@@blackpenredpen PLEASE respond..didnt you first think to replace sinxcosx by the double angle sine formula sin2x divided by 2 and replace the top by square root of cos2x..seems more logical and intuitive than what yiu did..inwould think a lot of people did that.this is why math is so damn infuriating and ridiculous so much..
Solve my exercise with bisector to get u sub which allows you to avoid inverse trig substitution i63.tinypic.com/rkvnkg.png In triangle ABG length of two sides are given Angle BAG is labelled as theta GL is bisector of angle complementary to theta Express the ratio GB/BL in terms of given length of sides
I was once given this exact problem, but with bounds on the integral. I exploited the symmetry of the bounds because I figured there was no elementary antiderivative. Fascinating video!
6:41. That's not tricky. I don't know why foreigners ( I'm indian) teach this integral separately. We have direct formula for integrals of the form x²-a²,a²-x²,root x²-a²,root a²-x², 1/x²-a² ,1/a²-x², 1/root x²-a², 1/root a²-x². We have direct formulas for these in the NCERT textbook. 😄
Would it not have been easier to solve the equation in the note for sin x cos x and substitute it back into the integral so one doesnt have to add and subtract the 1?. ;)...also cute kitten.
If, at 1:30 you had left the numerator as sqrt((cos x)^2 - (sin x)^2), couldn't you have then converted that sqrt(cos 2x)? Which would then allow you to convert the denominator into sqrt ( 1/2 sin 2x). Bringing that back into a single square root that would have given sqrt( (cos2x)/(1 /2 sin 2x) ), which equals sqrt (2 cot 2x). Wouldn't it have been easier to solve that way?
We can solve it by arctanhx function Let sqrt tanx = u So that integral can be 2^0.5 arctanh((sqrt tanhx + sqrt cothx)/2^0.5) This approach is like the way of integration of sqrt tanx
blackpenredpen, Hey, man extremely stunning approach of solution, I wonder how you easily launch your pure math imagiation and miracle creativity.Are you math professor of University or just olympiad winner student? I love your way of solution. Unfortunately, I can't put double thumb-up sign above your video, wish you further promotion as a blogger.
I may have found an error. I used Desmos to set f(x)= sqrt(2)*ln(abs(sinx+cosx+sqrt(sin2x))). Then I plotted f'(x). Then I plotted sqrt(cotx)-sqrt(tanx). But these functions don't line up. In a way, there may be a negative. See image of desmos screen here: i.imgur.com/y5FAEpK.jpg
I looked into this too, but I don't see it! They look identical to me when I graph the difference between the two (one solution sub the other). I can't get the computer algebra systems to integrate this integral to anything easy or anything that will simplify down, but that can just be down to how it is complicated to simplify. Numerically it seems correct.
And now ... you can make integral of sqrt(cot(x))+sqrt(tan(x)) ... and get by adding and substracting the both famous integrals of sqrt(cot(x)) and sqrt(tan(x)) ... and Voilà ... isn't it !
I still don't understand why we can just forget about the absolute value?? Because I think that y = sqrt(cot(x)) - sqrt(tan(x)) *is different with* y = (cos(x) - sin(x))/sqrt(sin(x)cos(x)), and it will produce a different integral
To do it completely right, you need to determine the range where the integrand is integrable, after which you can skip the absolute values if their arguments are always positive in the valid range. In the valid range near 0, sin x and cos x are both positive, for example.
Aw someone who never took Integral calculus, is the trick that it would have been easier doing it the other way? Was the double substitution unnecessary?
Please help me, I am in the limited developments and I know that the inverse of a sum isn't the sum of the inverse. But I wonder what is the inverse of a sum? Especially what is the inverse of 1+x^2/2!+x^4/4!+x^6/6!+...?
yes but what I really searched is the limited development of sech(x). I wonder if you can find this development just by doing the inverse of the one of the cosh(x), but that's not easy.
yo,i have a question for you regarding the integration of (1/(x^2-1))dx;in that manner,can't we think (-1) as being i^2(if we integrate in a complex plane),and the bottom becomes x^2+i^2,so we use the formula for arctan,and thus getting (1/i)*arctan(x/i) +c , is that a valid answer?
