I have a different approach here. First define u=cos(x), so du = -sin(x)dx and substitute it into the integral. Getting the result: integrate arccos(u) / (1 + u^2) du from -1 to 1. Then, eyeball that 1 / (1 + u^2) is actually the derivative of arctan(u), so we can do integration by parts. Getting the result: arctan(u) * arccos(u) + integrate arctan(u) / sqrt(1 - u^2) du from -1 to 1. Now we can notice that arctan(u) is an odd function, and sqrt(1 - u^2) is an even function, hence arctan(u) / sqrt(1 - u^2) is an odd function. Integrate an odd function from -1 to 1 is 0, so we are only left with the boundary term. Now evaluate arctan(u) * arccos(u) from -1 to 1, to obtain the result: pi^2 / 4.
The peasants have been trying to construct their own property as of late. Thank goodness nothing has come of it. If they prove successful, they may just leave the disintegration of the calculus aristocracy in their wake…
Integrating functions of the form x*f(sinx) from 0 to pi is a rather common integration exercise/problem. We can use this trick to prove that this more general integral is equal to the integral of f(sinx)*pi/2
I evaluated it before watching the video, and my method was to do a substitution x = pi - u and solve for the desired integral in terms of a similar one. I then multiplied the numerator and denominator by sec^2 u, changed it from an integral of 0 to pi to 2* an integral of 0 to pi/2, and did another substitution t = sec u. It was trivial to evaluate after that.
i literally did a question on the same property from an old video of yours !! it was a cambridge problem on int of xf(sinx) from 0 to pi which had 3 parts ! i was so excited for this one because i knew how to approach this !!
I just divided the top and bottom by cos^2. this left me with an easy ibp, with integrating tanxsecx/(sec^2(x)+1) as an easy you sub. the only part left is to find int from 0 to pi of arctan(sec(x)). let pi/2-x=u. this gives you the int from -pi/2 to pi/2 of arctan(csc(x)). since this is an odd function from -a to a it is 0. this leaves the whole problem as xarctan(sec(x)) (from IBP) so I=pi^2/4
You can also IBP and you will get pi²/4 minus the integral of a function (let's call it g(x) ) where g(x) = tan^-1(cosx) You then substitute u = cosx and you will obtain an integral between -1 and 1, and you just have to show that this integral is odd, so it's equal to 0 and then you find (pi²/4)-0 QED
i did use the same IBP as you did at the start, getting the int. to be pi squared over 4 + int from 0 to pi of arctancosxdx. Then I used the same trick as you to show that the remaining integral is 0, using cos(pi-x)=-cosx, the oddity of the arctan func., and the linearity of the integral operator.
@@Latronibus i actually started with the taylor series, but then i noticed each cosine integral was 0 and i said to myself 'there must be an easier way of seeing this' and went back.
I feel like I am animating some of the few neurons I still have back to life!! Thank you so much for making me remember the best moments of my high school.
Professor Leonard is a legend, hands down was really friendly, easy to understand, and has a lot of content for you to follow. I took calc 3 in this passing summer, and he was a big help for when I didn't understand the topics from my instructor. My favorite calc was calc 2, then 3, then 1 in that order, but I agree with what you said about the difficulty, calc 2 is the easiest, then calc 1, then calc 3. Just make sure you spend a lot of time practicing the material especially in the second half of the semester, because the class starts to build a lot more and it's easy to fall behind. I ended up getting a B+ because I didn't spend enough time remembering the formula for Stokes' theorem. I wish you the best of luck this semester though!
@@WingedShell82 thankfully I'm allowed a cheat sheet. Handwritten notes in an A4 paper to bring into the exam, double sided. It'll be easy because there's not much memory work involved
@@nvapisces7011 Oh that nice, well then I'd say just make sure your mechanical skills are in tip-top shape lol. Practice is really needed in that class, just so you know what you're doing.
When I first saw calc 3, I feel the exact same. You need to understand several rules of partial derivatives first, before even starting to learn del operator and how to do multi-dimensional integration. And about polar/cylindrical coordinates, and spherical coordinates conversions...memorize those equations, there is no shortcuts. This is probably the most difficult part, because you are not used to that. Once you master all basic of calc 3, you can try solving partial differential equations, but don't expect that you can solve all, because no mathematicians can.
