If you guys like this video, give it a like and share it with others! Let me know if I should do a part 2 for this? Here's the file: docs.wixstatic.com/ugd/287ba5_f1afe86da0454ef08c92eae9b9e9c18c.pdf Here's the playlist to the solutions to everybody ru-vid.com/group/PLj7p5OoL6vGwS5ilte_7fjRB2rQ_yE_Hj integral of ln(x)/x^3, 0:45 integral of sec^4(x), 2:29 integral of (2x+3)/(x^2-5x+4), 3:50 integral of x^2*tan(x^3), 5:50 integral of 1/(1+x^2)^(5/2), 7:17 integral of e^sqrt(x), 8:50 integral of sin^2(x), 10:37 integral of 1/(sqrt(x+1)-sqrt(x)), 11:20 integral of e^x/sec(x), 13:15 integral of 1/(1+cos(x)), 15:09 integral of (x-4)/(x^4-1), 17:44 integral of x^2/sqrt(1-x^2), 19:24
omg, thank you very much for this video, I just had this kind of lesson today at my math class, I can learn better now. Also I like everything you do so keep it up ^^
Please do another video, maybe going over the methods and their basics, my calculus teacher is really bad at explaining them and your videos help a lot!
This is probably the most important Calc II video ever produced. It focuses exclusively on strategy, which is the most important element in all of calculus: Learning strategies (creativity), and recognizing when multiple strategies are valid, and why some might be better. Bravo, Steve! It's similar to your D/I "When to stop" video. Ever since I've stumbled on your channel, I've been re-learning calculus just for fun!
Wholeheartedly agree. After developing a proper understanding of the theory behind integral calculus, students should focus on consolidating their integration techniques. As repetitive as these drills may seem to be, they can save a lot of time in the long run.
This has to be one of your best videos, it really made me stop and think. The ability to stop the video, think the problem through, then resume is far better than classroom teaching of the past.
not really. i have questions, who can answer them for me? ofc i would have to wait for responses. then if i dont understand the answer, im still waiting on a response. a teacher could explain instantly
For those that want more practice, I have several videos where I solve complicated integrals step-by-step, using all the methods explained here. :) It might help some of you out there!
This is great, my biggest struggle in calculus 2 has been that intuition you're supposed to have about what integration method to use. It does not come naturally to me, same reason the whole "guess and check" process of factoring quadratics was a nightmare way back in elementary school
If you have a quadratic like x² - 7x + 10 = 0 we know that the two roots r,s satisfy r + s = 7 rs = 10 so what you can do is systematically consider all factors of 10: 1 x 10 = 10 but 1 + 10 ≠ 7 -1 x -10 = 10 but -1 + (-10) ≠ 7 2 x 5 = 10 AND 2 + 5 = 7 if 2, 5 didn't work you would then check all the other pairs of factors. Of course, you might notice that 1+10 = 11 and realize that you don't even need to consider (-1)+(-10), or realize that once you checked 2x5 that's the same as 5x2. In the same way with integration techniques you can systematicaly consider them one after another, just in your head trying to do the first steps, and once you find something that you seem to be able to make progress with, do that.
In essence, try thinking about using trig sub: look out for single term square root denominators and 1+x^2 or 1-x^2 IBP: When you can easily differentiate one part and integrate one part PFD: When the denominator is a higher power polynomial or when the denominator can be factorized (pretty much the same thing, but not quite?) u sub: When the derivative of something can cancel out a troublesome part or when you're doing variations of a simple integration question but don't want to be careless (Feels like the most powerful and flexible tool in my elementary calculus toolbox, though, so probably has way more specific uses) I'm not in uni yet or anything so I'm not sure if my insufficient contact with calculus problems has led to some mistake of some sort in the summary above so please let me know if there are better ways or corrections that I would want to know about, thanks~
In the first question, we can rather make a u substitution Lnx= t and x = e^t 1÷x dx = dt So the integral becomes T÷e^2t dt Then we can use integration by parts.
Honestly, as fantastic as my TA is, you are amazing. Both of y'all have done an amazing job at explaining and writing out problematic scenarios that give us lower level students the biggest issues. However you present these scenarios in all of your videos so well. Thank you very much. (note: calc2 student)
This was perfect. I have my midterm coming up and this was the perfect way to study as you give subtle hints without revealing the whole answer. Thankyou very much
In the 10th question, you may write the denominator as 2cos^2(x/2) using the half angle identity and then write it as sec^2(x/2) and proceed from there.
i have been looking for a video like this for ever ! I have not found one video breaking down the strategies you should use, which is the most important and confusing aspect of integration ! TYSM !!
I’m so glad BPRP has such a big following, you always make me smile even when I see a very hard problem that I can’t fathom in your vids! I really want to become smart like this guy!
