You MUST Memorize the Process of this Evil Integral!!!

[@Silver-cu5up]

OLD UPDATE: I JUST FOUND A BETTER MOTIVATIONAL WAY of SOLVING THIS!! I'll make another video on this integral soon! It's similar method to the one in the video, but it's a more motivating and easier to remember approach!!!
RECENT UPDATE: Here's the video: th-cam.com/video/UVKyiXnBo-I/w-d-xo.html

[@Silver-cu5up]

Timing myself:
1st Attempt: (2 minutes, 12.72 seconds) = 2:12.72
2nd Attempt (after knowing the process): 1:32.08
3rd Attempt (knowing the process better): 55.60
4th Attempt (memorized some answers of the process): 38.24 sec.
Now if I have wait long, i should be slow again forgetting some answers in the process. Then remember it back and repeat until I fully memorized the shortcut to this integral.
**SPOLIER: ** I'll end up remembering that:
integral of sqrt(x+sqrt(x)) dx from 0 to 1 = 4sqrt(2)/3 - 1/8 [integral of cosh(2x)-1 dx from 0 to arccosh(3)].

[@ninadmunshi2879]

I think inverse jailbreak is the fastest and simplest method to solve this integral. You quickly get that I = ( rectangle of area sqrt(2) ) - int y^2 + 1/2 - sqrt(y^2 + 1/4) dy (yes, minus, you have to take the minus root in the quadratic equation to satisfy f(0) = 0) from 0 to sqrt(2)

[@Silver-cu5up]

@@ninadmunshi2879 idk if it would be the simplest. The quadratic is too tricky to deal with for some others (although very satisfying), and the negative can be easily missed. And then the sqrt(x^2+1/4) theyd have to deal with after.

[@ninadmunshi2879]

@Silver-cu5up I can understand the hesitation behind quadratic formula, but there's no way around this integral without some kind of "trick" and quadratic formula is a one and done instead of chugging through hordes of identities. Second, the integral of sqrt(x^2+a^2) is the area under a hyperbola. Either you can quickly do the cosh^2t integral or it's just muscle memory because this integral shows up all the time. But that means computation wise there's only two things to do that require work instead of multiple things.

[@cbmiami1646]

If you substitute (u+1/2)=w, the integrand will have one function of the form xsqrt(x^2 -a) which can be integrated in the obvious way. The other one, sqrt(w^2 - 1/4), can be solved by setting w = (sec(theta))/2 and solving integrals of sec^3 and sec. Solving this integral is better since it puts you directly into sec, tan and logs of sec and tan, and you can substitute directly for an exact answer without having to know a bunch of hyperbolic trig and inverse trig identities.

[@Silver-cu5up]

Oooooh, I will say, youre method is a bit scary but it does come out to what you said, and it does seem faster to those who knows the answer to sec^3(x) immediately.
I'mma keep this in mind because this method could be potential to those who are more comfortable with sec^3(x) than hyperbolic identities.

[@doozy-r1e]

this actually is the best method imo

[@maxvangulik1988]

sinh(2x)=2sinh(x)cosh(x)
sinh(arcosh(x))=sqrt(x^2-1)
sinh(2•arcosh(3))=12sqrt(2)
sinh^3(2•arcosh(3))=3456sqrt(2)
I=3456sqrt(2)/12-12sqrt(2)/16+arcosh(3)/8
I=1149/2sqrt(2)+2ln(1+sqrt(2))

[@holyshit922]

Substitutions
w = sqrt(x)
sqrt(w^2+w) = u - w

[@Silver-cu5up]

AINT NO WAY 0_0

[@doozy-r1e]

wait i dont get it, how do we proceed after making the required substitution where is the holy shit moment

[@doozy-r1e]

i got int_{0}^{1} (u-w)^2*2w/(u+0.5) du ...

[@holyshit922]

@@doozy-r1e
If you have sqrt(w^2+w) = u - w substitution
then solve for w , differentiate both sides,
calculate sqrt(w^2+w)
If you want to have integral of rational fuction that's your substitution
and in this integral calculations don't take long time
After another substitution v = 1+2u
you should get integral
\frac{1}{32}\int_{1}^{3+2\sqrt{2}}\frac{(v-1)^2(v^2-1)^2}{v^4}dv

[@Silver-cu5up]

@@holyshit922 sucks to Euler Dodge tho ;_;
This method is very advanced.
Finding du is annoying and then power rule spammin is also annoying.