WHHHHHHHHHHHHHHHATTTTTTT!?!?!?!?! Why the HELL does stewart's book not teach this in 7.1 ?!?!? Seriously? You just saved my life. does this only work with combinations of e and trig functions?
You've helped me on like 5 questions that have appeared from my textbook. You also offer one of the best explanations out of TH-cam. Thank you so much for this
You sir are amazing. Currently taking a class on ordinary differential equations and whenever I have trouble remembering an integration strategy you come to my rescue!
Your explanations on these topics are the best ones I've found so far....better than my professor's (he's a really good prof too) and he's been doing it for more than 30 years!!! Thank you very much.
Like your videos they really help. It my first time writing comment in youtube but you deserve a big thank. Please Continue doing videos from that book for chapter 6 and 7 and 8.
4 mounths after a soul apears In case u still wondering what was that, he just "passed it to the other side" , for example, if you have x + 1 = 0 , x = -1 , pretty logical right, but what you are actually doing is x+ 1 (-1) = 0 (-1) wich is equal x = -1 . Thats what he has done at that part, but with a fucking bigger thing haha
You can use it for any problem where standard integration by parts can be applied, simply by knowing which of the three stops to anticipate, and which function to assign to which column.
Given: integral t*e^(-t*(s + 1)) dt, from 0 to infinity Assume s is a constant, since it wasn't specified. Let k = -(s + 1), to simplify our writing. Thus we have: integral t*e^(k*t) dt Construct IBP table: S ___ D ___ I + ___ t ____ e^(k*t) - ____ 1 ___ 1/k*e^(k*t) + ____ 0 ___ 1/k^2*e^(k*t) Construct our result: t/k * e^(k*t) - 1/k^2*e^(k*t) Evaluate our results: At t = infinity: inf/k * e^(k*inf) - 1/k^2*e^(k*inf) = 0 At t = 0: 0/k * e^(k*0) - 1/k^2*e^(k*0) = -1/k^2 Definite integral result = 1/k^2 Recall how we defined k, substitute, and we have our result: 1/(s + 1)^2
Given a general case of integration by parts, where you are integrating f(a*x)*g(b*x) dx, suppose you are differentiating f(a*x) and integrating g(b*x). Every time you differentiate f(a*x), you accumulate a multiplication out in front by the constant a. Every time you integrate g(b*x), you'll divide by b every time out in front. Let G(x) be the integral of g(x), and let H(x) be the second integral of g(x). This builds the following table: S ___ D ___________ I + ___ f(a*x) _______ g(b*x) - ____ a*f'(a*x) ____ 1/b*G(b*x) + ___ a^2*f"(a*x) __ 1/b^2*H(b*x) Which constructs these terms: +f(a*x) * 1/b*G(b*x) - a/b^2 * f'(a*x)*H(b*x) + a^2/b^2 * integral f"(a*x)*H(b*x) dx For f(a*x) = e^(a*x), and g(b*x) = sin(b*x), this result becomes: e^(a*x) * 1/b*-cos(b*x) + a/b^2 * e^(a*x)*sin(b*x) - a^2/b^2 * integral e^(a*x) *sin(b*x) dx The first coef in this construction is 1/3, which comes from 1/b in this example. The second coef in this construction 2/9, which comes from a/b^2 in this example. Since we spot a constant multiple of the original integral, we call that I, and equate this construction to I as well. Then solve algebraically for I. I = e^(a*x) * 1/b*-cos(b*x) + a/b^2 * e^(a*x)*sin(b*x) - a^2/b^2 * I I*(1 + a^2/b^2) = e^(a*x) * 1/b*-cos(b*x) + a/b^2 * e^(a*x)*sin(b*x) Dividing by (1 + a^2/b^2), and simplifying, we get the general solution: [a*sin(b*x) - b*cos(b*x)]*e^(a*x)/(a^2 + b^2) + C And for this example: [2/13*sin(3*x) - 3/13*cos(3*x)]*e^(2*x) + C
For simple trig and exponentials, it is arbitrary which one you assign to D and which one you assign to I. They will both loop back on themselves, and construct the original integral across a row. You get the same answer either way.
