integral of sqrt(1+x^2)/x vs integral of x/sqrt(1+x^2)
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- เผยแพร่เมื่อ 3 ต.ค. 2024
- NOTE: it's possible to do u sub for the integral of sqrt(1+x^2)/x as well. Thanks to usuario0002hotmail for pointing out.
, trig substitution, calculus 2 tutorial.
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I used u=sqrt(1+x^2) -> x=sqrt(u^2-1) on the left side integral, and I got the answer after some division :P
Ah, yes you are right! I shouldn't be lazy to not give u sub a try for the left one. : )
=int(u^2/(u^2-1))du ?
Yeah that's the one, More Girl
@@moregirl4585 wouldn't it be integral(u/sqrt(u^2 -1))du
Instead of division I decided to try _wouldn't it be nice._
∫ u²/(u²-1) du
∫ (u²-1+1)/(u²-1) du
∫ 1 + 1/(u²-1) du
u - ∫ 1/(1-u²) du
u - tanh¯¹ (u) + _c_
Or
u - ½ln |(u+1)/(1-u)| + _c_
Or
u - ½ln |(u+1)/(u-1)| + _c_
Or
u + ½ln |(u-1)/(u+1)| + _c_
(and substitute back from u=√(1+x²), I'm not going to retype everything here ok)
To my surprise, there are a lot of variations for the second term.
Don't drink and derive!
Lol. You might spill your softdrink
My favourite math guy on TH-cam with his spherical microphone.
Love it.
You always save me from my calculus homework, thank you so much!
Hi! Same here
Btw from which country are u?
Thank you so much ❤️❤️❤️
You are my most favourite TH-camr right now!!!
thank you for this, really well explained :D
Wow u are verified
Thank you so much . Earlier i was doing it with 1+x²= t method . Thanks for teaching me this ❤️❤️
I've only had an introductory course on integration. We weren't taught any of the common integration techniques like u-sub, integration by parts, partial fractions, trig sub, you name it. We only had integrals like the integral of x/(1-x^2) which we did by the definition of the integral (you'll even see me employ this kind of thinking in this answer). Anyway, my point is that I haven't yet learned any of this trig sub business so I went the algebraic way. Here's how I did the integral on the left:
∫ √(1 + x²) / x dx
U-sub:
u = √(1 + x²)
u² = 1 + x²
x² = u² - 1
x = √(u² - 1)
dx = 1 / 2√(u² - 1) * 2udu
dx = u / √(u² - 1) du
∫ √(1 + x²) / x dx
= ∫ (u / √(u² - 1)) * (u / √(u² - 1)) du
= ∫ (u / √(u² - 1))² du
= ∫ u² / (u² - 1) du
= ∫ (u² - 1 + 1) / (u² - 1) du
= ∫ (u² - 1) / (u² - 1) + 1 / (u² - 1) du
= ∫ 1 + 1 / (u² - 1) du
= u + ∫ 1 / (u² - 1) du
= u - ∫ 1 / -(u² - 1) du
= u - ∫ 1 / (1 - u²) du
= u - ∫ (1 - u + u) / (1 - u²) du
= u - ∫ (1 - u + u) / ((1 - u)(1 + u)) du
= u - ∫ (1 - u) / ((1 - u)(1 + u)) + u / ((1 - u)(1 + u)) du
= u - ∫ 1 / (1 + u) + u / ((1 - u)(1 + u)) du
= u - ln|1+ u| - ∫ u / ((1 - u)(1 + u)) du
= u - ln|1+ u| - ∫ u / (1 - u²) du
We wanna have the nominator be multiplied by -2 -- you'll soon see why. In order not to change the question, let's also multiply the whole integral by -1/2:
= u - ln|1 + u| - (-1/2) * ∫ -2u / (1 - u²) du
Now you see, the nominator is the derivative of the denominator inside the integral. Recall that d/dx ln(f(x)) = f'(x) / f(x).
