How to integrate 1/(1+tan(x))

Рет қаралды 28,078

bprp calculus basics

Күн бұрын

Пікірлер: 67

@bprpcalculusbasics 3 жыл бұрын

Actually, I did this 9 years ago! kzbin.info/www/bejne/Z3iahGRrebuAZs0

@cicik57 3 жыл бұрын

nice, it is solvable with substitution tanx = t, but this is mutch shorter

@adandap 3 жыл бұрын

Alternatively, sin(x) + cos(x) = root(2) sin(x + pi/4) then u = x + pi/4 and it's easy from there.

@cartermikovich1369 3 жыл бұрын

that was my first thought as well

@baezmarklimuel7871 3 жыл бұрын

@adandap explain how sin(x) + cos(x) = sqrt(2)sin(x+45)

@adandap 3 жыл бұрын

@@baezmarklimuel7871 Sure, sin(x) + cos(x) = sqrt(2) [ sin(x) . 1/sqrt(2) + cos(x) . 1/sqrt(2)] = sqrt(2) [ sin(x) . cos(pi/4) + cos(x) . sin(pi/4)]. The quantity in the square bracket is sin(x + pi/4) because sin(a+b) = sin(a) cos(b) + cos(a) sin(b) and sin(pi/4) = cos(pi/4) = 1/sqrt(2)

@Z7youtube 2 жыл бұрын

@@adandap nice! thx for sharing this

@alberteinstein3612 3 жыл бұрын

You never fail to give me fun math problems to do! :)

@ta_helado 3 жыл бұрын

@rogerkearns8094 3 жыл бұрын

Wouldn't it be nice if I'd ever even think of doing that.

@sourabhdubey7518 3 жыл бұрын

Great technique, now I will use it in my term end examinations

@hetsmiecht1029 2 жыл бұрын

0:59 my immediate reaction was "sure it would be nice if we only had to integrate 1, but that's not gonna happen". Then you wrote exactly that on the board :)

@calculus280 2 жыл бұрын

Introduce this substitution: x = arctan(u^2). There will be partial fractions but it works out nicely in the end.

@sir.dinosuras4846 3 жыл бұрын

Just pure gold. Thanks sir for your support. Cheers 🥂.

@محمدمحمد-خ7ف2ب 8 ай бұрын

You can multiple and divide by sec ^2 x and make u = tanx then continue with partial fractions

@SeeTv. 3 жыл бұрын

An easier integral? integral cos(x) dx is easier

@not_vinkami 3 жыл бұрын

But it's not nice b/c it's basically irrelevant

@gurkiratsingh7tha993 3 жыл бұрын

Integral constant dx is easier

@AirshipToday 3 жыл бұрын

@@gurkiratsingh7tha993 integration of nothing(0) is easier

@hanumangupta607 7 ай бұрын

But you might remember it's value

@MohammedAli-ig2nu 3 жыл бұрын

Can you investigate Integral of (1+tanh x)/(1+tan x) ? Does it have a solution? And how can one decide if the antiderivative of some expression can be found?

@justinpark939 3 жыл бұрын

Had my own way and the antiderivative was x/2 + ...ln|cos(2x)|/4 + ln|sec(2x)+tan(2x)|/4 + c. Saw Mr Steve's answer and found that the trig stuff was equivalent to his trig stuff, after some manipulation. edit: wouldn't it be nice if you were good at finding complicated ways to write 1 or 0. I wish finding creative ways to rewrite the multiplicative and additive identity were my forte.

@FireFly_best_girl 3 жыл бұрын

but as I know I can do some thing else ,,, first step : u=tan(x) then du=sec^2(x)dx then dx=du/sec^2(x) so we will have integral of du/u+1(sec(x))^2 ,,,, alternatively u^2=tan^2(x) so u^2=sec^2(x)-1 so (u^2)+1=sec^2(x) so the final form will be integral of du/(u+1)((u^2)+1) then solve it by partial fractions if I am wrong please tell me (:

@anshumanagrawal346 3 жыл бұрын

Haha, not me using the t = tan(x/2) substitution without thinking and making this way more complicated than it needs to be

@abdullahimohamedalasow7783 3 жыл бұрын

It is so fun. every time new method, thanks.

@alextaunton3099 2 жыл бұрын

When the answer just popped out, I audibly laughed at the brilliant simplicity

@ahmaurya6187 Жыл бұрын

Hay

@ahmaurya6187 Жыл бұрын

India🇮🇳

@Mariosergio61 2 жыл бұрын

Very ingenious!

@beatrixwashere 3 жыл бұрын

thank you for erasing the tiny blue mark at the end

@littlelilly7480 3 жыл бұрын

Same qn arrived in our previous exam

@mathmancalc7753 3 жыл бұрын

Wait! You shaved? I thought this was an old video you uploaded to this channel.

@michelecastellani1 3 жыл бұрын

That's pratically born haber cycle applied to integrals... How fun😂😂😂

@gddanielk8491 3 жыл бұрын

I like integrals that aren’t hard they’re ok :)

@SimsHacks 3 жыл бұрын

Weierstrass sub u=tg(x/2) would work

@Blend33 3 жыл бұрын

Taking trig; on my way to calc 1, 2, 3

@kushaldey3003 3 жыл бұрын

Very nice trick.

@yoganandasp2085 3 жыл бұрын

Bro why can't it be rationalised? 1-tanx then put tanx=t 1-t/1-t^2 then it's easy

@fivestar5855 3 жыл бұрын

The work of genius god damn :p

@ronaldrosete4086 3 жыл бұрын

I miss your goatee.

@jagula 3 жыл бұрын

This is new video? Did you shave?

@petereziagor4604 3 жыл бұрын

Old good days

@joaquinlemus43 3 жыл бұрын

….mind blowing….🤷🏻‍♂️😱😱😱

@rithwikanand9451 3 жыл бұрын

Please do integral of sin(x^2).

@trayne5151 3 жыл бұрын

cos(x^2)*2x Derivative of the trig function with the chain rule for the x^2 on the inside

@deeznuts4333 3 жыл бұрын

@@trayne5151 integral

@rithwikanand9451 3 жыл бұрын

@@trayne5151 no that's not correct. If you differentiate cos(x^2).2x you will get 2cos(x^2) - 4x^2.sin(x^2)

@Ayush-yj5qv 3 жыл бұрын

@@rithwikanand9451 ya tbh

@xinpingdonohoe3978 3 жыл бұрын

I'm not certain if this will work, but because e^iz = cos(z) + i sin(z), then sin(z) = Im (e^iz) Therefore, the integral of sin(x^2) dx equals the imaginary part of the integral of e^ix² dx. A u substitution, maybe -u² = ix², might be able to produce something in terms of erf(x) or erfi(x). Take the imaginary part, add a constant and you might have your answer. I don't know that this is correct though. Try it.