Calculus 2: Integration - Trig Substitution (3 of 28) Integral of SQRT(x^2-x^2) Ex. 1

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In this video I will solve: integral of [(a^2-x^2)^(1/2)]dx.

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He is good... I was a math teacher for 35 years and still enjoy watching his videos!! Plus I donate money to his cause.

dpmike

It is good to see all of the work laid out in one place at the end. That really puts the problem in perspective.

garyward

I as about to change out of a math related major until I saw you made Calculus videos! I love your content Mike your such a great guy! I have watched and learned from so many of your physics videos

milekyle

U saved me..I tried doing it so many times and wasted hours on a single problem..u explained it in less than 10 MINUTEs thanks!!

vaiebhavpatil

Thank you - I am trying to find the area of a segment of an ellipse and this integral came up. Very nicely laid out.

lawrencewhitfield

Watching this again I understand it much better than I did before. Thanks so very much for all these excellent lectures.

valeriereid

you are the king of teaching, , professors should learn this skill from you

nohhem

This was such a useful video, thank you so much. You teach way better than my professor.

MrDeadshotHD

Excellent. Found many doing the trick, but nobody else explained why he was doing what. I guess that is the difference between knowing how to do it an knowing how to teach it :-)

karinbeyaert

Great explanation! I’m just in precalc and I was able to understand this!

markgross

Wow...above my head, but your steps were very clear. Thanks for showing all the steps.

Phatcub

Thank you for making this video. It is very well made and easy to follow. I appreciate that you took the time to make this.

nathanross

the best video on trig substitution! Thanks!

alenakrytskaya

Question. Why is it in this case not sufficient to calculate the integral the 'normal' way? So you pretend that sqrt(a) = a^(1/2), and apply the reversed chain rule, because a = f(x).

TimeTraveler-hkxo

Thank you, this video cleared all my doubts

ronakagarwal

Hi, Michel, can you help me find out why wolfram alpha gives the answer "= 1/2 (x sqrt(a^2 - x^2) + a^2 tan^(-1)(x/sqrt(a^2 - x^2))) + c" instead? I just couldn't figure out how they got the 2nd term with tan^(-1) after substituting a = x/sqrt(a^2 - x^2) / cos &theta and x = a sin &theta.

joeyborja

THANKS MAN! I LOVE YOU! YOU HELPED ME SO MUCH! ❤️

thomaslagast

what if we had sqrt a^2 - y^2 dy? In this case, would we let y= acostheta?

kevinle

thx very much! very good teaching, i learnt soemthing

peterchin

I understand it somewhat, but I notice you skip steps. Can you go over them

SwayLeen-grcw

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