Question

Can you help me find the following integrals (a) integration of sin 2x cos 4x dx (b)integration of cos 2x cos 4x dx (c) integration of sin 2x sin 4x dx? what method should I use?

Answer 1

Use product to sum formulas.

Explanation:

We know the sum and difference formulas:

`sin(u+v) = sinucosv+cosusinv`
`sin(u-v) = sinucosv-cosusinv`

Adding these we get:

`sin(u+v)-sin(u-v) = 2sinucosv`

This allows us to rewrite

`int sinunderbrace(2x)_u cos underbrace(4x)_v dx` as

`int 1/2[sin(2x+4x)-sin(2x-4x)]dx`

`= 1/2 int [sin6x-sin(-2x)] dx`

`= 1/2 int sin6x+sin(2x) dx`. `" "` (using `sin(-theta) = -sintheta`)

`= 1/2[-1/6cos(6x)-1/2sin(2x)]` `" "` (by substitution)

Using

`cos(u+v) = cosucosv - sinusinv`
`cos(u-v) = cosucosv + sinusinv`

We can get formulas for

`cosucosv` and for `sinusinv`.

Answer 2

I would use substitution and the trig identities for either `cos(2theta)` or `sin(2theta)` considering that `4theta=2(2theta)` as well!

Explanation:

I tried only the first 2. Have a look and try the third using the same approach...if it doesn't work let me know:

and: