Integral of the Day 6.18.24 | Medium Spicy Level! What do you think? | Math with Professor V

Here's your latest Integral of the Day! I promised a more straightforward integral after my most recent upload, so let me know if you were able to successfully solve it on your own. It did require some attention to detail with changing the limits of integration, especially with the trig sub portion, so if you need help with that check out the video below:

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Professor V

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[David-pqwt]

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.

[ETBTS]

doing this and then watching the video makes me so grateful for hyperbolics

[aaravshah]

I think if you rationalize the denominator at the end, it does help simplify. Great vid like always! Enjoyed it!

[siyabongashoba]

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

[siyabongashoba]

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.

[khansahab]

Can you also use inverse hyperbolic sine? I got arcsinh(4/3)-arcsinh(1/3)

[AFCOE]

Hello ma'am, hope you're doing good

[vamikagupta]

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)