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whenever you are integrating a sine or cosine squared, rewrite it with one of the double angle identities for cosine. From

cos 2x = 1 - 2sin²x

if you solve for sin²x, you get:

sin²x = (1-cos2x) / 2

sin²x = 1/2 - 1/2cos2x

So for your expression, you can multiply by 2 and substitute 4x for x to yield:

4sin²(4x) = 2 - 2cos8x

Integrate:

∫4sin²(4x) dx

=∫2 - 2cos8x dx

= 2x - 1/4sin8x + C

It's by integration by parts or the product rule depending on you text book.

remeber the squared function of 4sin^2 4x can also be broken down to 4 (sin4x)(sin4x).

Use the equation (uv)' = vu'+uv' ' mean derivative

Should be able to figure it out from there

Good Luck!

u r right its not as easy to integrate a squared 1

firsstly u have to put it in the form

4(sin4x)^2=4*0.5(1-cos(2*4x))

so it will be 2(1-cos8x)=2-2cos8x

and its then easy to integrate

i.e 2x-16sin8x

hope i helped

You know that cos(A +/- B) = cos(A)cos(B) -/+ sin(A)sin(B). Let A=B=4x. You will have:

cos(8x) = cos^2(4x) - sin^2(4x)

1= cos(0) = cos^2(4x) + sin^2(4x)

Substract the first equation from the second and you have:

1 - cos(8x) = 2sin^2(4x)

So you can simply integrate 2 - 2cos(8x), which is pretty trivial. 2x - (1/4)sin(8x) + C

sin²4x = (1/2).(1 - cos 8x)

I = 4 ∫ (1/2).(1 - cos 8x).dx

I = 2 ∫ (1 - cos 8x).dx

I = 2.( x - sin 8x / 8) + C

I = 2x - (1/4).sin 8x + C