C4 Q) Integrate sin2xsinx

Not really sure how to integrate sin2xsinx

Can you use integration by parts? Or would that keep going on? There must be many different methods, but which ones are easy to spot and quickest?

Thanks

Reply 1

User1686961

Original post by ~amethyst~

Not really sure how to integrate sin2xsinx

Can you use integration by parts? Or would that keep going on? There must be many different methods, but which ones are easy to spot and quickest?

Thanks

Replace the sin(2x) with 2sin(x)cos(x) then you should get 2sin^2(x)Cos(x). Take out sin(x) as U.
Hopefully you can continue from there.

Reply 2

davros

Original post by ~amethyst~

Not really sure how to integrate sin2xsinx

Can you use integration by parts? Or would that keep going on? There must be many different methods, but which ones are easy to spot and quickest?

Thanks

The solution given by ThatGuyRik is fairly fast.

Alternatively, your formula book should contain a formula for rewriting sinAsinB as the sum of two trig functions so you can use that and then integrate directly.

Reply 3

~amethyst~

Original post by davros

how do you use that? Can you explain please

Reply 4

username1200599

Original post by ~amethyst~

how do you use that? Can you explain please

Sin2x is a double angle trig term, and you can rewrite it in different ways which you should learn in C3.

The way given by Davros you would rewrite sin2x as sin(x + x) = sinxcosx + sinxcosx = 2sinxcosx

Reply 5

~amethyst~

Original post by ThatGuyRik

Replace the sin(2x) with 2sin(x)cos(x) then you should get 2sin^2(x)Cos(x). Take out sin(x) as U.
Hopefully you can continue from there.

urmm im still confused. I got the 2sinxcosx but how does taking sinx as U help? Can you explain please?

Also, can't you look at 2sin^2(x)cos(x) and see that when you differentiate sin(x) you get cos(x) so when you integrate does the cos(x) cancel out? So you add one to the power and divide by that number...giving 2/3sin^3(x)

Not sure if this is right

Reply 6

~amethyst~

Original post by Iridann

Okay i understand that but isn't that the same method the first person suggested? Using 2sinxcosx?

Reply 7

User1686961

Original post by ~amethyst~

okay so the integral is 2sin^2(x)cos(x) dx
U=sinx du/dx=cosx
replace the dx with du/cosx
the cosx cancel's out and you are left with 2u^2. Then just integrate normally and put sinx back in.

Yeah you've done it correct. Didnt read the whole post soz :smile:

Reply 8

username1200599

Original post by ~amethyst~

Okay i understand that but isn't that the same method the first person suggested? Using 2sinxcosx?

I think I may have misread the other comment, apologies.

I believe Davros was referring to use of sin A + sin B = 2 ((sinA + B)/2) x ((cosA − B)/2)

(edited 8 years ago)

Reply 9

User1686961

Original post by ~amethyst~

Okay i understand that but isn't that the same method the first person suggested? Using 2sinxcosx?

Sorry i didnt elaborate on using the trig formula. I assumed you knew it.

Reply 10

~amethyst~

Original post by ThatGuyRik

Yeah that makes sense! Thanks!! :smile:

Reply 11

davros

Original post by ~amethyst~

how do you use that? Can you explain please

You should have a formula in your book that says

sinAsinB = [cos(A-B) - cos(A+B)]/2

or something similar.

So in this case,

sin2xsinx = [cos x - cos3x]/2

which lets you integrate sin2xsinx easily :smile: