What is the integral of sin4xcos4x dx?

In this problem, it is easiest to substitute twice.

First, substitute 4x for U and 4dx for dU (remember to balance it out by adding a 1/4 to the resulting integral).

Afterwards, substitute sin(U) for W and cos(U)dU for dW and you should get the integral of 1/4*W*dW, after which solve like a normal integral, which should be
(1/8)*(W^2) + C and replace the W's and U's for the substitutions made above and you should end up with
(1/8)*((sin4x)^2) + C.

To find the integral of sin(4x)cos(4x) dx, we can use the trigonometric identity:

sin(2x) = 2sin(x)cos(x)

If we rewrite sin(4x)cos(4x) using this identity, we get:

sin(4x)cos(4x) = (1/2)sin(8x)

Now, we can rewrite the integral as:

∫ (1/2)sin(8x) dx

To solve this integral, we can use a simple substitution. Let's substitute u = 8x. Now, we need to find the differential of u.

du/dx = 8

Rearrange this equation to solve for dx:

dx = du/8

Substituting the new variables into the equation, we get:

∫ (1/2)sin(u) (du/8)

Pulling the constants out of the integral, we have:

(1/16) ∫ sin(u) du

Now, we integrate sin(u) with respect to u:

(1/16)(-cos(u)) + C

Finally, substitute back the value of u:

(1/16)(-cos(8x)) + C

Therefore, the integral of sin(4x)cos(4x) dx is (-1/16)cos(8x) + C, where C is the constant of integration.

simpler way:

recognize that sin4xcos4x = (1/2)sin8x
and the integral of that is

(-1/16)cos(8x) + c