What's the integral of (1+cos(x))^2?

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

To solve this integral, we can expand the expression (1+cos(x))^2 using the identity (a+b)^2 = a^2 + 2ab + b^2. This gives us:

(1+cos(x))^2 = 1 + 2cos(x) + cos^2(x)

We can then integrate each term separately. The integral of 1 is simply x, and the integral of cos^2(x) can be found using the identity cos^2(x) = (1+cos(2x))/2. This gives us:

∫cos^2(x) dx = ∫(1+cos(2x))/2 dx
= 1/2 ∫dx + 1/2 ∫cos(2x) dx
= x/2 + 1/4sin(2x) + C

To integrate the term 2cos(x), we can use the substitution u = sin(x), du/dx = cos(x), and rewrite the integral as:

∫2cos(x) dx = 2∫cos(x) cos(x) dx
= 2∫cos(x) d(sin(x))
= 2sin(x)cos(x) - 2∫sin(x) d(cos(x))
= 2sin(x)cos(x) + 2∫cos(x) d(sin(x))
= 2sin(x)cos(x) + 2sin(x) + C

Putting all the integrals together, we get:

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

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