Given: integral x^3 * e^(2*x) dx Construct IBP table, with x^3 in the D-column and e^(2*x) in the I-column. Keep constructing rows, until the D-column gets to zero, since this is an ender. S ___ D ________ I + ___ x^3 ______e^(2*x) - ____ 3*x^2 ___ 1/2*e^(2*x) + ____ 6*x ____ 1/4*e^(2*x) - ____ 6 _______ 1/8*e^(2*x) + ____ 0 ______ 1/16*e^(2*x) Connect each sign, with the corresponding D-column entry. Then connect to the I-column entry from the next row down. +1/2*x^3 * e^(2*x) - 3/4*x^2*e^(2*x) + 6/8*x*e^(2*x) - 6/16*e^(2*x) Simplify, and gather all e^(2*x) terms: (1/2*x^3 - 3/4*x^2 + 3/4*x - 3/8)*e^(2*x)
You get -9/4 (integral...) if you differentiated sin instead of e in the beginning of the integration by parts. But you end up with the same answer at the end. So you are wrong as usual.
sir can you please help me with this problem its says use the method of undetermined coefficient to solve the differential equation : y"+3y'=1+(x. e(^-3x))
Given: y" + 3*y'= 1 + x*e^(-3*x) First find the homogeneous solution, by assuming the RHS = 0: yh" + 3*yh' = 0 Assume an ansatz of yh = e^(r*t), thus: (r^2 + 3*r)*e^(r*x) = 0 r*(r + 3)*e^(r*x) = 0 r = 0, r = -3 yh = A + B*e^(-3*x) Now produce the particular solution. Ordinarily, this would be a linear combination of a constant, e^(-3*x), and x*e^(-3*x), based on the form of the RHS. However, there is overlap with the homogeneous solution, since B*e^(-3*x) is already a solution, and a constant is already part of the solution. This means we need to multiply by x until we get linearly independent solutions. Thus: yp = C*x*e^(-3*x) + D*x^2*e^(-3*x) + E*x Take derivatives of yp, and apply to original diffEQ: yp' = C*e^(-3*x) + (2*D - 3)*C*x*e^(-3*x) - 3*D*x^2*e^(-3 x) + E yp" = (2*D - 6*C)*e^(-3*x) + (9*C*x -12*D)*x*e^(-3 x) + 9*D*x^2*e^(-3*x) Simplify -6*D*x*e^(-3 x) + (2*D - 3*C)*e^(-3 x) + 3*E = 1 + x*e^(-3*x) Equate like coefficients: 3*E = 1 2*D - 3*C = 0 -6*D = 1 Solve for the variables: E = 1/3 C = -1/9 D = -1/6 Construct result: yp = -1/9*x*e^(-3*x) - 1/6*x^2*e^(-3*x) + 1/3*x Add homogeneous part, and we have our solution: y = A + B*e^(-3*x) - 1/9*x*e^(-3*x) - 1/6*x^2*e^(-3*x) + 1/3*x
Please guru solved indifinite integration e to the power x.sinx.sin2xdx e to power cos square xdx 2 to the power x.sindxdx any one sir solved questions.
