Integrating using t=tan(x/2) substitution - [The Weierstrass substitution]

In this video, I showed how to integrate a function of the form 1/(c + bsinx +acosx) .

[jan-willemreens]

... Newton, what I forgot to say is, keep doing your presentations on the blackboard with just a simple piece of chalk, your handwriting is excellent for this! Jan-W

[boydbanda]

Big man you're doing it more than anyone. I like the way you Tutor us. Keep up the good work manh.

[dusscode]

You can actually solve it much more simply by multiplying the numerator and denominator by the conjugate of the denominator, to get ∫ (1-sin(x))/cos^2(x) dx.

[robread-jones]

Thanks Newton, that was so good. I really enjoyed that.

Keep up the great videos.

👍

[BrianLewis-ri]

Newton.. sincerely speaking you have helped me alot..🙏

[willwill]

you are a perfect teacher, moreover a perfect human

[kinyerajoel]

I am a teacher, but whenever I watch your video, everything is just fine to teach

[romanmuller]

Wow, this is a great video. You have such an excitement inducing voice. You're really getting the beauty of maths across.

[halimsemihozcan]

Thanks for this lesson . You are a good teacher .

[georgesadler]

Professor Prime Newtons, thank you for the video. Calculus Two playlist on YouTube does not cover this topic. This topic is called special substitution, which is part of Techniques of Integration in Calculus Two. This is an error free video/lecture on YouTube TV with Professor Prime Newtons.

[utuberaj]

Nice way to get the answer.
But Mr Newton, I did it WITHOUT any substitution.
In fact all I did was multiply and divide by the conjugate of the DENOMINATOR ( 1 - Sin X) and there onwards, it is quite simple really for a good Calc 1 or Calc 2 level student.
Your method, though a nice U-sub, seems quite lengthy, if I may say, Mr Newton.
To make this a little harder, why not try to solve the same Integral as a DEFINITE integral from 0 to pi/2? It is quite an interesting one indeed, trust me❤❤

[mb-hvkf]

Thank you for the nice example and exposition. Added another tool to the toolkit.

[estevaocachiliva]

Great video professor👏🏽
I have two questions:

1 - can I use this t substituion whenever I have a integral of cosine and sine? And if I have another trigonometric identity can I rewrite this identity in terms of sine or cossine or both to apply this substituion?

2 - how can I apply this substituion in those cases I have in the answer a angle in radians summing the variable on the argument of some trigonometric identity. Like for exemplo how to apply t substituion on the integral of dx/ sen(x) + cos(x).
The answer of this integral is 1/sqrt 2 that multiply ln( csc( x + pi/4) - cot( x + pi/ 4)

How can I reach the same result with t substituion.

[StrangeQuark.]

Thank you so much :D
It’s such a great explanation

[EE-Spectrum]

Is there a relationship between t-substitution and the half-angle identities?

[joseantonioandrade]

Newton you can also solve that integral using the conjugate of 1+sinx, that is multiplying up and the bottom by 1- sinx, and the final result is tanx-secx, just another way to do it. greetings

[RaadoNoori-kmej]

That is great, but I have question could we use t= tanx
Or we have to make angle x/2

[mikedavis]

I multiplied the numerator and denominator by the conjugate, 1 - sin X, got 1 - sin x/1-sin²X, substituted Cos²X for 1-sin²X, split the fraction, took the integral and ended up with tan x - sec X + c

[jan-willemreens]

... A good day to you Newton, Right out of one of my " old and trusted " little math notebooks regarding integrals the following solution path in short: Given INT(1/(1 + sin(x))dx [ multiply top and bottom of the integrand by (1 - sin(x)) ] --> INT((1 - sin(x))/(1 - sin^2(x)))dx = INT((1 - sin(x))/cos^2(x))dx = INT(1/cos^2(x))dx + INT(- sin(x)/cos^2(x))dx [ u = cos(x) --> du = - sin(x)dx ] = INT(sec^2(x))dx + INT(1/u^2)du = tan(x) + INT(u^-2)du [ applying the good old REVERSE power rule, remember Newton? (lol) ] = tan(x) - 1/u [ u = cos(x) ] = tan(x) - 1/cos(x) + C = tan(x) - sec(x) + C = (sin(x) - 1)/cos(x) + C ... etc etc ... I leave the outcome to everyone's preference ... Thank you too Newton for your great performance; I really mean this, an eye opener for me; isn't it called the Weierstrass method? A pleasant weekend to you, Jan-W

[kurtecaranum]

If you remember all your derivatives, the integral could be solved as

(1 - sin x)/(cos^2 x) dx

(sec^2 x - sec x tan x) dx

tan x - sec x + C