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