Yes, that's a valid answer in a complex world (complex number system), but not in a real world. Because actually, when x is a real number, the i will be cancelled out. That's why it's not actually necessary to use "i" in the equation. For me, I like to solve it using hyperbolic trigonometry substitution giving : integral of (1/(x^2-1))dx = -arctanh(x) + C No imaginary number needed :)
I know that answer aswell,but i wondered if that would work,because it got stuck in my head during one class and i havent found anything related to that on the internet;thank you for your answer ;)
I tried to plot the graphs of sqrt(cot(x)) - sqrt(tan(x)) and of (cos(x) - sin(x)) /sqrt(sin(x) cos(x)), but they are different. Why this happend becouse I don't see any error in transforming the first expression in the second one. Someone help me
They are indeed slightly different. Probably has to do with the fact that you are "cheating" out of the square root in the numerator so those negative values for sinx and cos are gone. Try putting in |cosx|-|sinx|/(sqrt(sinxcosx). Then you should get the original function. In these integral problems very often we get rid of the absolute value since it simplifies problems and doesn't affect the result in the end too much.
int of √tanx dx tanx=u² 2udu=sec²xdx dx=2udu/(u²-1) again, int of √tanx dx = int of u×2udu/(u²-1) = 2[int of u²du/(u²-1)] =2[int of (u²-1+1)du/(u²-1)] =2[int of du + int of du/(u²-1)] =2[u+int of du/(u²-1)] =2[u+½log|(u-1)/(u+1)|] +k =2u+log|(u-1)/(u+1) +k =2√tanx + log|(√tanx -1)/(√tanx +1)| +k Therefore, Int of √tanx dx = 2√tanx + log|(√tanx -1)/(tanx +1)| +k
Hi!
hi ooon hann
Oon Han
I see factoreo!
lol pinned
H times i factoreo where H is e
lol
Literally never been more impressed with any maths video on UA-cam this is so well explained and clear and clever
Yay!!! Thank you!!
Yet another video demonstrating the beauty of calculus. Love it.
Yay!
Thanks so much.I had to integrate the conjugate of this but couldn't do it until today.It really was amazing!!!
Yay!!!
Every time I watch you do integrations my mind is just blown by the brilliance of how it works the first try every time!
He probably practices before recording...
Yea.
Lol
this is awesome :D and thanks for doing the trig sub part instead of saying "and from there on it's easy" :P
Yayyyy!
@@blackpenredpen PLEASE respond..didnt you first think to replace sinxcosx by the double angle sine formula sin2x divided by 2 and replace the top by square root of cos2x..seems more logical and intuitive than what yiu did..inwould think a lot of people did that.this is why math is so damn infuriating and ridiculous so much..
so you turned a trig integral into algebrical integral and then back to trig to finally solve it.
dafuq?
Exactly! Told you it's a clickbait already.
Solve my exercise with bisector to get u sub which allows you to avoid inverse trig substitution
i63.tinypic.com/rkvnkg.png
In triangle ABG length of two sides are given
Angle BAG is labelled as theta
GL is bisector of angle complementary to theta
Express the ratio GB/BL in terms of given length of sides
Well here's what I call it:The joyride effect.
Is there anything this channel doesn´t have? Thank god i found it! please never stop making videos
I was once given this exact problem, but with bounds on the integral. I exploited the symmetry of the bounds because I figured there was no elementary antiderivative. Fascinating video!
This is what I was waiting for!!!!!.... Thanks BPRP
Hey man, thanks for this video. And this is actually an iit jee classic. Your solution helped me solve a problem related to this format.
6:41. That's not tricky. I don't know why foreigners ( I'm indian) teach this integral separately. We have direct formula for integrals of the form x²-a²,a²-x²,root x²-a²,root a²-x², 1/x²-a² ,1/a²-x², 1/root x²-a², 1/root a²-x². We have direct formulas for these in the NCERT textbook. 😄
Great video, greetings from Colombia!!
Thank you!!!!
Surely you can quote integral of 1/sqrt(u^2 -1) = arcosh(u) ? So the answer can be written as sqrt(2)arcosh(cosx + sinx) +C.
yes
"clickbait" lmao I love this so much
yes! : )
I really liked this video but please don't stop the combinatory videos. They break the constant integral videos on your channel pretty nicely
Yayyy. I try to change flavors ones in a while. Glad that you like it!!!
As always, that's another great video!
Thanks for another great video!
: )
x, u and theta world
Show me your world...
Nice. Now split the problem and integrate them term by term
I did that already, even the crbt(tan(x))
I really enjoy watching your videos... congratulations
nice integral! blackpenredpen would you try solving the integral from -1 to 1 of x^2/(e^(x)+1)?
Wow, the answer to that is just 1/3, so clean. (thanks to WFA tho), I will think about the steps when I have time.
yes, such an unexpected answer. Whenever you have time :)
No clickbait, but a solid BPRP Performance as always!
You could also convert cotx to 1/tanx and then proceed further.
Would it not have been easier to solve the equation in the note for sin x cos x and substitute it back into the integral so one doesnt have to add and subtract the 1?. ;)...also cute kitten.
hmmm, I think it's about the same. And thank you!! : )
Very cool! Thanks for sharing!
