Step 1: Use half-angle identities.
$1+\cos x=2\cos^2\dfrac{x}{2}$ and $1-\cos x=2\sin^2\dfrac{x}{2}$.
Step 2: Substitute into the product.
\[ (1+\cos x)^3(1-\cos x)^4=\left(2\cos^2\tfrac{x}{2}\right)^3\left(2\sin^2\tfrac{x}{2}\right)^4=2^7\cos^6\tfrac{x}{2}\sin^8\tfrac{x}{2}. \]
Step 3: Note the symmetry over $[-2\pi,2\pi]$.
The function is even and the limits are symmetric, so the integral equals twice the integral over $[0,2\pi]$, which keeps the work clean.
Step 4: Use the standard power-of-sine-cosine integrals.
The integral over a full period of $\cos^6\tfrac{x}{2}\sin^8\tfrac{x}{2}$ reduces using known reduction values for even powers of sine and cosine (the Wallis-type results).
Step 5: Combine the constants.
Carrying through the constant $2^7$ together with these standard period integrals gives a single clean value.
Step 6: State the result.
The integral evaluates to \[ \boxed{\frac{5\pi}{4}}. \]