Evaluate the integral
\[ \int \sin^3(x) \cos^2(x) \; dx \]
Rewrite as
\[ = \int \sin^2(x) \cos^2(x) \sin x \; dx \]
Use trigonometric identity \( \sin^2 x = 1 - \cos^2 x \) and substitute
\[ = \int (1 - \cos^2 x) \cos^2(x) \sin x \; dx \]
Expand the the integrand
\[ = \int (\cos^2(x) - \cos^4(x)) \sin x \; dx \]
Use Integration by Substitution: \( u = cos x \) so that \( du = - \sin x \; dx \)
\[ = - \int (u^2 - u^4) \; du \]
Use the common integral \( \int u^n du = \dfrac{1}{n+1} u^{n+1}+ c\) to evaluate the above integral
\[ - \dfrac{1}{3} u^3 + \dfrac{1}{5} u^5 + c \]
Substitute back \( u = cos x \) to obtain the final answer
\[ \boxed { \int \sin^3(x) \cos^2(x) \; dx = - \dfrac{1}{3} \cos^3 (x) + \dfrac{1}{5} \cos^5 (x) + c } \]