Used the substitution u = cos(x), then replaced arccos(x) by π/2 - arcsin(x) and saw, that arcsin(x)/(1+x^2) is odd and being integrated over a symmetric domain so it has to be 0, the other part was the answer
I computed this integral before watching your video, but my method is different so I write it below: Let A be the integral we need to compute. Since the variable x is in [0, π] and cos : [-1, 1] --> [0, π] is a bijection, we can make the substitution t = cos(x) in A (hence x = cos^-1(t) and dt = -sin(x)dx), thus A = int_1^(-1) -cos^-1(t)/(1+t^2) dt = int_(-1)^1 cos^-1(t)/(1+t^2) dt. Noticing that t ⟼ 1/(1+t^2) is the derivative function of tan^-1, we will do an integration by parts in this last integral, hence we get : A = [tan^-1(1)cos^-1(1)-tan^-1(-1)cos^-1(-1)] - int_(-1)^1 -tan^-1(t)/sqrt(1-t^2) dt = π^2/4 + int_(-1)^1 tan^-1(t)/sqrt(1-t^2) dt. But the integrand of the last integral is odd and the center of [-1, 1] is 0, hence this last integral is equal to 0. Then A = π^2/4 + 0 = π^2/4.
3:45 why this example doesn't make sense to me? Can someone explain to me for example why if I put the angle to 120 degrees and then add π I will have a sin value which will be of the opposite sign! So why is this considered to be treated as equal? sin (π-a) = sin(a)
You totally misunderstood. You’re never adding pi to anything. pi-x is a reflection of x over the y axis. If you wanna think in degrees, it’s 180-x. sin(120) equals sin(180-120). Flipping along the y axis doesn’t change the y value. Of course pi-x isn’t the same as pi+x. So where did you get “adding pi” from?
@@GooogleGoglee Yes, that’s right. Works for any angle. When you think of an angle increasing, you start on the 0 line and go counter-clockwise, but pi-x is like starting on the 180 line and going clockwise
I see some comments of Indians bragging about that these questions are easy they did it in 12 grade , we know that you can do just don't brag about it , it becomes annoying, appreciate the teacher, don't be 'i know it all guy'
A more stupid way: expand it in powers of cos, getting a sum of x sin(x) (-1)^n cos(x)^(2n) integrals. By parts each antiderivative is (-1)^(n+1) x cos(x)^(2n+1)/(2n+1) + (-1)^n (integral of cos(x)^(2n+1)/(2n+1)). That looks like an annoying reduction formula thing but the integrand is odd wrt pi/2 now so these guys are just zero. You get left to sum pi (-1)^n /(2n+1) and get the same result by recognizing the Maclaurin series for arctan.
Not joking or bragging it was so easy that I solved it in my head seeing the thumbnail,pls make videos on complicated integrals,maybe vector integration or multiple integration,it would be really fun!
I used IBP and came up with ln[(pi)^2 + 1]/2, but that’s about four times too low. The integral calculator website that I use gave a numerical approximation around 2.5, about half the solution in the video. (pi)^2 / 4, the solution in the video, is around 4.9.
I worked out the integration by parts fully. Bc I didn’t know what exactly it was with that integral, I just said it was the following: -xtan^-1(cosx)]0 to pi + integration 0 to pi(tan^-1(cosx)) First part comes out to pi^2/4 Second part I thought of it in more fundamental terms of what’s happening… From 0 to pi, cos is between -1 and 1. So, it’s finding the area of tangent’s angle when it is between -1 and 1. So you’re just summing up an equal amount of positives and negatives, making it 0. Hence, pi^2/4 + 0 = pi^2/4. If there was some gap in my reasoning, pls point it out. Tho I do have to admit, cool use of kings rule 😎😮💨
I'm taking Calc2 right know, so I didn't know this clever witchcraft lmao But I got to int arctan(cosx) dx and, having the (pi-x) integral, I managed to solve it! Very nice :)
Integral 0 to π [f(x)] dx = integral 0 to (π/2) [f(x)+f(π-x)] dx Its based on the same property but its very cool tho. You can easily eliminate x here.
This question is like the staple of HKDSE M2 paper. Except the (pi-x) part is usually an part a question while the actual integral being a part b, using result of part a. Essentially nerfing the question as it provides hint.
I tried switching the sin and cos for the hell of it, but sadly the I from the other side canceled instead of adding and it came out to… -pi tan^-1(sin x)]0 to pi = 0 Which upon plugging in, checks out 💪🥴
Well, are there any specific conditions to apply "f(x)dx to f(a+b-x)dx" property? Because the region where I live, I have been told that I should use it via trial and error.