Given: integral e^x * sin(x)*sin(2*x) dx Convert the trig functions to reduced trigonometric form, using the trig product identity: sin(a)*sin(b) = 1/2*[cos(a - b) - cos(a + b)] sin(2*x)*sin(x) = 1/2*[cos(2*x - x) - cos(2*x + x)] = 1/2*[cos(x) - cos(3*x)] Thus, the integral becomes: 1/2* integral [cos(x) - cos(3*x)]*e^(x)] dx Construct IBP table for an exponential and a cosine with a generalized frequency, we'll call w. That is, cos(w*x)*e^x S ___ D __________ I + ___ cos(w*x) ___ e^x - ___ -w*sin(w*x) ___ e^x + ___ -w^2*cos(w*x) ___ e^x Build the result: cos(w*x)*e^(x) - w*sin(w*x)*e^x - w^2 * integral cos(w*x)*e^x dx Spot our original integral, and solve for it algebraically: I = cos(w*x)*e^(x) - w*sin(w*x)*e^x - w^2 * I I*(1 + w^2) = [cos(w*x)*e^(x) - w*sin(w*x)*e^x] I = [cos(w*x)*e^(x) - w*sin(w*x)*e^x]/(1 + w^2) Plug in w = 1 and w = 3, to determine both versions of this: Integral cos(x)*e^x dx = [cos(x)*e^(x) - sin(x)*e^x]/2 Integral cos(3*x)*e^x dx = [cos(3*x)*e^(x) - 3*sin(3*x)*e^x]/10 Add these together, and recall our leading coefficient of 1/2. Add +C, and we're done: [cos(x)*e^(x) - sin(x)*e^x]/4 + [cos(3*x)*e^(x) - 3*sin(3*x)*e^x]/20 + C
For 2^x * sin(x) dx: Recognize that 2^x is equivalent to e^(ln(2)*x). Using the general equation you can derive with the process I showed previously, we can derive the following general result for e^(a*x) * sin(b*x): integral e^(a*x) * sin(b*x) dx = [a*sin(x) - b*cos(x)]*e^(a*x)/(a^2 + b^2) + C For us: a = ln(x), and b = 1 Thus the result is: [ln(2)*sin(x) - cos(x)]*e^(ln(x)*x)/(ln(2)^2 + 1) + C
WHHHHHHHHHHHHHHHATTTTTTT!?!?!?!?! Why the HELL does stewart's book not teach this in 7.1 ?!?!? Seriously? You just saved my life. does this only work with combinations of e and trig functions?
Hi Kris, sorry for the super late reply. Here's the complete version of the DI method th-cam.com/video/2I-_SV8cwsw/w-d-xo.html
six years after and i'm asking the exact same thing lol
@@harpiasan2885 he6
You've helped me on like 5 questions that have appeared from my textbook. You also offer one of the best explanations out of TH-cam. Thank you so much for this
You are welcome! Please subscribe and share my channel/playlist with others. Thank you for doing so! :)
Thank you so much for not skipping the basic algebra parts
Hi! I want you to know that you're a great professor. You're always helping me with these videos. They're the best on youtube. Thank you so much!
Thank you for your nice comment Julia.
Why on earth is this not taught in the James Stewart book in chapter 7.1... This method is SO CLEAR and EASY TO FOLLOW. THANK YOU.
You sir are amazing. Currently taking a class on ordinary differential equations and whenever I have trouble remembering an integration strategy you come to my rescue!
Integral of e^(2x)*sin(3x)
Your explanations on these topics are the best ones I've found so far....better than my professor's (he's a really good prof too) and he's been doing it for more than 30 years!!! Thank you very much.
whitechalkredchalk ! This DI method is a revelation. Thanks again.
you're helping create more doctors and engineers in this world for the better
I love how he pronounces the word theta but this is a great video !! thx
Moi _ Thanks!
ur tutorials really help me a lot....this is the very first time I realize this technique to deal with these two damn transcendental functions
I don't think you realise how BRILLIANT you really- YOU'RE BRILLIANT
Phenomenal video. It hadn't even occurred to me to add to the other side of the equation like that. Thanks!
thanks a ton man! this helps a lot! sometimes when the teacher explains it I just don't understand but you do a fantastic job!
= (2/13)e^2¢sin(3¢) - (3/13)e^2¢cos(3¢) + C
I just saw a bunch of videos of the same equation withouth understanding a single thing, but you really make it easy. Thanks!!!
THANK YOU SO MUCH. My jaw dropped due to how easy you made it!
hooked to your lectures. In fact I am glued to your lectures
You saved me so much time. Thanks
This is such a classic I just memorized both e^axsin(bx) and e^axcos(bx)
Amazing explanation! Everything clicked when you wrote the ratio on the initial part of the problem :P
i love ur work, please stay on! we need u
This is awesome! Just realize the mistake i had been doing all the time!
Like your videos they really help. It my first time writing comment in youtube but you deserve a big thank. Please Continue doing videos from that book for chapter 6 and 7 and 8.
this method is a blessing. u helped alot
Loving this DI method. My book doesn't teach it but it should.
