Q8, OH MAN, THAT - 1 

you know things about to get mathy when he brings out the blue pen.

You're an enabler of my addiction to integrals lol. I love how you always pick integrals that aren't obvious but also don't require higher level techniques to do, which shows you don't need to bust out contour integration whenever the going gets tough

I wish you were my professor back in '87. I would have aced this class just based on your style of teaching.

That -1 just makes a mess of things! Fun integration, as usual!

This is one of the best math channels on RU-vid! Thank you! 😃👍 Keep up the good work!

Yay I figured it out on my own. Big accomplishment for me, very rewarding. Thank you for the video!

found a cool way for the second one: e^(x/2) = sec(u) substitution

This made my morning. Thanks as always.

Excellent video. To clear your mind a bit after an intense day. Greetings.

Wow! I did the second integral by letting sec^2(u)=e^x, it was very straight forward and I'm so happy the result was correct too (although in another form, 2sqrt(e^x-1) - 2arcsec(sqrt(e^x)) ).

Its truly amazing how a simple -1 suddenly introduces an inverse tangent somehow.

I just love the way you solve it 😊

Now, thanks to you I can do integrals correctly! #YAY

So, i'm an agricultural science student. I didn't study math at university, at least, not integrals. But I love them. So I've tried at the beginning of the video and I did the same thing you did! I'm so happy now. I learn everything I know about Integrals from your videos!

Really like your videos, full of good information and very helpful to me .Thank you very much 曹老师X）

Tried to solve the 2nd integrate by doing e^x-1 = y. Although I was able to reach the same result, I was still forced to do u = sqr(y) down the line. The way shown in the video is the fastest afaik. Thanks and keep it up my man.

This revived my old traumas from the calculus class... Man, what a teacher... He just dropped this one on a term final and we weren't ready for that, it still hurts me to this very day haha (he is a great teacher though, loved it...)

YAY!! I GOT THE ANSWERS TO BOTH WHEN YOU TOLD US TO PAUSE AND SOLVE!! Im soo happy!! But yeah geez a -1 makes a hugeee difference

For the second antiderivative, I perfromed integration by parts twice! In the end, the intergral simplified to a form that was the same as if I performed the u-substitution you proposed :^)

Chen lu always saving us!

A lot of videos in one day? YAAAAAAY

I remember my self when i was in university and solving things in mathematic and my friends seeing me as a genius when i see all that as a standard to me. I miss thoses days when enjoying solving some mathematic books

Second integral can be also be done as : Let e^x-1=t^2 ...I) e^xdx=2tdt dx=2tdt/t^2+1 ( from I) Then integral becomes I=2t^2/t^2+1 Solving integral we get same answer as yours

thanku sir

need more definite integrals! #yay

2:26 Drops shoulder rest lol Oh wait wrong channel

8:41 i love this smile!!

Great sir

God bless you

I got the second one on my test, thanks lol

You can also do with trig substitute by letting sqrt(e^x-1) =tan(u)

I missed integral battles 💕

Bravo mon ami

u=sqrt(e^x-1), x=ln(u^2+1), dx=2udu/(u^2+1)

Solved in my brain

for the first integral i used a quite different method, but it had the same solution. i wrote u=e^x and du=e^x. i got integral u^1/2 du/e^x. I knew that e^x=u and it meant that x= In(u). I got the integral (u^1/2)/e^In(u) du. Afterwards i got integral (u^1/2)/u du. I got integral 1/sqrt(u) du. using a basic formula, i concluded that the integral is equal to 2*sqrt(u) or 2*sqrt(e^x). Love your methods of teaching

You've already went through the second integral in a previous video...

For the second integral by the time you get it all in terms of u there is integral of u square divide by (u square + 1) then you can also use trig substitution. U = tan x, du = (sec x)^2 so you have (tan x)^2 * (sec x)^2/(sec x)^2. Then that’s just sec x squared - 1 which is tan x - x which is u - tan inverse u and then you get the same answer.

You are awesome and great

it always bothers me when i find a close relationship between exponentials and trigonometry with complex numbers completely out of the picture.

In the second integral a substituted u for the inside and it worked too. But I had to do 2 substitutions 🤣🤣🤣

opinię marvelous!

"If you think this is long wait until the Calc 2 integrals."