The DI method which you denied can actually work, what you need to prove is: for x ∈ [0,π/2], arctan(cos(π/2 + x)) = - arctan(cos(π/2 - x)) Follow the cosine curve, it is straightforward that cos(π/2 + x) = -cos(π/2 - x), and arctan is an odd function, so arctan(cos(π/2 + x)) = - arctan(cos(π/2 - x)) So the positive part equals to the negative part, like what you get when you integrate an odd function with lower bound -b and upper bound b.
You can also set u = x - pi/2, then you rewrite the integral as the integral of (u+ pi/2)sin(u + pi/2)/(1+ cos^2(u +pi/2)) du from -pi/2 to pi/2. Notice that you can rewrite the integrand as (u + pi/2)cos(u)/(1 + sin^2(u)). The integral over ucos(u)/(1 + sin^2(u)) will evaluate to 0 as we are integrating an odd function from -a to a. So we are just left with the integral over (pi/2)cos(u)/(1 + sin^2(u)) du from -pi/2 to pi/2. This is a fairly standard integral, as we can multiply by two and change the boundaries, so that we integrate from 0 to pi/2 (notice that the integrand is even). This evaluates to pi*(arctan(sin(pi/2)) - arctan(sin(0))), which is just pi^2/4, if you know that arctan(1) = pi/4 and arctan(0) = 0. (the integral of cos(u)/( 1 + sin^2(u)) could be solved via substituting z = sin(u) again, but I think just seeing that the integrand is the derrivative of arctan(sin(u)) shouldn't be a problem if you know the derrivative of arctan.) That would have been my solution but yours is awesome, keep up the good work! :)
Hi I know this is weird but can anyone help me solve this? yln(0.5y+0.05) = (|cos(1.7x)|-1) solve for y Edit: ill give some context this started as y(ln((y+0.1)/2)) =(|cos(1.7x)|-1) And I simplified it down to yln(0.5y+0.05) = (|cos(1.7x)|-1)
Remember when you have a x multiplied to your function which you want to integrate, and if you get the same function (except the multiplied x) on replacing x with (b+a -x) in f(x) , then just replace the X, and you get a integratable function. If not just use queen rule , you might get something with it
Hello this is Sayed Yousaf from Afghanistan. I found a difficult math question. If you help and solve me it would be your pleasure. The question is: The limit x approaches 0 (x^x^...^x-x!)/(x!^x!-1)
@@AbhayShakya-nk2tiThey do, its just really really annoying that these types of "I live in country x and solve this at age y" are under every video. They are the math channel equivalent of the sex bot comments.
@@AbhayShakya-nk2ti it is not the issue that people can't handle, just the fact that the schools didn't teach in that corresponding grade. Therefore knowing the technique in earlier stages isn't a privilege for those in certain countries😅.
This is common problem in India for -12 gred actuly very easy you give this problem to -infinity grader in India hel be doing the laughter on you and give you the anser because he is knowing the answer from the birth. Please give actual hard intergal next time
Learn calculus on Brilliant: 👉brilliant.org/blackpenredpen/ (20% off with this link!)
i will
Sir pleast this question slove intergration of( tan inverse x)^3/x dx please sir 😢
Howdy 👋👋
King's rule is a really beautiful trick!
Indeed!
So true
@@jayantgautam9273 Who is "King" of King's Rule?
@@carultch that would be me, my dear citizen 🧐
King rule name got so famous, its actually named by Shri VK Bansal sir of Bansal Classes KOTA.
I have a different approach here.
First define
u=cos(x), so du = -sin(x)dx
and substitute it into the integral.
Getting the result:
integrate arccos(u) / (1 + u^2) du from -1 to 1.
Then, eyeball that 1 / (1 + u^2) is actually the derivative of arctan(u), so we can do integration by parts.
Getting the result: arctan(u) * arccos(u) + integrate arctan(u) / sqrt(1 - u^2) du from -1 to 1.
Now we can notice that arctan(u) is an odd function, and sqrt(1 - u^2) is an even function, hence arctan(u) / sqrt(1 - u^2) is an odd function. Integrate an odd function from -1 to 1 is 0, so we are only left with the boundary term.
Now evaluate arctan(u) * arccos(u) from -1 to 1, to obtain the result: pi^2 / 4.
