How to integrate ∫sin^4(x)cos^2(x)dx - even powers

[@rajendramisir3530]

Thanks for sharing your technique to find the anti-derivative of this trigonometrical integral. A+Bcos(2x)+Ccos(4x)+Dcos(6x)+ ...

[@home8506]

Thank you, I'm peruvian and in my language there's wasn't any video with this particular problem !! I'm so thankful teacher 😁

[@annhak]

I'm from the future and I can confirm too. Greetings from Mexico

[@davidliu7125]

well, I have to say that this is a very complicated way to solve this problem. when we get to the step: int{1/8*[(sin2x)^2-(sin2x)^2*cos2x]}dx, we should take out the 1/8 and calculate two inside terms separately.

[@MasterWuMathematics]

+David Liu Yeah I know it takes quite a few steps to solve this problem. Your method is perfectly good also. As long as we can come to same answer, there's no right or wrong way.

[@socraticmathtutor1869]

Very useful video! Thank you.

[@Chingach]

Thanks for helping Ukrainian student

[@MasterWuMathematics]

It's my pleasure

[@gilma4328]

Good explanation! Thank you.

[@edheani]

Thank you so much. Greetings from Mexico

[@MasterWuMathematics]

You are welcome!

[@emmanuelalbazi8560]

Great video.

[@rafaellisboa8493]

thank you cupinxa

[@vaibhavshekhar314]

This is an awesome technique....but wallis formula is more helpful if you want any shortcut method..

[@holyshit922]

For mental calculation reduction is easier and faster Int(sin^4(x)cos^2(x),x) = Int(sin^4(x)(1-sin^2(x)),x)=Int(sin^4(x),x)-Int(sin^6(x),x) Int(sin^n(x),x)=Int(sin(x)*sin^(n-1)(x),x) Int(sin(x)*sin^(n-1)(x),x) = -cos(x)sin^(n-1)(x)-Int(-cos(x)(n-1)sin^(n-2)(x)cos(x),x) Int(sin^n(x),x) = -cos(x)sin^(n-1)(x) + (n-1)Int(sin^(n-2)(x)cos^2(x),x) Int(sin^n(x),x) = -cos(x)sin^(n-1)(x) + (n-1)Int(sin^(n-2)(x)(1-sin^2(x)),x) Int(sin^n(x),x) = -cos(x)sin^(n-1)(x) + (n-1)Int(sin^(n-2)(x),x) - (n-1)Int(sin^n(x),x) nInt(sin^n(x),x) = -cos(x)sin^(n-1)(x) + (n-1)Int(sin^(n-2)(x),x) Int(sin^n(x),x) = -1/n cos(x)sin^(n-1)(x)+(n-1)/n Int(sin^(n-2)(x),x) Int(sin^6(x),x) = -1/6cos(x)sin^5(x)+5/6(-1/4cos(x)sin^3(x)+3/4(-1/2cos(x)sin(x)+1/2x))+C_{1} Int(sin^6(x),x) = -1/6cos(x)sin^5(x) -5/24cos(x)sin^3(x)-5/16cos(x)sin(x)+5/16x+C_{1} Int(sin^4(x),x) = -1/4cos(x)sin^3(x)+3/4(-1/2cos(x)sin(x)+1/2x)+C_{2} Int(sin^4(x),x) = -1/4cos(x)sin^3(x)-3/8cos(x)sin(x)+3/8x+C_{2} Int(sin^4(x),x) - Int(sin^6(x),x) = 1/6cos(x)sin^5(x)-1/24cos(x)sin^3(x)-1/16cos(x)sin(x)+1/16x+C

[@Ryangosling144p]

Bro gave his voice in oppenheimer ...

[@yadarapunarsimlu5208]

How u written -cos(2x) +1÷2 cos(2x) = - 1/2 cos (2x)

[@eleftheriap.m.4701]

Very helpful thanks a lot!!

[@1238MHUT]

great explanation!

[@AL-wc8oy]

Thankyou so much

[@MasterWuMathematics]

Most welcome 😊

[@dianakyryna7637]

Thank you very very much❤️❤️❤️

[@AbhipsaPatra17]

Thank you Sir ❤

[@rajbawanna6265]

Thank you very much for the help

[@prasenjitmukherjee7210]

Thank you sir.

