# Integral of $\sin^3 x$

 

Evaluate the integral $\int \sin^3 x \; dx$ Write the integrand $\sin^3 x$ as a product of as $\sin^2 x$ and $\sin x$ $\int \sin^3 x \; dx = \int \sin^2 x \sin x \; dx$ Use of the trigonometric identity $\; \sin^2 x = 1 - \cos^2 x$ to write the integral as
$\int \sin^3 x \; dx = \int (1 - \cos^2 x) \sin x \; dx$
Expand $(1 - \cos^2 x) \sin x$ and write $\int \sin^3 x \; dx = \int \sin x \; dx - \int \cos^2 x \sin x \; dx$
Integration by Substitution : Let $u = \cos x$ and hence $\dfrac{du}{dx} = - \sin x$ or $du = - \sin x \; dx$ and substitute to obtain $\int \sin^3 x \; dx = \int \sin x \; dx + \int u^2 \; du$ Use integral formulas to evaluate the above integral and write $\int \sin^3 x \; dx = - \cos x + \dfrac{1}{3} u^3 + c$ Substitute back $u = \cos x$ to find the final answer $\boxed { \int \sin^3 x \; dx = - \cos x + \dfrac{1}{3} \cos^3 x + c }$