how the fuck is this even possible. wtf
How the fuck is this even possible. wtf
Samuel Morales
David Campbell
bump, my head hurts. someone pls help me understand the intuition
Lucas Garcia
[eqn]
\sqrt{a^2 - x^2} \\
= a * \sqrt{1 - \frac{x}{a}^2} \\
[/eqn]
Let [math]\frac{x}{a} = \sin(\theta)[/math], ie [math]x = a * \sin(\theta)[/math]
This means that [math]\theta = \arcsin(\frac{x}{a})[/math] ; use that to compute [math]\theta_0[/math] and [math]\theta_f[/math].
This also means that [math]\mathrm{d}x = \mathrm{d}(a * \sin(\theta))[/math], ie : [math]\mathrm{d}x = a * \cos(\theta)\mathrm{d}\theta[/math] ; replace [math]\mathrm{d}x[/math] by this expression in your integral.
You have effectively did everything needed for a change of variable.
Finally going back to where we started, we can write :
[eqn]
\sqrt{a^2 - x^2} \\
= a * \sqrt{1 - \sin(\theta)^2} \\
= a * \cos(\theta)
[/eqn]
There you go, nasty integral ain't so nasty after all.
t. engineer
Joseph Hall
[math]
\sqrt{a^2 - x^2} \\ = a * \sqrt{1 - (\frac{x}{a})^2}
[/math]
Forgot muh parenthesis
>/sqt/ for question like that lad
Nolan Williams
It's just writing the integral in terms of a new variable. It's the same thing as u sub but it makes these integrals come out easier.
Blake Garcia
idk it just seems like a crazy concept. Its hard to really grasp wtf how this is even possible
Jackson Hall
arctan' and stuff
Carson Sullivan
Change of variables isn't a crazy concept
Let's have a point, it's represented without ambiguity by both Cartesian and polar coordinates, and you can change the expression of those in any function involved them
[eqn]x = r * \cos \theta \\
y = r * \sin \theta \\
x^2 + y^2 = r^2 * (\cos^2 \theta + \sin^2 \theta) \\
= r^2
[/eqn]
These substitutions are one of the most basic tools in solving integrals, right after (((inspection)))
Gavin Perez
thanks for the comment
im thinking that im really over complicating this.
Noah Carter
Watch zieth chnalu
youtube.com
Carson Morales
how do you write in smart on Veeky Forums?
(by smart I mean formulas)
Hudson Collins
...
Ayden Myers
wait until you discover
[math]\begin{align}
\arcsin(z) &{}= -i \ln \left( iz + \sqrt{1-z^2} \right) \\
\arccos(z) &{}= -i \ln \left( z + \sqrt{z^2-1} \right) = \frac{\pi}{2} \, + i \ln \left( iz + \sqrt{1-z^2} \right) \\
\arctan(z) &{}= \tfrac{1}{2} i \left[ \ln \left(1 - iz \right) - \ln \left( 1 + iz \right) \right]
\end{align}[/math]
Parker Young
read the sticky
Elijah Sanchez
You definitely are, its the exact same thinking you applied to the most based geometry problems when you were 8 or some shit finding the volume of a cylinder but needing to substitute diameter for radius
Angel James
Why this is defined just in the range of - pi/2 to pi/2?
Hunter Martinez
it makes sense if toy draw it out
upload.wikimedia.org
Grayson Morris
*you
Connor Rodriguez
this, i dont understand that either
Dylan Sanders
Are you serious? That's the range where the sine function has unique outputs. If the range included something like 5pi/6 then the sine of it would be 1/2, a value already yielded by pi/6. Take a look at the inverse trig functions if you need a visual
Evan Baker
see what i dont get is that thetha is restricting the domain from those bounds but from the orginal integral the x values arent restricted? so how can that work out
Dominic Rivera
Same doubt
Robert Diaz
And calc II is when the plebs change majors to business or history.....
Mason Rivera
I can solve the problems no doubt. But I'd rather try to understand why is like that. This has been bothering me for a whike