[@SamuelGauya]

What if both sine and cosine are odd but not both

[@alidesiner3935]

Thanks

[@harisankarsharma2612]

But step marking Is false

[@sonaliswapna....6095]

thanks for this

[@manishadevi5834]

Helpfull video

[@leulassefa7708]

Super

[@tharunivijjigiri5484]

How did u wrote - cos 2x+1/2cos2x=-1/2cos 2x

[@biswajitbarman8270]

Thnx a lot bro

[@portiagonzales2513]

It should be 1/2 cos 6x

[@bijaybera6469]

`int(sin4x)/(cos^(4)x)dx` What is the solution of this question

[@MasterWuMathematics]

It took a while, but I've made a video for you here: kzbin.info/www/bejne/Z4vdg6dultp3mpY

[@edgardoconcepcion8308]

the right formula for d product to sum formula s cos (u) cos (v) = 1/2[cos(u-v) - cos(u+v)] . correct me f i'm wrong. reply plz. thx

[@MasterWuMathematics]

The left hand side of the formula you have stated should be sin(u)sin(v)

[@sarahkamal2585]

can I get the link for the video where u explained cos(u)cos(v)

[@Furapika]

Try to type this on math way its different

[@tharunivijjigiri5484]

Did u take lcm

[@tharunivijjigiri5484]

Of 1/2

[@candrahman8774]

hmmm, sir i am want to ask, what happen if the question is ∫cos^2(x)sin^4(x)dx, is the answer the same sir?

[@MasterWuMathematics]

Yes, the answer will be exactly the same.

[@ziaarfeen9992]

sir is there any video in which you tell half angle formulas please if you have then tell me. I did not want anyone to tell me that i sucks in math i already know. exams are coming i'm so worried

[@MasterWuMathematics]

Have confidence in yourself my friend. Understand as much as you can and you'll be fine for your exams. I derive the half angle formulas in this video: kzbin.info/www/bejne/h4LEh5Kld8-GZ8U

[@Furapika]

Oof but the answer in my book is different from your answer sir. Im confused

[@doworktomorrow]

mine is solved like this (1 − cos (2x)/2)^2 . (1 + cos (2x)/2) dx

[@Rafaelmiramadi]

This is also correct, But alot of books simplify alot further by using more trigonometry

[@santoshmathapati3089]

Thanks Sir

[@round2heaven792]

greeaaaattt

[@rohanjha9801]

Sir do u have any video for higher even powers of sin and cos if have suggeste me

[@MuhammadIrfan-nf9pb]

yes you can resolve them in (sin^2)^n sinx (cos^2)^N

[@PoonamGupta-uz7ct]

How -1/2cos 2x came

[@shrutiyadav7706]

At 1:40 what are A, B, C and D If any know plz tell

[@anamikayadav8416]

What if the angle of sin and cos both are different rather than same angle x

[@rajbawanna6265]

Anamika Yadav that will be more easy than this

[@tharunivijjigiri5484]

Hello masterwumathimatics plzzz clarify my doubt🙄🙄

[@shakirahahmadsuhaimi3306]

Great video! Thank you so much. But can you make a video on how to solve this problem? "integrate sin^5(x)(cos(x))^1/4". I really don't have any ideas on even how to start. Thank you in advance. ^_^

[@MasterWuMathematics]

Here you go... kzbin.info/www/bejne/nInCe35opr93iZI

[@JavierReyesAb]

Hi, I have that integrate with another result. Can you answer me to talk? Greetings from Mexico

[@MasterWuMathematics]

Yes. What result do you have?

[@JavierReyesAb]

Sorry for the delay. My result is : x/16 - sin4x/64 - sen^3 2x/48 + c

[@fiorelavargas8924]

@@JavierReyesAb me sale lo mismo

[@tharunivijjigiri5484]

Hello

[@glorytv1510]

so Woooow

[@clickityClank_]

At 3:27 you say sin(x)cos(x) = 1/2 sin(2x), where does the 1/2 come from? i though it was 2sin(x)cos(x) = sin(2x)

[@victorcorcini8305]

2sin(x)cos(x) = sin(2x) Multiplying both sides by 1/2 you find sin(x)cos(x) = 1/2 sin(2x)

[@victorcorcini8305]

2sin(x)cos(x) = sin(2x) dividing both sides by 2 you find sin(x)cos(x) = 1/2 sin(2x)

[@victorcorcini8305]

It's important to remember to multiply by 1/2 is the same that to divide by 2

[@ABHISHEKKUMAR-02048]

This is a simple algebraic calculation : We know that sin(2x) = 2sin(x)cos(x) We can also write this trigonometrical formula as : 2sin(x)cos(x) = sin(2x) Which implies sin(x)cos(x) = (1/2)sin(2x) You should understand this result with : 2x = y implies x = y/2 or, x = (1/2)y

[@nadiamedina957]

5:18 ?

[@shakeelkalsoehffgy6efyhrry179]

simply opened brackets , lol late af reply

[@sriabhijit]

Very long Need shortcut.

[@thegoldenstatus3691]

Ooo

[@tharunivijjigiri5484]

Plz reply

[@thegoldenstatus3691]

Hindi padha le na

[@basantahota538]

Thanks sir