Integral battlel#11, the theta world!

Integral battlel#11, the theta world!
integral of 1/sqrt(x^2-1) vs. integral of 1/(x*sqrt(x^2-1)),
integral of 1/sqrt(x^2-1), integral of 1/(x*sqrt(x^2-1)),

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[SatyaVenugopal]

To check the relationship between arcsec(x) and arctan[√(x^2 – 1)]:

Let y = arcsec(x)
∴ x = sec(y) –– (1)

And let z = arctan[√(x^2 – 1)]
∴ √(x^2 – 1) = tan(z)
x^2 – 1 = [tan(z)]^2
x^2 = [tan(z)]^2 + 1
x^2 = [sec(z)]^2
x = sec(z) –– (2)

From (1) and (2):
sec(y) = sec(z)
∴ y = z

∴ arcsec(x) = arctan[√(x^2 – 1)]

[AlgyCuber]

the end part :
arcsec(x) = arctan(tan(arcsec(x)))
=
= arctan(sqrt(x^2-1))

[johantorres]

You are the best teacher I've ever had

[meh]

The equality holds true only for positive values of x.

[amissdirge]

Wouldn't the integral of 1/sqrt(x^2-1) just equal arccosh(x)?

[imme]

In the final formula, they are equal if |x| >= 1.
Arctan(sqrt(x^2-1)) = y <==> sqrt(x^2-1) = tan(y) <==> x^2-1 = tan^2(y) <==> x^2 = tan^2(y) + 1 = sec^2(y) <==> x = sec(y)
<==> y = sec-1(x)

[Rick_Sanchez_Jr.]

3:40, you saved my life bro, thank you

[calyodelphi]

arctan(sqrt(x^2-1)) = arcsec(x) can be proven using the same "Trick Substitution" method you used on the integrations to put them in the "Theta World":
Let x = sec(theta)
arctan(sqrt(sec^2(theta)-1)) = arcsec(sec(theta))
arctan(sqrt(tan^2(theta))) = arcsec(sec(theta))
arctan(tan(theta)) = arcsec(sec(theta))
theta = theta; arctan(tan) and arcsec(sec) cancel each other out to leave just theta
Therefore, arctan(sqrt(x^2-1)) = arcsec(x)

[holyshit]

These integrals are of the form Int(x^m(ax^n+b)^p, x) or Int(R(x, sqrt(ax^2+bx+c)), x)
so two types of u substitution will work

u sub for first integral
u = x/sqrt(x^2-1)
and for both
sqrt(x^2-1) = u - x

[MrShyguyRS]

Ok but isn’t the derivative of arcsec(x) 1/(|x|*sqrt(x^(2)-1))? With absolute value around the lone x? Therefore wouldn’t the antiderivative have to be arcsec|x|+C?

[Ivan]

As far as the first formula, why don't we use hyperbolic substitution?

x=ch(u), and as ch^2(u)-1=sh^2(u), in the u-world the wormula is only 1*du.

[deeptochatterjee]

My girlfriend showed me this problem hey physics teacher gave her and it was essentially the same thing except it was 1/sqrt (x^2+25) instead of 1/sqrt (x^2-1). I just sat there looking at it, probably for 20 minutes until I remembered a little something called sec^2x -1 = tan^2x. Shows how terrible my Calc II teacher was that she didn't even teach us trig sub.

[sdbstar]

they gave us that first integral in an exam, i integrated it the same as you you did at the end of the video, 😉

[Jessica_Jessica_Jessica]

Is the triangle you drew constant? Like, tan is always square root of (x^2-1) no matter the given?

[joshuasorber]

For the first integral you did could you have substituted another trig function like sin(theta) and used the Pythagorean identity ?

[sauravdubey]

What happened to u r voice.? In this world

[wryanihad]

Can we solve the second one without trig

[Luca-vofk]

A simple solution for the first integral would be to say that the function is the differentiated function of arcosh(x), so the answer would be arcosh(x).

[July__-eupd]

I rewritten the whole fraction as tan of sin of x, and then I didn’t know how to derive it so idk

[StuartSimon]

Actually, the two results are not equivalent. Sqrt(x^2-1) is an even function, and so tan^-1(sqrt(x^2-1)) is an even function. The inverse secant operation that matches is sec^-1(abs(x)).