## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 13 - The Trigonometric Functions - 13.3 Integrals of Trigonometric Functions - 13.3 Exercises - Page 697: 10

#### Answer

$$\frac{{{{\sin }^5}x}}{5} + C$$

#### Work Step by Step

\eqalign{ & \int {{{\sin }^4}x\cos x} dx \cr & {\text{or we can write as}} \cr & = \int {{{\left( {\sin x} \right)}^4}\cos x} dx \cr & {\text{set }}u = \sin x{\text{ then }}\frac{{du}}{{dx}} = \cos x,\,\,\,\,\,\,\,\,\frac{{du}}{{\cos x}} = dx \cr & {\text{write the integrand in terms of }}u \cr & \int {{{\left( {\sin x} \right)}^4}\cos x} dx = \int {{u^4}\cos x} \left( {\frac{{du}}{{\cos x}}} \right) \cr & = \int {{u^4}} du \cr & {\text{integrate by using the power rule for integration}} \cr & = \frac{{{u^5}}}{5} + C \cr & {\text{write in terms of }}x \cr & = \frac{{{{\left( {\sin x} \right)}^5}}}{5} + C \cr & or \cr & = \frac{{{{\sin }^5}x}}{5} + C \cr}

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