Integral of sin^4x

[@TheOrganicChemistryTutor]

Integration - Formula Sheet: bit.ly/3XCT6oz Full-Length Exam Videos: www.patreon.com/MathScienceTutor Calculus Video Playlists: www.video-tutor.net/

[@jimcar53]

What ever your name is - You are the most informative U tube teacher i have listened to. GREAT JOB.

[@rubasham8241]

i've never failed to find a video of yours on a subject that i'm having trouble with, thank you sooo much!

[@snoopdoug4394]

sin^2 (x) = 1/2[1-cos 2(x)]

[@songxavier2452]

can’t imagine the world without u❤

[@AydinovaSabina]

Your videos are very useful. Greetings from Azerbaijan

[@prime2836]

from india

[@obww306]

from KZ

[@evelynluna8272]

You did an amazing job at teaching it thank you!

[@LukasKamin]

is the formula of sum squared, namely (a+b)^2=a^2+2ab+b^2 NOT taught in western schools ?????? that's not the first time I see this well-known formula from basic algebra neglected by this guy , it looks so weird to me. BTW I wonder does these formulas (of sum/difference squared as well as differnce of squares and the same with cubic powers) have some name in western mathematical tradition? in ex-USSR tradition we call them formulas of abbreviated/shortened multiplication (translated literally), for each of this formulas has one side that may be presented as the multiplication of the alike or same expressions in braces

[@siimeen11]

This guy is just trying to explain his actions as clearly as possible. I belive learning by understanding last much longer than learning by using formulas without knowing the origin... a^2+2ab+b^2 is not some magic russian trick, its basic math... There is a reason why alot of people understand this guy, he does not assume anything.

[@joerivera6156]

Awesome you made that very simple. Thank you

[@chaixaioqi3019]

Excellent sir

[@oxygen2671]

Can i solve it by using intigration by part ?

[@piusokolo2921]

Please do you have a video on Jordan normal form

[@spirosgrivas7848]

Nice work! Thank you very much! Interesting you call yourself The Organic Chemistry Tutor but you have so much Math and Physics in here. And what about Physical and Inorganic Chemistry? lol

[@apostolismoschopoulos1876]

spiros grivas ναι και εγώ το βρήκα περίεργο αυτό

[@spirosgrivas7848]

Apostolis Moshopoulos The Organic Chemistry Tutor does it well! Maybe he started with Organic Chemistry and then expanded to the other fields. He is helping a lot of people that way. Have a good day!

[@user-ii6jv1hf5n]

I used integration by part then u- substitution, but I got completely different answer

[@TheLukeLsd]

Qual foi a que vc achou?

[@sulav.108]

Thank uuuu ,you're the bessst ❤️‍🔥❤️‍🔥❤️‍🔥

[@piusokolo2921]

How about cos^4x and sin^5x

[@sonokochan2321]

u r always helpful thankyou so much

[@gayandesilva3180]

Thanks

[@pradeepkathait1149]

really helpful.. thanks a lot..

[@SW-nx4jz]

Omg thank you so much!

[@aileen5639]

thank you!

[@upeecb]

There is no other method instead of it

[@Sai1ence]

The method in this video is a valid way of solving the integral, but it's not the only method available. Another one, specifically used for higher ordered trigonometric function derivatives, includes the usage of the reduction formula of the respective trigonometric function. There is the reduction formula for sine, and it goes like this: Int(sin^n(x)dx) = -1/n * cos(x)sin^[n-1](x) + (n-1)/n * Int(sin^[n-2](x)dx) Reduction formulas exist for other trigonometric functions as well. Here's a video that blackpenredpen made on the reduction formula for sine: kzbin.info/www/bejne/r16cmoWGerOLpbc - - - *Int denotes an integral

[@theowillis6870]

This is L from death note. for stem students/Ppl who do stem for fun/genuises

[@who_we_are______5926]

Dude what's your name? Also will you ever do a face reveal?

