Integral of \( \sin^3(x) \cos^2(x) \)

Derivation and Evaluation

Evaluate the integral:

\[ \int \sin^3(x) \cos^2(x) \, dx \]

Rewrite the integral by splitting off a single sine factor:

\[ = \int \sin^2(x) \cos^2(x) \sin x \, dx \]

Use the trigonometric identity \( \sin^2 x = 1 - \cos^2 x \) and substitute:

\[ = \int (1 - \cos^2 x) \cos^2(x) \sin x \, dx \]

Expand the integrand:

\[ = \int (\cos^2(x) - \cos^4(x)) \sin x \, dx \]

Use Integration by Substitution: let \( u = \cos x \), which gives \( du = -\sin x \, dx \) or \( \sin x \, dx = -du \). Substituting this yields:

\[ = - \int (u^2 - u^4) \, du \]

Use the common integral formula \( \int u^n \, du = \dfrac{1}{n+1} u^{n+1} + c \) to evaluate the integral:

\[ - \left( \dfrac{1}{3} u^3 - \dfrac{1}{5} u^5 \right) + c = - \dfrac{1}{3} u^3 + \dfrac{1}{5} u^5 + c \]

where \( c \) is the constant of integration.

Substitute back \( u = \cos x \) to obtain the final answer: