Integrals of the product of the powers of sine and cosine come in 4 permutations:

1. The powers m and n are both even

2. The powers m and n are even and odd respectively

3. The powers m and n are odd and even respectively

4. The powers m and n are both odd

In this video, we explore case 1 where both powers are even. In this case, our aim is to reduce the powers to the first power of cosine, so that we have…

sin^m(x)*cos^n(x) = A + B*cos(2x) + C*cos(4x) + D*cos(6x) +…

We can get the integrand into this form using the power reducing half-angle formulas and the product-to-sum formulas.

We look specifically at the example of the integral of sin^4(x)*cos^2(x).

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Hindi padha le na

Ooo

At 1:40 what are A, B, C and D

If any know plz tell

Thank you so much. Greetings from Mexico

Very long

Need shortcut.

It should be 1/2 cos 6x

Good explanation! Thank you.

Thank you sir.

Thank you very very much❤️❤️❤️

Try to type this on math way its different

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

thanks for this

great explanation!

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

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

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

Very helpful thanks a lot!!

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

Thanks sir

Helpfull video

Thnx a lot bro

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

How -1/2cos 2x came

5:18 ?

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

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

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. ^_^

greeaaaattt

6:41 there is a mistake it should be 1/2cos(6x)

Thanks Sir

Thank you very much for the help

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

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

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)

so Woooow

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.

Great video.