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

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