My pleasure!!
I watched the Chinese version, I still understood it though. :-) Well 'explained'. Those substitutions are really smart.
If, at 1:30 you had left the numerator as sqrt((cos x)^2 - (sin x)^2), couldn't you have then converted that sqrt(cos 2x)? Which would then allow you to convert the denominator into sqrt ( 1/2 sin 2x). Bringing that back into a single square root that would have given sqrt( (cos2x)/(1 /2 sin 2x) ), which equals sqrt (2 cot 2x). Wouldn't it have been easier to solve that way?
This is a very late reply, but remember, sqrt((cos(x))^2)-sqrt((sin(x))^2) is not the sqrt((cos(x))^2-(sin(x))^2).
8:45 "we are ready to integrate"
Yesssss the fematika shoutout!!
Yup!
We can solve it by arctanhx function
Let sqrt tanx = u
So that integral can be 2^0.5 arctanh((sqrt tanhx + sqrt cothx)/2^0.5)
This approach is like the way of integration of sqrt tanx
when you put u=√(tan(x))
you'll get the integral is equal to
√(0.5)ln([tan(x)+√(2tan(x))+1]/[tan(x)-√(2tan(x))+1])+C
holy shit this is such a good quality video
cotx =1/tanx and use substitution u^2=tanx
I'm actually disappointed about the fact that Fematika really has less subscribers.
Thank you so much for the compliment!
He definitely deserves more! : )
Yep definitely the most underrated UA-cam channel
You could have used standard result that integral 1/(x^2 - 1) dx = arccosh(x) + C
And arccosh(x) = ln(x + sqrt(x^2 - 1))
I tried this integral by PI and using the DI method for help. I gave up after 15 minutes because the number of integrals kept increasing lol.
I used u=sinh(theta) and got arctanh(sin(x)+cos(x))*sqrt(2)
It's correct.
Very nice sir!!!
blackpenredpen, Hey, man extremely stunning approach of solution, I wonder how you easily launch your pure math imagiation and miracle creativity.Are you math professor of University or just olympiad winner student? I love your way of solution. Unfortunately, I can't put double thumb-up sign above your video, wish you further promotion as a blogger.
Won't it be sqrt(2)(inverse cosh(sinx + cosx)?
Yes.
so good!
hehehe
dear master... thank you!
Beautiful!!
I may have found an error. I used Desmos to set f(x)= sqrt(2)*ln(abs(sinx+cosx+sqrt(sin2x))). Then I plotted f'(x). Then I plotted sqrt(cotx)-sqrt(tanx). But these functions don't line up. In a way, there may be a negative. See image of desmos screen here: i.imgur.com/y5FAEpK.jpg
I looked into this too, but I don't see it! They look identical to me when I graph the difference between the two (one solution sub the other). I can't get the computer algebra systems to integrate this integral to anything easy or anything that will simplify down, but that can just be down to how it is complicated to simplify. Numerically it seems correct.
And now ... you can make integral of sqrt(cot(x))+sqrt(tan(x)) ... and get by adding and substracting the both famous integrals of sqrt(cot(x)) and sqrt(tan(x))
... and Voilà ... isn't it !
I still don't understand why we can just forget about the absolute value??
Because I think that y = sqrt(cot(x)) - sqrt(tan(x)) *is different with* y = (cos(x) - sin(x))/sqrt(sin(x)cos(x)), and it will produce a different integral
To do it completely right, you need to determine the range where the integrand is integrable, after which you can skip the absolute values if their arguments are always positive in the valid range. In the valid range near 0, sin x and cos x are both positive, for example.
CLICKBAIT has to get a like^^ (like all other videos anyway)
Yayy!!!
so beautiful
Aw someone who never took Integral calculus, is the trick that it would have been easier doing it the other way? Was the double substitution unnecessary?
Now you should try to integrate cbrt(cot(x) )-cbrt(tan(x)) dx
Nice bro!
Can u plz solve this.....|x| + |y| + |x+y| less than equal to 2...find the area of the region in xy plane which satisfies this inequality...
I will have to think about it.
AWESOME!
Yay.
I got as far as deciding the original was the same as sqrt(cot(x) + tan(x) - 2) but then gave up.
Why didn't you draw a box when you have finished the solution
Why not simplly mutiply and divide by sqrt(tan(x)) and usub rt(tan(x))
Please help me, I am in the limited developments and I know that the inverse of a sum isn't the sum of the inverse. But I wonder what is the inverse of a sum? Especially what is the inverse of 1+x^2/2!+x^4/4!+x^6/6!+...?
yes but what I really searched is the limited development of sech(x). I wonder if you can find this development just by doing the inverse of the one of the cosh(x), but that's not easy.
yo,i have a question for you regarding the integration of (1/(x^2-1))dx;in that manner,can't we think (-1) as being i^2(if we integrate in a complex plane),and the bottom becomes x^2+i^2,so we use the formula for arctan,and thus getting (1/i)*arctan(x/i) +c , is that a valid answer?
