Integral of sqrt(x^2-1)

In this video, the last one of the integral-trilogy, I find the antiderivative of the square root of x^2 - 1. The method is surprisingly similar to the one of the square root of x^2 + 1, and I recommend you see that one first, because it uses the formula for the integral of sec^3, which I derive in that video. Enjoy!

[TheRedfire]

for some reason I didn't understand why one could get rid of the absolute value and you just explained it to me in less than 4 seconds. best teacher ever.

[JJCUBER]

I’m in AP Calc AB, and we were given a handout to learn about u-sub for the FIRST time (we just recently started actual solving of definite and indefinite integrals). One of the questions was this exact one, and since I already learned a bit ahead, I tried integration by parts 3 different ways to the 4th iteration, until I gave up. I completely forgot that trig sub was a thing, but I wouldn’t have gotten it regardless since I never actually learned how to do it. Funnily enough, our teacher told us that our answer was supposed to be “Can’t solve this, since the material isn’t within the scope of AP Calc AB.” I feel like it’s kind of dumb to include a question where you are supposed to give a non-answer, especially since this is the first time this whole school year we were expected to answer without an answer (ofc I’m excluding DNE, that’s different since it means there is NO solution, as opposed to “there is a solution, you just can’t get it”).

[wryanihad]

Can we solve it without trig trick?

خيلي زيابست

[amankataria]

oh ye bro you taught this good i was stuck at the point of tan^2theta sectheta
thanks

[bamdadshamaei]

You could've done a u sub of x = cosh(u), integrated sinh2(u), and turned back into terms of x at the end which becomes arccosh(x)/2 - x*sqrt(x2-1)

[nickstenerson]

Can you teach all of my courses? I'm an engineering student with a huge interest in mathematics. I want to go to grad school when I graduate undergrad in a year but I don't know if I should go for engineering or math... I'm so interested in your fractional Calculus videos. I actually bought a book and your few videos have definitely helped segment some of the concepts. Seriously though, I hope you get a really good position in a very good university. (if that's what you want to do of coarse) I'm just curious where you went to undergrad. I seriously need some of your enthusiasm lol. Seriously though, your videos are amazing and if you ever stop making them I might cry!

[sansamman]

can you make a video on (integral of) tan^1/4(x) dx
please?

[selfdiagnosedhappiness]

Can this be a solution?
*Factor out i from the function
= (x^2 - 1)^(1/2)
= ((-1)(1-x^2))^(1/2)
= ((-1)^2)(1-x^2)^(1/2)
= i(1-x^2)^(1/2)
since i is a constant, the integral can be evaluated as the integral of (1-x^2)^(1/2) times i
=i((1/2)(arcsinx) +(1/2)(x)(1-x^2)^(1/2)) + c)
=(iarcsinx / 2)+(ix(1-x^2)^(1/2) / 2 + c

[jeromesnail]

Ahah non kidding I said "is have some fun" in the same time as you 😂
Your sense of humour is contageous ;(

[MrRyanroberson]

okay.... so i wonder, if blackpenredpen did derivative of g(y, infinity) for g(x, k+1)=sqrt(x+g(x, k))
can you somehow do the integral of g(x, infinity)?

[srpenguinbr]

I tried it and I found a result proportional to sqrt(-1). Is that wrong? Maybe it works when x is in the interval (-1, 1).

0.5i(arcsin(x)+x•sqrt(1-x²)).

I took the derivative, and, after simplifying a bit, I got
i•sqrt(1-x²)

[holyshit]

This example is good for Euler substitution
sqrt(x^2-1)=u-x
sqrt((x-1)(x+1))=(x+1)u
First substitution is better because after this substitution we will have to reverse power rule
Second substitution also works but Ostrogradsky method will be probably method of choice after this substitution
(Steve recorded video about isolation of rational part of integral by Ostrogradsky method)
Subistitutions written above should cover all integrals of the form Int(R(x, sqrt(ax^2+bx+c)), x) where R(x, y) is rational function of two variable

[robertherbert]

....why don't you clean the board so that people can read what you have written ?

[MilanPrajapati-oe]

i struggle that untill i tool the antiderivative of tan^2(x) itself in that way i got the right answer heres prof and onto ∫tan^2(x)sec(x)dx i get the DI Method where(D, I)(sec(x), tan^2(x)), (-sec(x)tan(x), tan(x)-x). => then D, I metod again (D, I)(x, sec(x)tan(x)), (-1, sec(x)), then we know t=sec(x), x=arcsec(t) i may have replace x with t but use t so to solve for arctan(u)=arcsec(t) we can rewrite arcsec(t)=arccos(1/t) => ADJ then HYP as 1 and OPP as √(1-1/t^2) so we know tan(x)=OPP/ADJ so so since we done it in x instead of t we can say its