integral of sqrt(x^2+2x), trig substitution, calculus 2 tutorial.
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- Опубликовано: 14 фев 2015
- integral of sqrt(x^2+2x), trig substitution, calculus 2 tutorial.
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I’m amazed how you can start with a fairly simple integrand and end up with a rather complicated solution.
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@Amos Royal Instablaster =)
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@Amos Royal Glad I could help xD
that triple variable substitution was handled like a boss
Shaikh Gaming thanks
It's so cute and adorable when you say "u-world", "x-world", and "theta world" lol. Btw, you and PatrickJMT would make a great team in making math vids :)
+Jem Celespara Happy New Year Jem! I know the PatrickJMT channel but unfortunately we dont know each other personally!
+blackpenredpen that's so nice of you!! And happy new year too!! Oh and I want to ask when are you going to make Differential Equations related videos? I'll be looking forward :)
You made a small copying error at the end of the computation (which I do all the time). The last term of the answer should be ln|x+1+sqrt(x^2+2x)|, since
x+1+sqrt(x^2+2x)
that was the most straight answer I've ever seen in a math tutorial, just what I needed thanks :)
Your way of teaching is highly appreciated.
You can not drop the absolute value sign for the ln because when x is negative it is not always a positive output. For instance when -500 is plugged into (x+1+sqrt(x^2 +2x)) the output is -.001
Im always amazed by what we can do in maths just through the application of rules and formulas, without understanding the meaning of any of the intermediate steps.
You are Great man! Thanks, really apreciate your help.
What will you get if you bisect theta and \frac{pi}{2}-theta
and then calculate tangents and cotangents in new right triangles
I know I am very late to the party on this one, but if I'm not mistaken don't you need to put an absolute value inside the natural log in the final solution? (x + 1 + sqrt(x^2+2x)) will be negative for all x < -2.
If you are late so what am i ?
Bless your soul
I really like your videos and learn a lot from them. I have a comment to make and a question to ask....
Comment: Using the Euler Substitution would be a lot easier especially as your quadratic expression inside the square root has real roots. Also it doesn't assume knowledge of the trigonometric expressions table you have on the board.
Question: Is there a method to integrating the cube root of quadratic functions?
When you Euler Substitute, you end up with something like 8T^2/(T^2-1)^3 to integrate, and that's quite a challenge.
Interesting,so what will be the solution to the integral of √ (1+4x²)
When x=-2, ln(x+1+sqrt(x^2+2x)) = ln(-2+1+sqrt(4-4))= ln(-1) so you actually need the abs. value
Flammable the best !✌
I always preferred the hyperbolic trig method because there's one less substitution.
Awesome videos!
Thanks sir you are good teacher for young boys
Well done but why didn't you just start by x-1 = tan(u)?
thank you very much my good sir.
This was a great problem!
this guy is the goat
I could be missing something (I haven't actually really done integration since a while, and trig-wise I never used sec and the likes, only sin cos and tan), but at the 3:50 mark, if sec²(theta)-1 is equal to tan²(theta), taking the square root seems like it would return the absolute value of tan(theta).
Maybe it was for the sake of the explanation and it actually is tan(theta)?
Anybody else just remember from highschool having learnt integration of √u²-1=u/2√u²-1 - 1/2ln|u+√u²-1| +C 😂
There is a simplified version of it:
= (x+1)/2 * sqrt(x^2 + 2x) - 1/2 * cosh^-1(x+1)
Hem... x+1+sqrt(x²+2x) > 0 only in IR{+} world. So the abs value is necessary in the final solution if you're working in entire IR world (if you erease it, you're juste assuming complexe values and so working in C world).
Can you solve this with euler substitution?
thank you :''''')
when you get the Sqrt[sec^2(θ)-1], you should not have considered the absolute value of sec(θ) ??? whats happen there, can you explain me please? :(
I'm not far in math but I know any real number squared is positive
u substitution can be derived in two ways
1.