[@ntwisisochauke3495]

5 years later and we still waiting for the reveal

[@h_kmack4132]

Excellent!!

[@victoriaowiti1545]

Thanks ❤

[@Melisa-oz6kb]

thank youuu

[@reecegathright150]

I love u so much

[@TheLukeLsd]

I tried the hard way 😅 ʃsin²(x)sin²(x)dx Let u=sin²(x) dv=sin²(x)dx ʃsin²(x)sin²(x)dx=ʃudv=uv-ʃvdu -> du= 2sin(x)cos(x)dx v=ʃsin²(x)dx v=ʃsin(x)sin(x)dx v=ʃrds r=sin(x) dr=cos(x)dx ds=sin(x)dx s=-cos(x) v=rs-ʃsdr = sin(x)(-cos(x))-ʃ(-cos(x))cos(x)dx = -sin(x)cos(x)+ʃcos²(x) = -sin(x)cos(x)+ʃ(1-sin²(x))dx= sin(x)cos(x)+ʃdx - ʃsin²(x)dx = ʃsin²(x)dx -> 2ʃsin²(x)dx=sin(x)cos(x)+ x -> ʃsin²(x) =(sin(x)cos(x) + x )/2 ->ʃsin⁴(x)dx = ʃsin²(x)sin²(x)dx = ʃudv = uv-ʃvdu = u(rs-ʃsdr)-ʃ(rs-ʃsdr)du = sin²(x)(rs-ʃsdr) - ʃ(rs-ʃsdr)2sin(x)cos(x)dx = sin²(x)[sin(x)(-cos(x))-ʃ(-cos(x))cos(x)dx] - ʃ[sin(x)(-cos(x))-ʃ(-cos(x))cos(x)dx]2sin(x)cos(x)dx = sin²(x)[-sin(x)cos(x)+ʃcos²(x)dx] - ʃ[-sin(x)cos(x)+ʃcos²(x)dx]2sin(x)cos(x)dx Let cos²(x)=1-sin²(x) & ʃsin²(x) =(sin(x)cos(x) + x )/2 sin²(x)[-sin(x)cos(x)+ʃ(1-sin²(x))dx] - ʃ[-sin(x)cos(x)+ʃ(1-sin²(x))dx]2sin(x)cos(x)dx = sin²(x)[-sin(x)cos(x)+ʃ1dx-ʃsin²(x)dx] - ʃ[-sin(x)cos(x)+ʃ1dx-ʃsin²(x)dx]2sin(x)cos(x)dx = sin²(x)[-sin(x)cos(x)+x-(sin(x)cos(x) + x )/2] - ʃ[-sin(x)cos(x)+x-(sin(x)cos(x) + x )/2]2sin(x)cos(x)dx = sin²(x)[-sin(x)cos(x)+3x/2-sin(x)cos(x)/2] - ʃ[-sin(x)cos(x)+3x/2-sin(x)cos(x)/2]2sin(x)cos(x)dx = sin²(x)[-3sin(x)cos(x)/2+3x/2] - ʃ[-3/sin(x)cos(x)/2+3x/2]2sin(x)cos(x)dx = 3/2×sin²(x)[-sin(x)cos(x)+x] +3/2×ʃ[sin(x)cos(x)+x]2sin(x)cos(x)dx = 3/2×{sin²(x)[-sin(x)cos(x)+x] +ʃ[sin(x)cos(x)+x]2sin(x)cos(x)dx} = 3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×ʃsin²(x)cos²(x)dx + 2ʃxsin(x)cos(x)dx} Let dn=sin²(x)cos(x)dx & m=cos(x) ʃsin²(x)cos²(x)dx = ʃmdn = mn-ʃndm = sin³(x)cos(x)-ʃsin³(x)(-sin(x))dx = sin³(x)cos(x) + ʃsin⁴(x)dx Then 3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×ʃsin²(x)cos²(x)dx + 2ʃxsin(x)cos(x)dx} = 3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×(sin³(x)cos(x) + ʃsin⁴(x)dx) + 2ʃxsin(x)cos(x)dx} Let p=x & dq=sin(x)cos(x)dx ʃxsin(x)cos(x)dx = ʃpdq = pq-ʃqdp = xsin²(x)/2-ʃsin²(x)/2×1dx = ½xsin²(x)-½ʃsin²(x)dx = ½xsin²(x)-½(sin(x)cos(x) + x )/2 = ½xsin²(x)-¼(sin(x)cos(x) + x ) Then 