= u - ln|1 + u| - (-1/2) * ln|1 - u²| + C
= u - ln|1 + u| + 1/2 * ln|1 - u²| + C
Let's convert the answer back into terms of x:
= √(1 + x²) - ln|1 + √(1 + x²)| + 1/2 * ln|1 - (1 + x²)| + C
The function inside the first natural logarithm is always positive so we may remove the absolute value signs from there:
= √(1 + x²) - ln(1 + √(1 + x²)) + 1/2 * ln|1 - 1 - x²| + C
= √(1 + x²) - ln(1 + √(1 + x²)) + 1/2 * ln|-x²| + C
= √(1 + x²) - ln(1 + √(1 + x²)) + 1/2 * ln(x²) + C
Minding that √(x²) = |x|, let's simplify the second natural logarithm:
= √(1 + x²) - ln(1 + √(1 + x²)) + ln|x| + C
= √(1 + x²) + ln|x| - ln(1 + √(1 + x²)) + C
= √(1 + x²) + ln(|x| / (1 + √(1 + x²))) + C
This is essentially the same function that bprp answers with, you may check yourself.
You've always been my life saver in exams
I like the U-world best. The best part of this video is that the pen fell on the floor twice.
I really love you YOU SAVED MY LIFE!!!
I like U world cause it can also transform into V world and W world and those are my favorites. Theta world will get messy if you take it to another world
Love your videos, thanks for making them!
You have the BEST explanations!! Thank you so much for all of the math help ^_^
Fantastic explanation
Thanks for allowing us to watch an integral battle turn into a three markers battle!
I don't prefer any world over the other. I enter them when I assume they might help me solve an integral.
Btw: The theta world also helps for the integral on the right side.
i was fighting for my life over the problem on the left. gracias carnal que te llegue todos los bendiciones del mundo
Thx for your vids they are awesome
5:46 I seriously wrote this as sine and cosine and did integration by parts💀 what you have done would have been much simpler
It's 2 in the morning here on Brazil, worth it!
For the first integral i put √1+x²=t² and boom . But its a simple pattern you can observe that if the derivative of denominator is coming in numerator the answer is denominator itself
The theta world is becoming my favorite! It’s satisfying to see the problem come full circle
Thnku sir 😊
sorry, sir. why is the answer of integral csc x = ln (cscx-cotx) instead of -ln(cscx+cotx) like in your other video?
Well the integral on the right side could've been done in a more easier way by making the substitution u=sqrt(1+x^2)
How so
@@skateboarddude8260 hi
yess that wt I think, however, the result was ln(x) hhhhhhh
What a explaination sir ,jai bharath.
The theta world generally becomes more triggy than the u.
#YAY
8:50 Oh hell no! Mind totally blown. I had to watch that part at least 5 times.
Excellent professor. Easy to understand.
0:00 "I swear, only one drink."
Effects at 3:41 and 6:42.
Thank you, bprp, very cool!
Thanks
For the left one:
Let u = x^2 +1, du is 2xdx.
We get .5 integral(sqrt(u)/(u-1))du
And then some trig can be introduced
No, that would get very messy, and might not even work.
Final exam is tomorrow.... I m still struggling with trig identities 😢😢
Tq ...love from India...🇮🇳🇮🇳🇮🇳 Indiawale🎉🎉
Thanks 😮
on todays maths test our teacher gave us the function y=ln(x) and we had to calculate its length over some interval, that example was too powerful for me.
I used integration by inverse for the left one. I got the inverse function 1/sqrt(x^2-1), whose integral was arcosh(x). I get
sqrt(1+x^2)-arcosh(sqrt(1+x^2)/|x|)+c
thanks man this really helps
4:51 I did integration by parts and it somehow worked, lol
I differentiated cscθ and integrated sec²θ
In there was a cotθ*tanθ, which nicely canceled out into a 1
You can also use hyperbolic Sub
In fact both integrals can be calculated using the same u-sub
u^2=1+x^2
1:14 Maybe x is in a wrong place but we can multiply top and bottom by x
If you look at the result then second Euler substitutiion will look good
Second Euler substitutiion is hidden in log function
We could let u be equal the argument of log
Right
Got sqrt(1+x^2)-arctanh(sqrt(1+x^2)) for the second integral
Best any easiest method.. I tried 3 to 4 other methods but this' great!
The integral of csc(x) is not -ln|csc(x) + cot(x)|?
I think it is a mistake on the video.
Ingl cosecx is log[cosecx-cotx], correct
@@anishmathew7593 no,, that is not correct. You can differentiate it to proof it
@@alvindrajaya3878 yes, it is log(cosecx-cotx) . You simply differentiate this by chain rule you wl get it as *cosecx*
@@alvindrajaya3878
[1/(cosex-cotx)] × (-cosecx. cotx+ cose^2x)
ie, cosecx( -cotx+cosecx)/(cosecx-cotx)
ie, *cosecx*
@@anishmathew7593 both work, -ln|csc(x)+cot(x)| is also correct
love you💕 from Cambodia!