Thank you for helping me finally get that part.
studying for my calc final and could not solve this! Thank you so much!
YESSS!!! I might actaully be able to pass single dimensional analysis!! :D Thanks!!!
you are like julioprofe english version
i know that you dont know what mean that but, its a great compliment...
Thank you very much, it was extremely helpfull!
Great explanation. Helped a bunch!!
Thank you so much ,easy and effective ways .
WOW! you are awesome!! thank you so much
Thank you for this. You have really helped me understand this!
You are the Man!!!!!!
Saving alot of lives here.
You are a math god
sir thank you very much great explanation sir go on why don't colleges teach this method
thank you for a great explanation and helpful D I method
jenasin You're welcome. I also have a complete version of the DI method th-cam.com/video/2I-_SV8cwsw/w-d-xo.html
Helped a lot! Thanks!
glad to help!
thank you so much you are the best
Thanks!
This one was particularly tricky haha
Thank you. Great explanation.
man, you're fucking godsend. Thanks man.
Hy me sadqy tery theeta ty❤❤
sir it helped me a lot but can u make vid for inverse trigonometric function of integration by parts
I have those videos already. YOu can check my site here: blackpenredpen.com/math/Calculus.html
4:28 how did you add that ?
i didnt understand that part.
:(
4 mounths after a soul apears
In case u still wondering what was that, he just "passed it to the other side" , for example, if you have x + 1 = 0 , x = -1 , pretty logical right, but what you are actually doing is x+ 1 (-1) = 0 (-1) wich is equal x = -1 .
Thats what he has done at that part, but with a fucking bigger thing haha
Is the U substitution version of this problem available?
original function: f(t)= t.e^-t below equation has come from Laplace transform formula.
very helpful, thank you
Can you use the DI method for any problem or are there some restrictions?
You can use it for any problem where standard integration by parts can be applied, simply by knowing which of the three stops to anticipate, and which function to assign to which column.
thanks mate! it's really help :D
why the- 1/9 does not change the symbole?
May Allah bless you
thank you so much!
perfeito! me ajudou muito!
If I have e^3x then how many times integrate and differentiate
Integral 0 to infinite t.e^-t(s+1) dt ,using DI method..?
Given:
integral t*e^(-t*(s + 1)) dt, from 0 to infinity
Assume s is a constant, since it wasn't specified. Let k = -(s + 1), to simplify our writing.
Thus we have:
integral t*e^(k*t) dt
Construct IBP table:
S ___ D ___ I
+ ___ t ____ e^(k*t)
- ____ 1 ___ 1/k*e^(k*t)
+ ____ 0 ___ 1/k^2*e^(k*t)
Construct our result:
t/k * e^(k*t) - 1/k^2*e^(k*t)
Evaluate our results:
At t = infinity: inf/k * e^(k*inf) - 1/k^2*e^(k*inf) = 0
At t = 0: 0/k * e^(k*0) - 1/k^2*e^(k*0) = -1/k^2
Definite integral result = 1/k^2
Recall how we defined k, substitute, and we have our result:
1/(s + 1)^2
I don't understand why there is a 2/9 part is there, is it part of a rule?
Given a general case of integration by parts, where you are integrating f(a*x)*g(b*x) dx, suppose you are differentiating f(a*x) and integrating g(b*x).
Every time you differentiate f(a*x), you accumulate a multiplication out in front by the constant a.
Every time you integrate g(b*x), you'll divide by b every time out in front. Let G(x) be the integral of g(x), and let H(x) be the second integral of g(x).
This builds the following table:
S ___ D ___________ I
+ ___ f(a*x) _______ g(b*x)
- ____ a*f'(a*x) ____ 1/b*G(b*x)
+ ___ a^2*f"(a*x) __ 1/b^2*H(b*x)
Which constructs these terms:
+f(a*x) * 1/b*G(b*x) - a/b^2 * f'(a*x)*H(b*x) + a^2/b^2 * integral f"(a*x)*H(b*x) dx
For f(a*x) = e^(a*x), and g(b*x) = sin(b*x), this result becomes:
e^(a*x) * 1/b*-cos(b*x) + a/b^2 * e^(a*x)*sin(b*x) - a^2/b^2 * integral e^(a*x) *sin(b*x) dx
The first coef in this construction is 1/3, which comes from 1/b in this example.