This questions can also be done using substitution method... For the 1st question use substitution e^x=t^2 and for the second use e^x-1=t^2

IN THE INTEGRAL //SQRT (E^X - 1)DX, best to just let sort(e^x) = u^2 and 2 e*x dx = du. Then we have //sqrt(u*2) 2 u du / ( u^2) which is accomplished also by dividing numerator and denominator by e^x. Then we have the integral equal to //2 du or 2u + c or 2 sort(e^x) . I use the double slash as the integral sign which makes it easy to distinguish from the single slash used in division, etc. This is a good video the way it is. No good answers. Each approach that is correct is good!! (Second integral can also be done using the same approach).

Great!

I got as result arctan(1/sqrt(e^x -1) in The second integral. I Just divided numerator and divider by u² and find 1/1+(1/u)²

Substituting x = log(sec^t) also does it for the second integral (and is easier I think) :d

Why have I only discovered this channel after my a-levels...

Podrías subir unos cuantos videos de ecuaciones diferenciales porfavor 😀😀👌👌

i should start learning integral/derivation tables for the trig functions ^^

I like your videos ☺☺☺

I wish I knew how to do integrals to solve them before you solve them 😰

This can also be done by contour integration ( branch point integration)

On first problem better to start out and multiply numerator and denominator by e^x so we then have the integral // ((sqrt(e^x)) * (e^x)) dx / (e^x) which is now // (u * 2u du) /(u^2) du = 2 // du = 2 u + C = 2 sqrt(e^x) + C. A bit shorter me thinks. I use the double slash ( // ) to denote the integral to differentiate it from the slash “/“.

I solved for x in u=sqrt(e^x-1), leaving x=ln(u^2+1) and dx=2udu/(u^2+1)

U r great

I used trig sub by letting e^x/2 into sec(y)

I totally forgot the easier way in the second one and went all the way to substitute e^x = sec^2@ I got the same answer though the process was a little longer..

un crack, demasiado pro eres :v +100000000pro

Awesome

This can be done very simply by using 2 simple substitution..u=e^x-1..this will give integral √u/(u+1)du..again use v=√u.. you'll get integral v^2/(v^2+1)dv..and tralala..it's a child's play from there ...hope it helps..

1.substitute u=sqrt{(e^x-1)}, 2. du= [e^x/(2Xsqrt(e^x-1))].dx----rewrite as----->> du=(2u/(u^2 +1)).dx by writing e^x= u*u+1 , 3. original integrand is now 2(u*u)/(u*u+1) 4. Solve easier integral to get I= 2[sqrt(e^x -1)-arctan(sqrt(e^x-1))]

this is such a hard integral

Sir please solve this limit from 0 to 1 x½(x-1)½dx

Can the first sum could be done in this way e^x=t e^xdx=dt dx=dt/t (√t/t) dt Multiply &divide by √t We get( 1/√t) dt That is 2√t+c Ans 2√e^x+c

I did the second integral with complex numbers. First, I used u=e^x-1 and got Sqrt(u)/(1+u) (Sqrt(u)+i-i)/(1+u) I splitted the integral to get -i int of 1/(1+u)+ int of (sqrt(u)+i)/(1+u) In the second one I multiplied top and bottom by (sqrt(u)-i) and solved the first one -i ln(1+u)+ int of 1/(sqrt(u)-i) Using t=sqrt (u)-i: -i ln(1+e^x-1)+2 int of (t+i)/t -i ln(e^x)+2 int of t/t +2 int of i/t -ix+2t+2i ln (t) Or -ix +2sqrt(e^x-1)-2i+2i ln(sqrt(u)-i) Solving the logarithm and ignoring the -2i (it is a constant), we get: 2sqrt(e^x-1)+2arctan(1/(sqrt(e^x-1))+C It is just off by a constant

BPRP... why do you write fractions out in front of variables ( 1/2 * x ) instead of doing it the simpler way ( x/2 )?? I understand there are situations when a helpful technique "requires" unsimplified forms. What do you/we gain by writing e ^ ( 1/2 * x ) as opposed to e ^ ( x/2 )??

because of that -1, an inverse tangent appears

Still, indeed, with the u-substitution.

