# Integral of $\cos^3 x$

  

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