Integral of the Day: 7.17.23 | Calculus 2 | Math with Professor V

Happy Monday to all my lovely starfishes (aka internet students)! :) Here's an integral that I thought would be inspiring to you all as you start a fresh new week. Who is taking summer classes? How are they going? Excited for a new week ahead!

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Professor V

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[georgesadler]

Professor V, thank you for another great Integral of the Day. Professor V, for this problem, I let u equal to the ln(Cosx) and let dv equal to Sin x dx and using the classical Integration by Parts gives me the exact solution that you computed. This is an error free video/lecture on YouTube.

[k_wl]

integral from 0 to pi/3 of sinXln(cosX), let u = cosX, -sinXdX = dU
turns into integral from 1/2 to 1 of lnUdU after changing the bounds. {switched the order of 1 and 1/2 as theres a negative from the dU)
integral of lnU can be solved by IBP and i get UlnU - U, evaluating it from 1/2 to 1 is:
-1/2 - ln(1/2)/2 ... and thats the answer


after watching the video i see that it can be simplified further, thanks professor V.