Damn man I really expected you would have found the indefinite of it only to see you used odd fxn rule 😕 🥲
@@ThorfinnBus Tks for pointing that out, I guess I will loss some points for didn't notice that.
King and queen property are indeed very useful in solving definite integrals.
What's the queens property
The peasants have been trying to construct their own property as of late. Thank goodness nothing has come of it. If they prove successful, they may just leave the disintegration of the calculus aristocracy in their wake…
Bro Watching Ashish Aggarwal Sir's Lecture. This is a easy question.
@@anonymous29_lol I'm from there too
@@darcash1738 ever heard of ramanujans integral???.......go back to doing what u all do best..scrolling tiktoks
Integrating functions of the form x*f(sinx) from 0 to pi is a rather common integration exercise/problem. We can use this trick to prove that this more general integral is equal to the integral of f(sinx)*pi/2
Michael Penn already talked about this on one of his videos so I already knew how to approach this integral.
You can generalize it to f(sin x, cos^2 x)
@@TheEternalVortex42 cos^2 = 1-sin^2 a function of sin
This guy is an absolutely masterful teacher.
I evaluated it before watching the video, and my method was to do a substitution x = pi - u and solve for the desired integral in terms of a similar one. I then multiplied the numerator and denominator by sec^2 u, changed it from an integral of 0 to pi to 2* an integral of 0 to pi/2, and did another substitution t = sec u. It was trivial to evaluate after that.
Very subtle solution - great👍
i literally did a question on the same property from an old video of yours !! it was a cambridge problem on int of xf(sinx) from 0 to pi which had 3 parts ! i was so excited for this one because i knew how to approach this !!
I just divided the top and bottom by cos^2. this left me with an easy ibp, with integrating tanxsecx/(sec^2(x)+1) as an easy you sub. the only part left is to find int from 0 to pi of arctan(sec(x)). let pi/2-x=u. this gives you the int from -pi/2 to pi/2 of arctan(csc(x)). since this is an odd function from -a to a it is 0. this leaves the whole problem as xarctan(sec(x)) (from IBP) so I=pi^2/4
You can also IBP and you will get pi²/4 minus the integral of a function (let's call it g(x) ) where g(x) = tan^-1(cosx)
You then substitute u = cosx and you will obtain an integral between -1 and 1, and you just have to show that this integral is odd, so it's equal to 0
and then you find (pi²/4)-0
QED
You are amazing man ❤
i did use the same IBP as you did at the start, getting the int. to be pi squared over 4 + int from 0 to pi of arctancosxdx. Then I used the same trick as you to show that the remaining integral is 0, using cos(pi-x)=-cosx, the oddity of the arctan func., and the linearity of the integral operator.
Very nice observation
This is equivalent to what I did with Taylor series but with significantly less algebra overhead.
@@Latronibus i actually started with the taylor series, but then i noticed each cosine integral was 0 and i said to myself 'there must be an easier way of seeing this' and went back.
It is the same as the Cambridge integral that u do it previously
U can use the identity of xf(sinx)
I feel like I am animating some of the few neurons I still have back to life!!
Thank you so much for making me remember the best moments of my high school.
Thanks PROF as always very interesting 👍
I'm currently starting my calc 3 and it's crazy hard. I found calc 2 to be the easiest, followed by calc 1, with calc 3 the hardest
go to professor leonard.
Professor Leonard is a legend, hands down was really friendly, easy to understand, and has a lot of content for you to follow. I took calc 3 in this passing summer, and he was a big help for when I didn't understand the topics from my instructor. My favorite calc was calc 2, then 3, then 1 in that order, but I agree with what you said about the difficulty, calc 2 is the easiest, then calc 1, then calc 3. Just make sure you spend a lot of time practicing the material especially in the second half of the semester, because the class starts to build a lot more and it's easy to fall behind. I ended up getting a B+ because I didn't spend enough time remembering the formula for Stokes' theorem. I wish you the best of luck this semester though!
@@WingedShell82 thankfully I'm allowed a cheat sheet. Handwritten notes in an A4 paper to bring into the exam, double sided. It'll be easy because there's not much memory work involved
@@nvapisces7011 Oh that nice, well then I'd say just make sure your mechanical skills are in tip-top shape lol. Practice is really needed in that class, just so you know what you're doing.
When I first saw calc 3, I feel the exact same.