Actually you CAN solve this without trig sub! I wondered why you did it with the trig sub , you can let u=sqrt(1+x^2) and everything will go well!
Fady Omari that is exactly what I thought about
Absolute value of ln is not necessary as the hypotenuse is longer than any other side.
thank u guy
isnt the integral of csc equal to -ln(cscx+cotx)
Bien explicado, gracias!
Thank you so much
As long as u (you) are in the theta world; Iike them both.
I like the theta world, because trig substitutions are so intuitive and flexible; everything makes perfect sense there. Don't get me wrong, the u world is good too, but much less intuitive (from my pov)
5:27 I multiplied the top and bottom by tanθ and then set u = secθ.
du = secθtanθdθ
There’s already secθtanθdθ in the numerator of the fraction so I subbed in du for that.
I was left with:
Int u^2 / (u^2 - 1) du
Int (u^2 - 1 + 1) / (u^2 - 1) du
Int 1 + (1 / (u^2 - 1)) du
Then, I did partial fraction decomposition.
It was quite messy from there.
lol i thought you meant the left one was the easy one so i started with left and had to derive sqrt(1+x^2) at one point and got the answer for the right one by accident :D
you can solve it without trigonometric functions by multiplying with x/x you can find a very nice and fast soltioun
This is why he's not Blackpen Redpen Bluepen :D
So if you compare the answers, shouldnt you be able to get the ln|...| value to zero somehow? If it was a constant term then it would disappear into the c but it has x so...?
I was actually able to do this, yay!
I always thank you for your good lectures. But, integral of cosec x has wrong p/n sings. In other videos, you have s right solution.
You'd better use integration by parts in integral of squrt (1+x^2)/ x dx
4'16'': I was wondering: why not considering absolute value of sec theta ?
Why do we stil have the sec yet we had replaced it with 1/cos which I has been crossed....?
Why can’t we do a U-Sub on the left?
great videos!!!
To be fair u could get rid of the (1/x) that is inside the Ln
But still awesome
Thanks for helping me out in calculus 2
Thanks man....I tried item a) so hard and didn't find the right answer.
Nicely explained 🙏
Superb buddy
Jhu used a bluePen
U world forever
hey correct me if im wrong but im pretty sure sqrt(1+x^2)/x - 1/x is always positive so i dont think you need the absolute value
Ah yes.
For the first one, use u=sqrt(1+x^2)
Thank you for explaining. 🤓
I prefer to memorize the integral of csc t as log tan(t/2) + C.
Thnks
So very nice but sir you put wrong value of cos x in last step .You put 1/x but the true value is (1/√(x²+1).......
We should do by partial integration also
Wolfram Alpha gives the same answer for the left integral, but without the absolute value bit. Which is more correct?
That's amezing isin't it
eu te amo
Does the video brightness vary for everybody or just me. It’s very distracting.
Can you do a video on the integration x-1/√x^2-x
how do we know the integral of csc theta? remember or caculate? tnx
why can I consider ((secx) ^ 2) ^ (1/2) is sec (x) and not Abs [sec (x)]?
5:10 sec/tan=cosec. So I used D-I method on cosec * sec^2.
Don't worry BPRP, you'll be able to hold 3 markers in no time. Practice!
Left would be really easy with a hyperbolic trig sub
Antiderivative of cscx is -ln|cscx + cotx|
You are great
Math exam in 2 weeks and the first equation just made me confused
LoL RiP
buenaaaa :D
Just multiply both sides with zero. You'll get the answer quick 🔥💀
Post another unedited video with no cuts
A much more simpler solution can be to just put product rule
yooooo I had so much momentum going with the trig sub that I didn't even think of doing u-sub on the integral on the right hahah.
hahah hey, I got it right tho.
Op
My initial reaction is that the one on the left would be harder since there was a possibility of dividing by zero there, but the right equation’s denominator couldn’t be less than 1. Is there any validity to that line of reasoning, or did I just get lucky?
Lucky, since with these integrals we can't use direct substitution in order to answer them, so you wouldn't be able to just plug in a 0 and have that issue. Good question though!
Jack Daly Thanks. Kinda thought so, just because I don’t recall the profs saying anything about that in math class.
Nice nice
May be same result ٠
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