Integral of (cos(x)-1)/(cos(x)+1)

[@IntegralsForYou]

Hi! I started writing all the solutions in a PDF file 📄. You can get it in my Patreon page: www.patreon.com/integralsforyou/shop ❤

[@paulogabriel1251]

Gostei muito do caminho que sua resolução tomou! Eu parti para as identidades de seno e cosseno, mas foi um caminho mais longo haha preferi o seu. Parabéns pelo vídeo! :)

[@IntegralsForYou]

Glad you liked it! Another way to solve it is by multiplying by conjugate, cos(x)-1, on the numerator and denominator: Integral of (cos(x)-1)/(cos(x)+1) dx = = Integral of [(cos(x)-1)/(cos(x)+1)]*[(cos(x)-1)/(cos(x)-1)] dx = = Integral of (cos(x)-1)^2/[(cos(x)+1)*(cos(x)-1)] dx = = Integral of (cos(x)-1)^2/(cos^2(x)-1) dx = = Integral of (cos(x)-1)^2/(-sin^2(x)) dx = = - Integral of (cos(x)-1)^2/sin^2(x) dx = = - Integral of (cos^2(x)-2cos(x)+1)/sin^2(x) dx = = - Integral of (1-sin^2(x)-2cos(x)+1)/sin^2(x) dx = = - Integral of (-sin^2(x)-2cos(x)+2)/sin^2(x) dx = = - Integral of ( -sin^2(x)/sin^2(x) - 2cos(x)/sin^2(x) + 2/sin^2(x) ) dx = = - Integral of ( -1 - 2cos(x)/sin^2(x) + 2/sin^2(x) ) dx = = Integral of 1 dx + 2*Integral of cos(x)/sin^2(x) dx - 2*Integral of 1/sin^2(x) dx = = Integral of 1 dx + 2*Integral of 1/sin^2(x) cos(x)dx - 2*Integral of 1/sin^2(x) dx = Substitution: u = sin(x) du = cos(x)dx = Integral of 1 dx + 2*Integral of 1/u^2 du - 2*Integral of 1/sin^2(x) dx = = Integral of 1 dx + 2*Integral of u^(-2) du - 2*Integral of 1/sin^2(x) dx = = x + 2*u^(-1)/(-1) - 2*(-cot(x)) = = x - 2/u + 2cot(x) = = x - 2/sin(x) + 2cot(x) + C = = x - 2/sin(x) + 2cos(x)/sin(x) + C = = x - (2-2cos(x))/sin(x) + C = = x - 2(1-cos(x))/sin(x) + C = = x + 2(cos(x)-1)/sin(x) + C ❤

[@Aurelius7777]

Thanks for teaching :)

[@IntegralsForYou]

Thanks for watching! ❤

[@quark67000]

Please respect the rules. "c" and "o" are not variables and "s" (in "s(x)") is not the name of non standard function. However, cos is a standard function, so it must be displayed in upright font. For this, you have in LaTeX the \cos command. Please correct the thumbnail of your videos. Some authors also displays "d" in upright, as it's a standard "operator" instead of a variable. Variables are always in italic. Thanks for correct this. Because as it is yet, we read the product of some variables ("c" and "o"), and the value s(x) of a non standard function "s" (which is not defined). The rule is standardized in ISO 80000-2 (see Wikipedia). Hint: if LaTeX has a command named \cos, it's not for the decoration.

[@IntegralsForYou]

Thanks for your comment! I 100% agree with you, I want to change all my thumbnails and I will do it very soon 💪

[@lgl_137noname6]

1:26 @IntegralsForYou @IntegralsForYou from 1min to 2:min5, you do the substitution with 3 identities. I'm lost with the identity where you equate cos(x) to (1-t^2)/(1+t^2). I don't understand the origin,mechanic and logic of that substitution . Would you please kindly explain how that is ? Thank you for your favorable consideration . Regards

[@IntegralsForYou]

Hi! I'm sorry, maybe it is not clear... I am performing the Weierstrass substitution which consists of doing the substitution t=tan(x/2). When we do this substitution, we need to calculate sin(x), cos(x) and dx in terms of "t" and I am sorry because I forgot to put the link to the video where I calculate it. Here it is: kzbin.info/www/bejne/oZ7Jfpyhiap8rq8 Thanks for asking! ❤

[@lgl_137noname6]

@@IntegralsForYou Thank You !

[@IntegralsForYou]

@@lgl_137noname6 My pleasure! 💪

[@СнежныйБарс-х8ь]

Умножить числитель(top) and знаменатель ( down) на сопряженное ( cosx-1). Ответ: x- 2(cosx-1)/(sinx)+C.