The second coef in this construction 2/9, which comes from a/b^2 in this example.
Since we spot a constant multiple of the original integral, we call that I, and equate this construction to I as well. Then solve algebraically for I.
I = e^(a*x) * 1/b*-cos(b*x) + a/b^2 * e^(a*x)*sin(b*x) - a^2/b^2 * I
I*(1 + a^2/b^2) = e^(a*x) * 1/b*-cos(b*x) + a/b^2 * e^(a*x)*sin(b*x)
Dividing by (1 + a^2/b^2), and simplifying, we get the general solution:
[a*sin(b*x) - b*cos(b*x)]*e^(a*x)/(a^2 + b^2) + C
And for this example:
[2/13*sin(3*x) - 3/13*cos(3*x)]*e^(2*x) + C
shouldnt the +C be 9/13C?
Can you answer me the integration of e^x×sin^2x
Thank you!
Sina Ahmadi u r welcome
Hello,
Can this method be done with definite integrals?
I have a similar problem like this be it is a definite integral.
for example the definite integral from 0 to 2pi. (e^-2x)(cos(jx))dx
A definite integral is easily calculate from its undefinite form. Just find the integral and plug the values!
int^a_b(f) = int(f)(b) - int(f)(a)
Thank you
Thanks
You cannot use ilate formule
Here 1st function is sinx and 2nd function is e^x
For simple trig and exponentials, it is arbitrary which one you assign to D and which one you assign to I. They will both loop back on themselves, and construct the original integral across a row. You get the same answer either way.
Thaks ;)
very good.
Very helpful :)
how do we know when to use DI or IBP?
DI is a shorthand way of doing IBP.
Baia baia. Como 6 horas pensando esta integral y saber que solo tocaba restar la parte cíclica jajajaj
Can you please integrate x^3 times e^2x ??? Please.
Given:
integral x^3 * e^(2*x) dx
Construct IBP table, with x^3 in the D-column and e^(2*x) in the I-column. Keep constructing rows, until the D-column gets to zero, since this is an ender.
S ___ D ________ I
+ ___ x^3 ______e^(2*x)
- ____ 3*x^2 ___ 1/2*e^(2*x)
+ ____ 6*x ____ 1/4*e^(2*x)
- ____ 6 _______ 1/8*e^(2*x)
+ ____ 0 ______ 1/16*e^(2*x)
Connect each sign, with the corresponding D-column entry. Then connect to the I-column entry from the next row down.
+1/2*x^3 * e^(2*x) - 3/4*x^2*e^(2*x) + 6/8*x*e^(2*x) - 6/16*e^(2*x)
Simplify, and gather all e^(2*x) terms:
(1/2*x^3 - 3/4*x^2 + 3/4*x - 3/8)*e^(2*x)
Who is this and why is he not using a black and red marker?
I think his name is WhiteChalkRedChalk.
it should be -9/4(integral...) correct me if im wrong please
You get -9/4 (integral...) if you differentiated sin instead of e in the beginning of the integration by parts. But you end up with the same answer at the end. So you are wrong as usual.
