# Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 11

$$\sin x - \frac{{{{\sin }^3}x}}{3} + C$$

#### Work Step by Step

\eqalign{ & \int {{{\cos }^3}x} dx \cr & {\text{split off }}\cos x \cr & = \int {{{\cos }^2}x} \cos xdx \cr & {\text{pythagorean identity}} \cr & = \int {\left( {1 - {{\sin }^2}x} \right)\cos x} dx \cr & u = \sin x,{\text{ }}du = \cos xdx \cr & = \int {\left( {1 - {u^2}} \right)} du \cr & {\text{evaluate the integral}} \cr & = u - \frac{{{u^3}}}{3} + C \cr & {\text{replace }}u = \cos x \cr & = \sin x - \frac{{{{\sin }^3}x}}{3} + C \cr}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.