Let's cut the curve y^2=ax^2+bx+c with secant line ets
2.
Lets draw a right triangle with sides labeled like in inverse trigonometric substitution
Bisect the angle complementary to our angle theta
After bisection we will get new right triangle Calculate cotangent of angle created after bisection in that new right triangle
You can solve the integral of sqrt(a*x^2+b*x+c)...
Why did you allow yourself to write the square root of tan^2 theta as tan theta without the absolute value?
This is the only problem in the exam that I cannot solve 😥
Legend
Me: *does 300 substitutions and finally gets to the answer*
Me: Yes! Finally!
The 299 worlds I have to go back to: Are we jokes to you?
I hope you don't get offended, bprp, but this haircut is "the most Chinese" one :P
I don't say you look bad though
It is possible to do this substitution instead?: u=sen(theta). In this case the integral became: cos^2(theta). Right?
Yeah you can do it that way as well
Integral of sec^3 theta isn’t a known.
why does √(sec^2 θ - 1) = tanθ but not |tanθ| ?
Can you do this with an Euler substitution too?
Yes you can
and what if the function says int sqrt( -x^2+2x) ? how can i integrate this ?
thanks man !
You can construct a binomial series for this function and integrate term by term. At least, that's how I'd do it. To do so, you need to realize that the inside can be factored as (1-(x-1)^2), and with an exponent of 1/2 (the square root) can be expressed as a power series, where the kth term is (1/2Ck)x^k. (1/2Ck) is the best I can do for the binomial coefficient for 1/2. You would replace x with -(x-1)^2 in this situation. From there, integrating is easy, it's just the power rule.
Love from Bangladesh ❤
It is not the only possibility we can also have as answer 1/4(sh 2u)-1/2u
Where u is the inverse of ch(x+1)
But the approach is also great
sec(x)^3 by parts
That was fucking epic
Actually x+1+√[(x^2)+2x] can be negative
Why do you do sooooo easy question?
I see what I had done wrong now, integral of sec is ln of sec+tan, I did not know this
i had the exact same question on webassign and its marking me wrong i did it the exact same way u did and i have the exact same answer but its still marking me wrong wtf
+brave ali It's usually the input problem or just the problem on webassign itself...
+blackpenredpen lol it was actually the absolute sign and i was putting brackets instead of it
It looked so harmless 😫
web assign?
You are so cute explained, but I'm want the shot way. Anyway thanks for the video
Leo Stark thank you.
shot way?
You teach long cut
My God, someone save me from this accursed subject.
How's the school world going bro?
Now solve this:
Integral SQRT ((x^2+1)^2-1) dx
:D
That's simple, let x^2+1=u and 2xdx=du.
If you plug them in in the integral then you get the integral of (1/2)sqrt(u+1)du.
After doing the integral you'll get this answer: (1/3)[(x^2+2)]^(3/2) + C.
If you know the formulas, it can be done in 20 seconds..
I like when you say theta
teytah
honestly... you is absolutely smart.... but... for pity's sake... there's a short way of solving... :D... thanks anyway... great!
No sharing that way?
i think its by substituting x=coshu bcz then dx=senhudu and you get int (senh²u)du and you can use senh²u=(1+senh2u)/2 (if im not mistaken) and then youre good to go
cosh(arccosh x) = x
sinh(arccosh x) = umm idk
@@blackpenredpen You could've used the Special Integrals forms...Its in the third set out of three sets of three special integrals. No need of substitution, direct answer. I think you know the special integrals, but if you dont, i can email you if you provide your id.
@@AlgyCuber if cosh(u)=x then
cosh²(u)=x² that is nothing but 1+sinh²(u)=x² (due to the hyperbolic identity
cosh²(θ)-sinh²(θ)=1 )
So sinh(u) = sinh(arcosh(x))= √(x²-1)
i dont no how many time should i press like button
Mohamad ald Lol thank you
9:53 HaHa