Jim Coroneos' 100 Integrals ~ 029 ~ ∫1/(5 + 3cosx).dx

[@MarkFromMullion]

Thank you very much - I'd been struggling with this integral and you made it so clear.

[@CrystalClearMaths]

I am glad that the video was a help to you, Mark. Best wishes for your further studies!

[@arshantv3579]

Salute u master (sir) My question is what isRichard Feynman’s Integral Trick and how he invented this trick And my 2nd question is Why u r not making new videos for your lovely students Im waiting for your new video plzzz for god sake

[@CrystalClearMaths]

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!

[@MrVoayer]

Swift but quite enjoable presentation of a solution to an example integral. A technique that involves knowledge and final solutuon of some other integral examples in the series, presented so far. Obvioulsy, expressing cos x in termes of t still requires some further elaboration.Thank you for presenting us more and more discoveries, dear Grame!

[@CrystalClearMaths]

+MrVoayer I am glad that you enjoyed the video, friend. I am always exercised over how much information/explanation to share in each video, and how much knowledge is to be assumed. Some viewers will find my explanations over detailed and too slow ... others will be perplexed by my assuming knowledge that they do not have. I will, of course, be explaining about cosx and t-formulae in future videos, much sooner now that Kristyn has asked for such material (see her comment below). Warm regards to you from the antipodes.

[@physicsbys.k.tripathi1590]

wonderful explanation sir

[@CrystalClearMaths]

You are welcome, Shiv. Kind regards to you.

[@aayushkhatri4641]

Thanks mate!

[@CrystalClearMaths]

You are welcome, Aayush.

[@shootmegaming5161]

Sir Bansi integrated 1 by 4+secx please down my answer

[@CrystalClearMaths]

Hello Kamlesh. I do not understand. Do you wish to know how to integrate 4 + secx? I do not know who Bansi is. Please reply and let me know. Kind regards to you.

[@andamariadanciu3592]

What if it is definite on 0-2pi? In this case, the change of the variable is going to change the interval in 0-0. Is the final result 0?

[@CrystalClearMaths]

What a perceptive question! One way to resolve this is to recognise that the substitution, t = tan(x/2) has an asymptote at x = π. This means that the integral is better resolved in two parts, taking the absolute value of each part. I.e. 0∫2π 1/(5 + 3cosx).dx = ABS{{0.5arctan[tan(x/2)/2]} from 0 to π} + ABS{{0.5arctan[tan(x/2)/2]} from π to 2π} = ABS(π/4) + ABS(-π/4) = π/4 + π/4 = π/2

[@cricketinfinity-wj7qr]

Ty

[@CrystalClearMaths]

You are welcome, Amit.

[@titoariza3298]

how is ∫ 1/5+3sinx?

[@CrystalClearMaths]

Hi tito. ∫1/(5+3sinx).dx would be resolved in the same way, except that sinx = 2t/(1 + t²) using the same notation. The integral is considerably more difficult to resolve and I do not have the time or space to do it here, but I did track down this page for you ~ www.chegg.com/homework-help/questions-and-answers/dx-5-3sinx-q25270303. You could also try Wolfram Alpha (www.wolframalpha.com/). Unfortunately, you will have to subscribe (pay) to use either site. It is a sufficiently interesting challenge that I may create a video about it when I resume producing videos in another six months or so. Thank you for your enquiry.

[@aniketmehar1410]

Muze all internal lesson ke all video chahiye

[@CrystalClearMaths]

I do not understand, Aniket. I am sorry.

[@holyshit922]

1/(4+t^2) = 1/4 * 1/(1+t^2/4) 1/(4+t^2) = 1/4 * 1/(1+(t/2)^2) 1/(4+t^2) = 1/2 * (1/2)/(1+(t/2)^2)

[@CrystalClearMaths]

Absolutely!