. In the case of mathematical problems, I do not leave them alone (at first). If they struggle and are on the verge of 'giving up,' I ask a QUESTION. Good leading questions lead to discussions that allow them to make discoveries, so that they can take another step. Finally, when they reach the goal and solve the difficult problem, I am full of PRAISE for their achievement. I will tell others and tell their parents (and, sometimes, though rarely, provide a extra reward). You will not win with everyone (as not everyone will develop a passion for mathematics), but your successes will sustain you ... and you will profoundly help and motivate some students.