Final answer:
To evaluate the integral ∫sin² x cos x In(sin x)dx, we can use the Table of Integrals. The integral of sin² x cos x is (1/3)sin³ x and the integral of In(sin x)dx is (-1/2)sin² x + (1/2)∫cos² x (1/sin x) cos x dx. Combining both parts of the integral, the final result is (1/3)sin³ x - (1/2)sin² x + (1/2)∫cos² x (1/sin x) cos x dx + C.
Step-by-step explanation:
To evaluate the integral ∫sin² x cos x In(sin x)dx, we can use the Table of Integrals. From the table, we can see that the integral of sin² x cos x is equal to (1/3)sin³ x + C, where C is the constant of integration. However, the integral also includes In(sin x), which does not have a direct entry in the table. To evaluate this part, we can use integration by parts. Let's say u = In(sin x), then du = (1/sin x) cos x dx and dv = sin² x dx. Integrating both u and dv, we can find v = (-1/2)cos² x. Applying the integration by parts formula, ∫ u dv = uv - ∫ v du, we get the integration of In(sin x) as (-1/2)sin² x + ∫ (1/2)cos² x (1/sin x) cos x dx. Combining both parts of the integral, we have ∫sin² x cos x In(sin x)dx = (1/3)sin³ x - (1/2)sin² x + (1/2)∫ cos² x (1/sin x) cos x dx + C.