\( \int _{}^{}\sin (2x)\cos (x)(\cos (x)) d x \)

To solve the integral \[ \int \sin(2x) \cos(x) \cos(x) \, dx, \] we can simplify the expression first. Notice that \(\cos(x) \cos(x) = \cos^2(x)\). Thus, we can rewrite the integral as: \[ \int \sin(2x) \cos^2(x) \, dx. \] Next, we can use the identity for \(\sin(2x)\): \[ \sin(2x) = 2 \sin(x) \cos(x). \] Substituting this into the integral gives: \[ \int \sin(2x) \cos^2(x) \, dx = \int 2 \sin(x) \cos(x) \cos^2(x) \, dx = 2 \int \sin(x) \cos^3(x) \, dx. \] Now, we can use the substitution method. Let: \[ u = \cos(x) \quad \Rightarrow \quad du = -\sin(x) \, dx \quad \Rightarrow \quad -du = \sin(x) \, dx. \] Substituting these into the integral, we have: \[ 2 \int \sin(x) \cos^3(x) \, dx = -2 \int u^3 \, du. \] Now, we can integrate: \[ -2 \int u^3 \, du = -2 \cdot \frac{u^4}{4} + C = -\frac{1}{2} u^4 + C. \] Substituting back \(u = \cos(x)\): \[ -\frac{1}{2} \cos^4(x) + C. \] Thus, the final result for the integral is: \[ \int \sin(2x) \cos^2(x) \, dx = -\frac{1}{2} \cos^4(x) + C. \]