Evaluate the integral of sin^4(x) dx.

The integral of sin^4(x) dx is (3x/8) - (1/4)sin(2x) + (1/32)sin(4x) + C.

To evaluate the integral of sin^4(x) dx, we can use the identity sin^2(x) = (1/2)(1 - cos(2x)) to rewrite sin^4(x) as (1/4)(1 - cos(2x))^2. Then, we can expand the square using the binomial formula to get sin^4(x) = (1/4)(1 - 2cos(2x) + cos^2(2x)).

Now, we can integrate each term separately. The integral of 1 dx is x, and the integral of cos^2(2x) dx can be found using the identity cos^2(2x) = (1/2)(1 + cos(4x)), which gives us the integral of cos^2(2x) dx as (1/2)(x/2 + (1/4)sin(4x)).

For the integral of cos(2x) dx, we can use the substitution u = 2x, du = 2 dx to get the integral of cos(u) du, which is sin(u) + C. Substituting back, we get the integral of cos(2x) dx as (1/2)sin(2x) + C.

Putting it all together, we have:

∫sin^4(x) dx = ∫(1/4)(1 - 2cos(2x) + cos^2(2x)) dx
= (1/4)(x - ∫2cos(2x) dx + ∫cos^2(2x) dx)
= (1/4)(x - sin(2x) + (1/2)(x/2 + (1/4)sin(4x))) + C
= (3x/8) - (1/4)sin(2x) + (1/32)sin(4x) + C.

Therefore, the integral of sin^4(x) dx is (3x/8) - (1/4)sin(2x) + (1/32)sin(4x) + C.