Actually in the first one, 'u' substitution is going to work. We substitute 1/x as u and we integrate to get (dx×-1/x²)=du. Then we take the natural log on both sides of the 1/x=u and get -ln(x)=ln(u). Then we can substitute all the values to get u×ln(u)du as the integrand. Cheers
for the first one you can just multiply and divide the whole thing by -2 x -2 and then shift one of the -2s into the power of x in lnx, after which u sub becomes very easy (u = 1/x^2, du = -2/x^3 dx) then youre left with 1/4 int(lnu du) which simplifies thereon
IN the beginning of the video he said that there will be some calculus 2 in this so I said I will concentrate on things I learn in calc 1 since I didn’t take calc 2. To my surprise I took every thing he explained in the calc 1 course in my uni. Now I really believe my uni is super super hard, I wonder what they’ll teach me in calc 2
For the 6th question, Using u = e^sqrt(x) would give a faster answer. For the 10th one, using half angle formula and converting 1+cos(x) as 2cos^2(x/2) makes it very easy to get the solution.
21:15 "People don't like to put down a d-whatever. If you don't put down anything, that's like driving without your seatbelt on: very dangerous, so don't do that."
i throw the term lifesaver around a lot. buy me a morning coffee? lifesaver. drive me to campus when i'm out of fuel? lifesaver. but you my friend, you are the only one for whom the term lifesaver takes on its truest meaning
I really appreciate and follow up ur tutor regularly. Wow i've got a lot of knowledge in easy way and surprised ur broaden mathematical wisdom. Mathematical difficulties of both derivatives & Integration become easy with ur lovely way of teaching methodology. Many thanks indeed. NA from Ethiopia
for the 10. integral you could use substitution u=tan(x/2), it's a beautiful substitution that works every time in integrals where you have sin,cos,constants and their addition
Can you show an integration where you have to use all these techniques together? For example: first you have to integrate by parts, then make u-sub, then trig-sub? It would be very nice.
Hi, I love your videos. Ii used to doubt my ability to do integrals but you make it so easy. Thank you for sharing your knowledge and techniques. You are an amazing person to do this vlogs.
Here's a very powerful method for integrating 1/1+Cosx . Simply use of t=tan(x/2) (btw if sin50x was present use half of the angle i.e t=tan(25x) ) and perform integration by substitution. Extremely powerful you can integrate 1/(sin99x + cos99x) , 1/1+cos100x you name it. Then sinx = 2t/1+t². And cosx=1-t²/1+t².
Going off a comment from another video, how come you don't take the time to explain that in the last one, the fact that 1 - x^2 is under a square root makes -1
|.xx dx/¥(1-xx)=(ax+b)¥(1-xx)+C|dx/(1-xx) Permisalan tsb dideferensialkan untuk menemukan besarnya konstanta q&b Dan suku terahur kan menghasilkan arcsin
For number 10 I used the power reduction formula in reverse: 1/(1+cosx) = 1/(2*½(1+cos(2*(x/2))) = 1/(2cos²(x/2)) Now set u = x/2, 2*du=dx, so you get the integral of (2*du)/(2*cos²(u)), equals the integral of du/(cos²(u)), which all in all is equal to tan(u)+c, plug in x/2 for u: tan(x/2)+c
for 10 task i have another way of solution, you divide cos x into cos^2(x/2) - sin^2(x/2), then 1 into cos^2(x/2) + sin^2(x/2). Then in the denominator will leave 2cos^2(x/2), plug in 2 in dx => d(x/2). Finally you get sec^2(x/2) which of the integral is ..... :)
I really learned a lot in your videos, especially the 100 integrals you solve. Aside from the fact that I like your speaking tone and explanation and lastly your eyes. So cute.
Nice revision! Ex.10 with 1/(1+cos x) use trig formula cos2A = 2(cosA)^2 -1 with x=2A to get rid of the 1 and leave just cos^2(x/2) on bottom.... basically now a sec^2 integral which is standard.
Where tf were you when I was in calc and div eq? I got my EE engineering degree and don't rely on much of this, so it doesn't weigh much on me, but your lessons, along with 3BLue1Brown, have still helped me understand things better. Hopefully many find your channel for their well being and yours. I have one recommendation, though. Find a microphone you can keep with your head. Maybe a headset or a chest piece closer to your mouth. Whenever you turn away, your voice drops and throws off the audio of the video. It would also give you more consistent audio playback in your editing and make things easier to follow.
I used another technique for the 10th First, observe that cos(x) can be written using Euler's formulas. Now you can use u substitution twice. If u = e^ix, you're left with the inverse of a polynomial, 1/(u²+2u+1). The result of this integration method is (1-e^ix)i/(1+e^ix), which's in fact equal to sin(x)/(1+cos(x)) (The last step needs some clean up using Euler's formulas and a readjustment of the integration constant to get a function that goes from R to R)
