That was a good one. So satisfying when my answer matches yours. I have been doing series problems and watching your series videos but trig sub is my favorite integral technique so when that comes up I have to give it a try lol.
Dear Prof.V good luck with your summer school ! At least I have something to work on. you did an outstanding work with creating videos it does help a lot in university . anyways please enjoy the rest of your day
Dear Prof.V I hope you are doing well, for this one I let u = e^t and du= e^t*dt ,u(ln(4)) = 4 and u(0) = 1 and get a integral of 1/sqrt( u^2 +9) from 1 to 4. and apply u-sub by letting u = 3*tan(x) and du = 3*(sec(x))^2dx when I put everything back to the original from I got integral of sec(x) which is equal to ln|tan(x) + sec(x)| but now change from x to u which was equal to ln( (sqrt(9+u^2)/3 + u/3 ) and the deference was ln(9) - ln(sqrt(10) + 1) hopefully it is right. let me enjoy the video.
is ln(sqrt(10)-1) also an acceptable answer? I was able to simplify ln(9/sqrt(10)-1) down to ln(9(sqrt(10)-1)/10-1)) which is just ln(9(sqrt(10)-1)/9) which then simplifies to ln(sqrt(10)-1)
That was a good one. So satisfying when my answer matches yours. I have been doing series problems and watching your series videos but trig sub is my favorite integral technique so when that comes up I have to give it a try lol.
Yayyy so glad you liked it! I’ll do more trig sub since I love them too! ☺️
Easy peasy😄
doing this and then watching the video makes me so grateful for hyperbolics
I think if you rationalize the denominator at the end, it does help simplify. Great vid like always! Enjoyed it!
Oooh now I’m mad I didn’t! I’ll have to try it to see. Thanks for the support and your comment! ☺️☺️
Can you also use inverse hyperbolic sine? I got arcsinh(4/3)-arcsinh(1/3)
Dear Prof.V good luck with your summer school ! At least I have something to work on. you did an outstanding work with creating videos it does help a lot in university . anyways please enjoy the rest of your day
Thanks, you too! I always love your comments. ☺️
Dear Prof.V I hope you are doing well, for this one I let u = e^t and du= e^t*dt ,u(ln(4)) = 4 and u(0) = 1 and get a integral of 1/sqrt( u^2 +9) from 1 to 4. and apply u-sub by letting u = 3*tan(x) and du = 3*(sec(x))^2dx when I put everything back to the original from I got integral of sec(x) which is equal to ln|tan(x) + sec(x)| but now change from x to u which was equal to ln( (sqrt(9+u^2)/3 + u/3 ) and the deference was ln(9) - ln(sqrt(10) + 1) hopefully it is right. let me enjoy the video.
🙌🏻🙌🏻🙌🏻
Hello ma'am, hope you're doing good
You too!
is ln(sqrt(10)-1) also an acceptable answer?
I was able to simplify ln(9/sqrt(10)-1) down to ln(9(sqrt(10)-1)/10-1)) which is just ln(9(sqrt(10)-1)/9) which then simplifies to ln(sqrt(10)-1)
Yes! 🙌🏻 In fact; I wish I simplified the final answer as much as you did since it’s cleaner. Well done! 👍🏻
@@mathwithprofessorv Thank you!!
Lavaguyn DASAXOS@ MATEMATIKAI🙂