You need to understand several rules of partial derivatives first, before even starting to learn del operator and how to do multi-dimensional integration.
And about polar/cylindrical coordinates, and spherical coordinates conversions...memorize those equations, there is no shortcuts. This is probably the most difficult part, because you are not used to that.
Once you master all basic of calc 3, you can try solving partial differential equations, but don't expect that you can solve all, because no mathematicians can.
Thats craaaaazzyyy. A function that just contains itself? This is so weird. I hope I can one day solve an integral like that myself without help!
Used the substitution u = cos(x), then replaced arccos(x) by π/2 - arcsin(x) and saw, that arcsin(x)/(1+x^2) is odd and being integrated over a symmetric domain so it has to be 0, the other part was the answer
这也是大陆这边大学一年级课后习题里的一道很基础，也经典的题目，我们管这种方法叫区间再现公式。用这种方法衍生出的公式，当遇见积分，形如从0-pai integral x f（sinx），可以将x提出积分号变成pai/2
I computed this integral before watching your video, but my method is different so I write it below:
Let A be the integral we need to compute. Since the variable x is in [0, π] and cos : [-1, 1] --> [0, π] is a bijection, we can make the substitution t = cos(x) in A (hence x = cos^-1(t) and dt = -sin(x)dx), thus A = int_1^(-1) -cos^-1(t)/(1+t^2) dt = int_(-1)^1 cos^-1(t)/(1+t^2) dt.
Noticing that t ⟼ 1/(1+t^2) is the derivative function of tan^-1, we will do an integration by parts in this last integral, hence we get : A = [tan^-1(1)cos^-1(1)-tan^-1(-1)cos^-1(-1)] - int_(-1)^1 -tan^-1(t)/sqrt(1-t^2) dt = π^2/4 + int_(-1)^1 tan^-1(t)/sqrt(1-t^2) dt.
But the integrand of the last integral is odd and the center of [-1, 1] is 0, hence this last integral is equal to 0. Then A = π^2/4 + 0 = π^2/4.
No one cares m8
@@danielotoole9053 i do
@@danielotoole9053 the point of this comment section isn't about what you care.
3:45 why this example doesn't make sense to me? Can someone explain to me for example why if I put the angle to 120 degrees and then add π I will have a sin value which will be of the opposite sign! So why is this considered to be treated as equal? sin (π-a) = sin(a)
You totally misunderstood. You’re never adding pi to anything. pi-x is a reflection of x over the y axis. If you wanna think in degrees, it’s 180-x. sin(120) equals sin(180-120). Flipping along the y axis doesn’t change the y value. Of course pi-x isn’t the same as pi+x. So where did you get “adding pi” from?
@@hydropage2855 so sin (π-200) is the same of sin (200) ?
@@GooogleGoglee Yes, that’s right. Works for any angle. When you think of an angle increasing, you start on the 0 line and go counter-clockwise, but pi-x is like starting on the 180 line and going clockwise
@@hydropage2855 got it, thank you for your time. It is hard for me, sometimes, to see it not in a visual form. I have got it now. Thank you
As a HK DSE student,I have seen this type of integration for millions times
Literally done it orally ....well i know anyone could do that🙂 easy question
I see some comments of Indians bragging about that these questions are easy they did it in 12 grade , we know that you can do just don't brag about it , it becomes annoying, appreciate the teacher, don't be 'i know it all guy'
Spot on, it's a tutorial. And the problem is for that age range as well, so they aren't doing anything special in India when they're bragging.
A more stupid way: expand it in powers of cos, getting a sum of x sin(x) (-1)^n cos(x)^(2n) integrals. By parts each antiderivative is (-1)^(n+1) x cos(x)^(2n+1)/(2n+1) + (-1)^n (integral of cos(x)^(2n+1)/(2n+1)). That looks like an annoying reduction formula thing but the integrand is odd wrt pi/2 now so these guys are just zero. You get left to sum pi (-1)^n /(2n+1) and get the same result by recognizing the Maclaurin series for arctan.
Not joking or bragging it was so easy that I solved it in my head seeing the thumbnail,pls make videos on complicated integrals,maybe vector integration or multiple integration,it would be really fun!
I used IBP and came up with ln[(pi)^2 + 1]/2, but that’s about four times too low.
The integral calculator website that I use gave a numerical approximation around 2.5, about half the solution in the video.
(pi)^2 / 4, the solution in the video, is around 4.9.