3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×(sin³(x)cos(x) + ʃsin⁴(x)dx) + 2ʃxsin(x)cos(x)dx} = 3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×(sin³(x)cos(x) + ʃsin⁴(x)dx) + 2[½xsin²(x)-¼(sin(x)cos(x) + x )]} = 3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×(sin³(x)cos(x) + ʃsin⁴(x)dx) + xsin²(x)-½(sin(x)cos(x) + x )} = 3/2×{sin²(x)[-sin(x)cos(x)+x] + 2sin³(x)cos(x) + 2ʃsin⁴(x)dx + xsin²(x)-sin(x)cos(x)/2 - x/2 } = 3/2×{-sin³(x)cos(x)+xsin²(x) + 2sin³(x)cos(x) + 2ʃsin⁴(x)dx + xsin²(x)-sin(x)cos(x)/2 - x/2 } = 3/2×{2xsin²(x) + sin³(x)cos(x) + 2ʃsin⁴(x)dx -sin(x)cos(x)/2 - x/2 } = 3xsin²(x) + 3sin³(x)cos(x)/2 + 3ʃsin⁴(x)dx -3sin(x)cos(x)/4 - 3x/4 =ʃsin⁴(x)dx -> 3xsin²(x) + 3sin³(x)cos(x)/2 -3sin(x)cos(x)/4 - 3x/4 = -2ʃsin⁴(x)dx -> 2ʃsin⁴(x)dx= -3xsin²(x) - 3sin³(x)cos(x)/2 +3sin(x)cos(x)/4 + 3x/4 -> ʃsin⁴(x)dx =3x/8 + 3sin(x)cos(x)/8 - 3sin³(x)cos(x)/4 - 3xsin²(x)/2 + C. -> ʃsin⁴(x)dx =3x/8 + 3sin(x)cos(x)/8 - 6sin³(x)cos(x)/8 - 12xsin²(x)/8 + C. -> ʃsin⁴(x)dx =3x(1-4sin²(x))/8 + 3sin(x)cos(x)(1-2sin²(x)/8 + C.

[@Salamanca-joro]

Daaaamn - Chris tucker

[@aguilarkurtlaurenceralfg.5527]

Im so confusee. Can someone explain it to me why sin^2 (x) = 1/2 (1-cos 2 (x)) because i thought sin^2 (x) = 1-cos^2 (x). Hehe. Thank you. Im just too dumb for this.

[@SKyrim190]

There is a formula for the cosine of the addition of angles cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) If we take a=b=x cos(2*x)=cos^2(x) - sin^2(x) If we use the Trigonometric Identity (that thing you are already aware of, that when you sum up cosine and sine squared you get one); we show: cos(2*x) = 2*cos^2(x) - 1 = 1 - 2*sin^2(x) If we isolate the cosine and sine squared, we get the two formulas he used. You are right that sin^2(x)=1-cos^2(x), but that won't help us, because the trigonometric terms are still raised to a power. Using the double-arch identity we can reduce the power of the exponent, eventually getting to an integral we are familiar with. I believe this strategy always works for even powers of trigonometric functions

[@victoriaowiti1545]

He used that because the question has no cos,,,,,1_cos^2 is used when both sin and cos is in the question,,,,,

[@satvikyadav3242]

wtf i thought it was gif on thumbnail