sir can you please help me with this problem
its says use the method of undetermined coefficient to solve the differential equation : y"+3y'=1+(x. e(^-3x))
Given:
y" + 3*y'= 1 + x*e^(-3*x)
First find the homogeneous solution, by assuming the RHS = 0:
yh" + 3*yh' = 0
Assume an ansatz of yh = e^(r*t), thus:
(r^2 + 3*r)*e^(r*x) = 0
r*(r + 3)*e^(r*x) = 0
r = 0, r = -3
yh = A + B*e^(-3*x)
Now produce the particular solution. Ordinarily, this would be a linear combination of a constant, e^(-3*x), and x*e^(-3*x), based on the form of the RHS. However, there is overlap with the homogeneous solution, since B*e^(-3*x) is already a solution, and a constant is already part of the solution. This means we need to multiply by x until we get linearly independent solutions. Thus:
yp = C*x*e^(-3*x) + D*x^2*e^(-3*x) + E*x
Take derivatives of yp, and apply to original diffEQ:
yp' = C*e^(-3*x) + (2*D - 3)*C*x*e^(-3*x) - 3*D*x^2*e^(-3 x) + E
yp" = (2*D - 6*C)*e^(-3*x) + (9*C*x -12*D)*x*e^(-3 x) + 9*D*x^2*e^(-3*x)
Simplify
-6*D*x*e^(-3 x) + (2*D - 3*C)*e^(-3 x) + 3*E = 1 + x*e^(-3*x)
Equate like coefficients:
3*E = 1
2*D - 3*C = 0
-6*D = 1
Solve for the variables:
E = 1/3
C = -1/9
D = -1/6
Construct result:
yp = -1/9*x*e^(-3*x) - 1/6*x^2*e^(-3*x) + 1/3*x
Add homogeneous part, and we have our solution:
y = A + B*e^(-3*x) - 1/9*x*e^(-3*x) - 1/6*x^2*e^(-3*x) + 1/3*x
Nice sir !! :D
Thaaaank youuuuu!
You help me so mutch. Tks from Brazil
god bless you
thank you...
thanks
love you
tu e tu der tuu dayta~
When you add the integral gets eliminated
Please guru solved indifinite integration
e to the power x.sinx.sin2xdx
e to power cos square xdx
2 to the power x.sindxdx
any one sir solved questions.
Given:
integral e^x * sin(x)*sin(2*x) dx
Convert the trig functions to reduced trigonometric form, using the trig product identity:
sin(a)*sin(b) = 1/2*[cos(a - b) - cos(a + b)]
sin(2*x)*sin(x) = 1/2*[cos(2*x - x) - cos(2*x + x)] = 1/2*[cos(x) - cos(3*x)]
Thus, the integral becomes:
1/2* integral [cos(x) - cos(3*x)]*e^(x)] dx
Construct IBP table for an exponential and a cosine with a generalized frequency, we'll call w. That is, cos(w*x)*e^x
S ___ D __________ I
+ ___ cos(w*x) ___ e^x
- ___ -w*sin(w*x) ___ e^x
+ ___ -w^2*cos(w*x) ___ e^x
Build the result:
cos(w*x)*e^(x) - w*sin(w*x)*e^x - w^2 * integral cos(w*x)*e^x dx
Spot our original integral, and solve for it algebraically:
I = cos(w*x)*e^(x) - w*sin(w*x)*e^x - w^2 * I
I*(1 + w^2) = [cos(w*x)*e^(x) - w*sin(w*x)*e^x]
I = [cos(w*x)*e^(x) - w*sin(w*x)*e^x]/(1 + w^2)
Plug in w = 1 and w = 3, to determine both versions of this:
Integral cos(x)*e^x dx = [cos(x)*e^(x) - sin(x)*e^x]/2
Integral cos(3*x)*e^x dx = [cos(3*x)*e^(x) - 3*sin(3*x)*e^x]/10
Add these together, and recall our leading coefficient of 1/2. Add +C, and we're done:
[cos(x)*e^(x) - sin(x)*e^x]/4 + [cos(3*x)*e^(x) - 3*sin(3*x)*e^x]/20 + C
For 2^x * sin(x) dx:
Recognize that 2^x is equivalent to e^(ln(2)*x). Using the general equation you can derive with the process I showed previously, we can derive the following general result for e^(a*x) * sin(b*x):
integral e^(a*x) * sin(b*x) dx = [a*sin(x) - b*cos(x)]*e^(a*x)/(a^2 + b^2) + C
For us:
a = ln(x), and b = 1
Thus the result is:
[ln(2)*sin(x) - cos(x)]*e^(ln(x)*x)/(ln(2)^2 + 1) + C
When i switch D = sin(3θ) & I = e^(2θ) I got 1/4 [
Thx :"v
thumbnail's wrong, it has the stuff as a function of theta and he put dx
A mi me gusta como pronuncia x
You're better than my math teachers who have big named degrees.