I substituted e^x=sec square theta in the second one

Uma solução possível também seria: Integral de (e^x-1)^(1/2) = 2*((e^x-1)^(1/2)-arcsec(e^(x/2))+c

Would an Euler substitution work on the integral of Sqrt(e^x-1)?

Can we divide by u^2 And we get tan^-1(1/u)=cot^-1(u)

Nice

Sir can you please explain why inverse trigonometric function does not lies third quadrant

Интересное видео спасиба

5:52 what if I keep on substituting for du and dx and just increasing the equation?

I didn't learn about that inverse tangent at the end, could somebody explain me how it works?

1:27 "Whats the integral of e to the ... Sanzei , well e to the sanzei is just e to the sanzei"

Ah, yes, the adding zero to the integrand trick.

I have an integral for you, tough as nut! Integrate sqrt(1+c/x^4) over x, c is an arbitrary constant form R, the + 1 is here the part that makes me hate it!

At 4:21, why not rewrite du = e^x dx as du = (u+1) dx - ie, dx = du/(u+1). then the integral becomes sqrt(u)/(u+1). this can be done easily with t=sqrt(u).

If we multiply and divide by e^x and take e^x-1=t and e^xdx=dt Still we get something similar I solved like that Just saying :)

Instead of adding +1 and substracting -1, isn't it easier to just divide the numerator by the denominator?

Set the second one’s u-sub to the entire square root. Easier derivatives

For the second integral, you can also do a trig substitution: sec^2(u) = e^x u=arcsec(sqrt(e^x)) x=ln(sec^2(u)) dx=2tan(u)du The integral then becomes: 2 ,/' sqrt(sec^2(u)-1) * tan(u) du = 2 ,/' tan^2(u) du = 2 ,/' sec^2(u) - 1 du = 2(tan(u) - u) + C Substitute back = 2(tan(arcsec(e^x)) - arcsec(e^x)) + C Draw a right triangle and simplify tan(arcsec(e^x)) into sqrt(e^x - 1). = 2sqrt(e^x - 1) - 2arcsec(sqrt(e^x)) + C **Note:** arcsec(sqrt(e^x)) is the same thing as arctan(sqrt(e^x - 1))

He didn't use c1 and c2 as constants REEEEEEEEEEEEE

Hi. Why not use trigonometric substitution? We have the same answer. :)

Can you please explain how you substituted tan^(-1) for 1/(1+u^2) I’m having trouble nailing down what trig sub that is...

Could you perhaps change that -1 to e^i.pi?

Hello, I was wondering if you could help me with one really simple question. What is the graph of the function x^(2/3)? The problem is the negative part of the function, does it exist or not? Try (-1)^(2/3)

Before watching video I think integration 2e^x/2 and that of under root e^x -1 would be e^x +cosec^(-1)(e^x)+C Edit: First right The second question I got wrong Anyways did anybody get it right before video?

Dunno why noone commented this, but why didn't you write U as sqrt(e^x-1) on the second line, and save multiple steps?

if you take sec^2(u)=e^x it works out very nicely! 2 sec^2(u) tan(u) du = e^x dx = sec^2(u) dx 2 tan(u) du = dx sec(u) = e^(x/2) cos(u) = e^(-x/2) u = cos^-1(e^(-x/2)) 2 ∫ tan(u) sqrt(sec^2(u) - 1) du 2 ∫ tan^2(u) du 2 ∫ (sec^2(u) - 1) du 2 tan(u) - 2 u + C 2 sqrt(sec^2(u) - 1) - 2 u + C 2 sqrt(e^x - 1) - 2 cos^-1(e^(-x/2)) + C It looks slightly different, but has the same value!

How that -1 made a mess.

Sir how to integrate ∫(e ^x ^3) dx=?

This was in my test 2 days ago and i couldn't do it Aaaaagggghhhhhhhh

If they think that was long, they should have taken my quiz in calc 2 where the teacher messed up and gave us a super ugly integral, did all of the trig sub stuff and ended up with cos^4(x) which needs to do the half angles identities like 3 times and have multiple arguments with the cosine and I'll dm you the integral I'm Twitter when we get the quiz back

For the second integration if you suppose u=e^x - 1 , and the do integration by parts you get the result . -3sqrt(e^x - 1 ) is that a wrong method or what?