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

 

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 $\boxed { \int \sin^3(x) \cos^2(x) \; dx = - \dfrac{1}{3} \cos^3 (x) + \dfrac{1}{5} \cos^5 (x) + c }$