No, you are mistaken here. π² is roughly 10, so π²/4 is indeed approximately 2.5.
Its also plausible with the residue theorem to solve this (complex analysis)
I missed this gentleman-guy too much.
I worked out the integration by parts fully. Bc I didn’t know what exactly it was with that integral, I just said it was the following:
-xtan^-1(cosx)]0 to pi + integration 0 to pi(tan^-1(cosx))
First part comes out to pi^2/4
Second part I thought of it in more fundamental terms of what’s happening…
From 0 to pi, cos is between -1 and 1. So, it’s finding the area of tangent’s angle when it is between -1 and 1. So you’re just summing up an equal amount of positives and negatives, making it 0.
Hence, pi^2/4 + 0 = pi^2/4.
If there was some gap in my reasoning, pls point it out. Tho I do have to admit, cool use of kings rule 😎😮💨
Been a sec since I practiced my integrals. Need to try this myself!
Mind blowing 🤯🤯🤯
Awesome!
I'm taking Calc2 right know, so I didn't know this clever witchcraft lmao
But I got to int arctan(cosx) dx and, having the (pi-x) integral, I managed to solve it! Very nice :)
Integral 0 to π [f(x)] dx = integral 0 to (π/2) [f(x)+f(π-x)] dx
Its based on the same property but its very cool tho.
You can easily eliminate x here.
Semplicemente geniale!
Awesome bprp👍
Hey BPRP I started Calc 2; wish me luck I will be using your videos for help!
As a Taiwanese, I didn’t know Taiwan had their own version of Reddit
Please derive voulme of pyramid using inttegral
Whenever you see trig and some pis, you know it’s kings rule.
This question is like the staple of HKDSE M2 paper. Except the (pi-x) part is usually an part a question while the actual integral being a part b, using result of part a. Essentially nerfing the question as it provides hint.
I tried switching the sin and cos for the hell of it, but sadly the I from the other side canceled instead of adding and it came out to…
-pi tan^-1(sin x)]0 to pi = 0
Which upon plugging in, checks out 💪🥴
Blackpenredpen, how do you find the derivative of the gamma function?
Well, are there any specific conditions to apply "f(x)dx to f(a+b-x)dx" property?
Because the region where I live, I have been told that I should use it via trial and error.
No conditions, Its always true
Integral is just an area, by applying this property, We are just shifting the origin, Area will still be same
@@AnshTiwari11 Thanks
solved it in my head, please do solve some more complicated integrals if you get the time, thanks
Sir is ilate rule universal in integration?....if not what is counter example?
I thought NTU you typed referred to Nanyang Technological University in Singapore which I might be interested in getting into lol
It's simple write -sin²x from 1+cos²x and after that it will become - integral of xdx/sinx and then it's very easy to solve
-sin^2x is equal to 1-cos^2x
The DI method which you denied can actually work, what you need to prove is:
for x ∈ [0,π/2], arctan(cos(π/2 + x)) = - arctan(cos(π/2 - x))
Follow the cosine curve, it is straightforward that cos(π/2 + x) = -cos(π/2 - x),
and arctan is an odd function, so arctan(cos(π/2 + x)) = - arctan(cos(π/2 - x))
So the positive part equals to the negative part, like what you get when you integrate an odd function with lower bound -b and upper bound b.
derivative of cotang is 1/(1+x^2).. how you did for sin x ?
What should I think about myself if I was able to get to the answer in my head? I don't think this was a fairly tough question tho
HAIL king,queen ,jack,ace
You can also set u = x - pi/2, then you rewrite the integral as the integral of (u+ pi/2)sin(u + pi/2)/(1+ cos^2(u +pi/2)) du from -pi/2 to pi/2. Notice that you can rewrite the integrand as (u + pi/2)cos(u)/(1 + sin^2(u)). The integral over ucos(u)/(1 + sin^2(u)) will evaluate to 0 as we are integrating an odd function from -a to a. So we are just left with the integral over (pi/2)cos(u)/(1 + sin^2(u)) du from -pi/2 to pi/2. This is a fairly standard integral, as we can multiply by two and change the boundaries, so that we integrate from 0 to pi/2 (notice that the integrand is even). This evaluates to pi*(arctan(sin(pi/2)) - arctan(sin(0))), which is just pi^2/4, if you know that arctan(1) = pi/4 and arctan(0) = 0. (the integral of cos(u)/( 1 + sin^2(u)) could be solved via substituting z = sin(u) again, but I think just seeing that the integrand is the derrivative of arctan(sin(u)) shouldn't be a problem if you know the derrivative of arctan.) That would have been my solution but yours is awesome, keep up the good work! :)
Hi I know this is weird but can anyone help me solve this?
yln(0.5y+0.05) = (|cos(1.7x)|-1) solve for y
Edit: ill give some context this started as
y(ln((y+0.1)/2)) =(|cos(1.7x)|-1)
And I simplified it down to yln(0.5y+0.05) = (|cos(1.7x)|-1)
Could Feynman Technique be used?
i think it will make it more harder to solve
Why go nuclear when dynamite is enough?
Just asking if there is a second way to solve.
THIS IS OUR 12TH STANDARD BASIC STUFF OF INTEGRAL CALCULUS.❤
In India?
@@Anmol_Sinha ya ofcourse
yes brother . and i saw the thumbnail of this video i get it that it is simple question hehehe even by 10yrs old bro can solve this
please help me solve integral of 0 to inf ( sinhx/sinx dx )
The integral does not converge, because of the divergent integrals near the vertical asymptotes in 1/sin(x).
Too late, i had that integral in June on exam xd
Some one give me a hint to solve this integral xsec^2x/2+tanx
if the question is (x(sec^2(x)))/2 + tanx use integration by parts
@@MrDKJha Since the OP didn't snare the denominator, the way it is written, it would imply:
(x*sec(x)^2/2) + tan(x)
i really did that in my mind
Remember when you have a x multiplied to your function which you want to integrate, and if you get the same function (except the multiplied x) on replacing x with (b+a -x) in f(x) , then just replace the X, and you get a integratable function. If not just use queen rule , you might get something with it
如果我財富自由～ 我也要常常來這玩數學！
Hmm... I think ill just use simpsons rule XD
I like the white board white out!!!!!
右邊照片是曹老師和老婆嗎 哈哈
I guess I am not getting into the University.
Hello teacher ❤
I'm not understand clearly
nice
I did all my calculus already and I'm in my masters,so why am I torturing myself by watching this? Idk xD
Hello this is Sayed Yousaf from Afghanistan.
I found a difficult math question.
If you help and solve me it would be your pleasure.
The question is:
The limit x approaches 0 (x^x^...^x-x!)/(x!^x!-1)
🎉
🌷ورده
This was given in my ncert text book of maths for 12th grade
Me thinking why you cna't trig identityit :/
The 1 method Is ok..
Meanwhile us Indians having this shit in highschool 💀
Yeah.
Evaluate the integration of the function from zero to infinity
(Sinx/x)⁶
😏😏 I challenged you
Indians do it in their intermediate
If you gave this problem to an Avg student of class 12th in India, At first, He will be going to laugh on you then give you the answer in next minutes
And then he’ll learn English hopefully
@@hydropage2855 nice bro 😂
Classic shit by Indians
d card ？ lmao
This property is known as King rule. bansal sir named this property
its ans is Pie sq divided by 4
got it in 30 sec 😅
Hindi mein boliye
hello
This is an easy question
As an Indian I can confirm that this is the easiest one we are taught in just school only
Average CBSE question 🗿
yes ncert 😂
@@AbhayShakya-nk2tiwe believe you, we just don’t give two shits, you’re all so boring
@@AbhayShakya-nk2tiThey do, its just really really annoying that these types of "I live in country x and solve this at age y" are under every video. They are the math channel equivalent of the sex bot comments.
@@AbhayShakya-nk2ti it is not the issue that people can't handle, just the fact that the schools didn't teach in that corresponding grade. Therefore knowing the technique in earlier stages isn't a privilege for those in certain countries😅.
@@AbhayShakya-nk2tiwe just don't care
We 16 year olds in India have to solve this in 6 minutes in a fking exam
And your point is what? Do they have to write it on a board and explain it in 6 minutes as well.
@@CrYou575These people really get under my skin
Basic class 12th Question in India
Why wasn't I ever taught this property
This is common problem in India for -12 gred actuly very easy you give this problem to -infinity grader in India hel be doing the laughter on you and give you the anser because he is knowing the answer from the birth. Please give actual hard intergal next time
Is this satire
@@arxalier2956 Yes. I hate these people
sir nerd virgin is back !!!
🎉
nice