Yes, that's a valid answer in a complex world (complex number system), but not in a real world.
Because actually, when x is a real number, the i will be cancelled out. That's why it's not actually necessary to use "i" in the equation.
For me, I like to solve it using hyperbolic trigonometry substitution giving :
integral of (1/(x^2-1))dx = -arctanh(x) + C
No imaginary number needed :)
I know that answer aswell,but i wondered if that would work,because it got stuck in my head during one class and i havent found anything related to that on the internet;thank you for your answer ;)
You're welcome
I have a different answers. I have seen even another answer in a book. What's going on?
This question is from which book ?
Hi! Can I use this trick for Integral of sqrt(cot(x))+sqrt(tan(x))? Please do let me know.
Halo...Nice entry..
i never clicked faster!
Really interesting, thanks!
: )
I tried to plot the graphs of sqrt(cot(x)) - sqrt(tan(x)) and of (cos(x) - sin(x)) /sqrt(sin(x) cos(x)), but they are different. Why this happend becouse I don't see any error in transforming the first expression in the second one. Someone help me
They are indeed slightly different. Probably has to do with the fact that you are "cheating" out of the square root in the numerator so those negative values for sinx and cos are gone. Try putting in |cosx|-|sinx|/(sqrt(sinxcosx). Then you should get the original function. In these integral problems very often we get rid of the absolute value since it simplifies problems and doesn't affect the result in the end too much.
Why exactly is this clickbait again?
Diego Tejada as long as click on it, it's a clickbait
Those are wise words. I'd like to start a petition to start every single title with "CLICKBAIT" except on April 1st, because it's a holiday
I will see what I can do. : )
Omg bprp you farging troll XD
I was expecting clickbait, I got clickbait. Bravo, good sir.
Nice integral! Where are you finding these integrals?
Eightc from other viewers
You failed to use this in finding integral √tan x,
Make a pdf compilation of cool integrals and leave it in the description. uwu
Btw, you already knew from previous steps that
tanθ = √(u²-1)
And that
√(u²-1) = √(2sinθcosθ)
😉🙂
Sir..5th step is direct formula
Good.
R u attention to me ....ohhh thanks sir 😚😚
The twist is that its actually just a NCERT example from class 12th 🤣
so whats the difference between NCERT and JEE?
Is JEE a special exam of some sort that people have to take to get into IIT?
Thanks sir
oh, I get it...
Yay!
....sneeky solution, but smart !
thanks
Hi, I was just wondering,
How can you just say that sqrt(cos^2(x)) = cos(x)? Isn't that |cos(x)|?
It is! But we ignore that for a second for integration purpose.
when it's just an indefinite integral
I see, thank you!
Bravoooo😚
jus use integral linearity? Wut?
sure.
What is sqrt(sqrt(e^e))
Meme Master 69
1.9730300492
e^(e/4)
Mistake in line 3. There should be modules. Cos and sin can be negative. Am I right?
Nice
Where is oreo? :"v
He will be up in the other videos later : )
He is also Caro's cat.
Yayyyyy!!
1,400 views 140:0 like:dislike ratio
yay!!!
Please make a video of checking of its derivative if it is the same from the original question. Thank you
hard one
: )
It's time for you to integrate them separately, and then simplify it, so that your video can be longer XD
How we show this equation physically?
How triangle change according to this equation ?
Simple
NUUU, CLIKBAITUUUUU!!! :)))
yay!!!!!!
Very cool! More Clickbait :-)
OK!!!
Integral clickbait, wew!
Yay!!!!!!!!!!
int of √tanx dx
tanx=u²
2udu=sec²xdx
dx=2udu/(u²-1)
again,
int of √tanx dx
= int of u×2udu/(u²-1)
= 2[int of u²du/(u²-1)]
=2[int of (u²-1+1)du/(u²-1)]
=2[int of du + int of du/(u²-1)]
=2[u+int of du/(u²-1)]
=2[u+½log|(u-1)/(u+1)|] +k
=2u+log|(u-1)/(u+1) +k
=2√tanx + log|(√tanx -1)/(√tanx +1)| +k
Therefore,
Int of √tanx dx = 2√tanx + log|(√tanx -1)/(tanx +1)| +k
Brpr caan you answer your email
What did you send me?
blackpenredpen an integral q
😶
: )
oh shet
Its nice that you put clickbait on title. Great ;)
Hehehe!