[@IntegralsForYou]

Oh, nice, good idea! I got x+2(cosx-1)/(sinx)+C instead of x-2(cosx-1)/(sinx)+C, I don't know if I did a mistake somewhere but I checked it with the derivative and it seems everything is ok. Let's see: Integral of (cos(x)-1)/(cos(x)+1) dx = = Integral of [(cos(x)-1)/(cos(x)+1)]*[(cos(x)-1)/(cos(x)-1)] dx = = Integral of (cos(x)-1)^2/[(cos(x)+1)*(cos(x)-1)] dx = = Integral of (cos(x)-1)^2/(cos^2(x)-1) dx = = Integral of (cos(x)-1)^2/(-sin^2(x)) dx = = - Integral of (cos(x)-1)^2/sin^2(x) dx = = - Integral of (cos^2(x)-2cos(x)+1)/sin^2(x) dx = = - Integral of (1-sin^2(x)-2cos(x)+1)/sin^2(x) dx = = - Integral of (-sin^2(x)-2cos(x)+2)/sin^2(x) dx = = - Integral of ( -sin^2(x)/sin^2(x) - 2cos(x)/sin^2(x) + 2/sin^2(x) ) dx = = - Integral of ( -1 - 2cos(x)/sin^2(x) + 2/sin^2(x) ) dx = = Integral of 1 dx + 2*Integral of cos(x)/sin^2(x) dx - 2*Integral of 1/sin^2(x) dx = = Integral of 1 dx + 2*Integral of 1/sin^2(x) cos(x)dx - 2*Integral of 1/sin^2(x) dx = Substitution: u = sin(x) du = cos(x)dx = Integral of 1 dx + 2*Integral of 1/u^2 du - 2*Integral of 1/sin^2(x) dx = = Integral of 1 dx + 2*Integral of u^(-2) du - 2*Integral of 1/sin^2(x) dx = = x + 2*u^(-1)/(-1) - 2*(-cot(x)) = = x - 2/u + 2cot(x) = = x - 2/sin(x) + 2cot(x) + C = = x - 2/sin(x) + 2cos(x)/sin(x) + C = = x - (2-2cos(x))/sin(x) + C = = x - 2(1-cos(x))/sin(x) + C = = x + 2(cos(x)-1)/sin(x) + C Check with derivative: Derivative of x + 2(cos(x)-1)/sin(x) + C = = Derivative of x + 2cot(x) - 2(sin(x))^(-1) + C = = 1 + 2(-1/sin^2(x)) - 2*(-1)(sin(x))^(-2)*cos(x) + 0 = = sin^2(x)/sin^2(x) - 2/sin^2(x) + 2cos(x)/sin^2(x) = = (1-cos^2(x))/sin^2(x) - 2/sin^2(x) + 2cos(x)/sin^2(x) = = (1 - cos^2(x) - 2 + 2cos(x))/sin^2(x) = = (-cos^2(x) + 2cos(x) - 1)/sin^2(x) = = -(cos^2(x) - 2cos(x) + 1)/(1-cos^2(x)) = = -(cos(x)-1)^2/(1-cos(x))(1+cos(x)) = = -(cos(x)-1)(cos(x)-1)/(1-cos(x))(1+cos(x)) = = (1-cos(x))(cos(x)-1)/(1-cos(x))(1+cos(x)) = = (cos(x)-1)/(1+cos(x)) Thank you very much for your comment, I really appreciated it! And very happy to see you around again! ❤

[@СнежныйБарс-х8ь]

@@IntegralsForYou 👋

[@СнежныйБарс-х8ь]

Then , uselly , it's can use would сопряжённый множитель для дроби .

[@СнежныйБарс-х8ь]

😉

[@lgl_137noname6]

wrong time stamp

[@quark67000]

I don't know the Cos function. But perhaps do you ask about the cos function? Be accurate. Also, I see nothing about cos(x) on 3:30. Do you speak about 1:30? For cos(x)=(1-tan²(x/2))/(1+tan²(x/2)), go to the Wikipedia article List of trigonometric identities and on the subsection Double-angle formulae, in the displayed formula for cos(2𝜃), replace 𝜃 by 𝜃/2.

[@IntegralsForYou]

Hi! I don't see where the time stamp is wrong... could you be more specific? I will correct is as soon as possible 💪

[@quark67000]

@@IntegralsForYou The user @lgl_137noname6 has modified his first comment. His first comment stated that at 3:30 he doesn't understand the change with the "Cos(x)" (he really wrote "Cos(x)'" instead "cos(x)"). I have give him some explanations. It's about cos(x)=(1-t²)/(1+t²), with t=tan(x/2). It's a known formulae, but perhaps forgiven by @lgl_137noname6, or not yet seen by him.

[@IntegralsForYou]

Oh, ok! Then I think I answered in the other comment posted by @lgl_137noname6 where I share the video to the calculations of sin(x), cos(x) and dx when t=tan(x/2): kzbin.info/www/bejne/oZ7Jfpyhiap8rq8 Thanks! ❤