yes Prof.V thank you so much for the new trick! I let cos(t) = u because I was afraid of the integral of 1/sin(t) but now you have opened my eyes I won't run away from such again. thank you so much.

@mathwithprofessorv

21 күн бұрын

Yay you’re so welcome! Don’t be scared 😂😉

Nice that one was fun. U-sub trig sub an natural logs all stuff I enjoy lol

@mathwithprofessorv

21 күн бұрын

Yay, me too!!! Glad you liked it. ☺️

Spice it up with chain rule of the day

Dear Prof.V I hope you are doing well thank you so much for the integral of the day. i let u = ln(x) and du = 1/x*dx and the was in the form sqrt( 1 - (u)^2 )/(u) and did trig sub by letting cost = u and -sint*dt=du and then simplify and break it up into two integral 1: integral sect and 2: integral of cost . the endpoint was -ln(sect + tant ) + sint + C by changing back to x it is -ln( (sqrt(1-(lnx)^2 ) + 1 ) /(lnx) ) + sqrt( 1 - (lnx)^2 ) + C let me enjoy the video

@mathwithprofessorv

21 күн бұрын

Excellent!

Yay I did it!!! 😊😊

@mathwithprofessorv

22 күн бұрын

Yayyyyyyyy 🎉🎉🎉🎉

¡¡Super bonita integral!! 🙂

@mathwithprofessorv

22 күн бұрын

¡Me alegra que te haya guatado!

Hi Professor V, l used u=sq root1-t^2 for my second sub. And I have something different.

@mathwithprofessorv

22 күн бұрын

Hi Tony: I don’t think that will work…did you remember to use the chain rule when finding du?

@tony413chow

22 күн бұрын

@@mathwithprofessorv Professor V, thanks for your quick response. Yes, I have used the chain rule to differentiate the u=sq root of 1-t^2. After elimination and substitution, I have got the integral of -u^2/1-u^2. I then used partial fractions to get integral 1 and -1/1-u^2. After this step, I have the result u-1/2*(ln(1+u)-ln(1-u))+C. The ln brackets are the absolute values of the ln. I am not sure if I have made mistakes in some steps.

@mathwithprofessorv

20 күн бұрын

Hi Tony! Sorry for the delay, been so busy teaching summer school. My class is taking an exam right now so I had some time to sit down and work out the problem using the method you used, and I arrived at the same result! It does look very different compared to the answer using trig sub, so I’m going to try to use log properties etc and manipulate them to match; but as far as the steps you followed I found no errors! 🙌🏻

I got the same answer but I wrote mines a bit different. -ln((1 + root(1 - (ln(x))^2))/ln(x)) can be rewritten as ln((ln(x))/(1 + root(1 - ln(x)^2)) and then broken down with properties of logarithms. I thought it looks a little cleaner so my final answer ended up being ln(ln(x)) - ln(1 + root(1 - (ln(x))^2)) + root(1 - (ln(x))^2) + C

@mathwithprofessorv

20 күн бұрын

Nice! I like it! 👍🏻👍🏻☺️

First try!

@mathwithprofessorv

22 күн бұрын

Woot woot 🙌🏻