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

More References and Links

1. Table of Integral Formulas
2. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
   
3. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
   
4. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8