## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises: 13

#### Answer

$${\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{dx}}{{{{\left( {16 - {x^2}} \right)}^{1/2}}}}} \cr & {\text{substitute }}x = 4\sin \theta \cr & = \int {\frac{{4\cos \theta d\theta }}{{{{\left( {16 - 16{{\sin }^2}\theta } \right)}^{1/2}}}}} \cr & = \int {\frac{{4\cos \theta d\theta }}{{4{{\left( {1 - {{\sin }^2}\theta } \right)}^{1/2}}}}} \cr & {\text{pythagorean identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & = \int {\frac{{4\cos \theta d\theta }}{{4{{\left( {{{\cos }^2}\theta } \right)}^{1/2}}}}} \cr & = \int {d\theta } \cr & {\text{integrating}} \cr & = \theta + C \cr & x = 4\sin \theta ,{\text{ then}} \cr & = {\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + C \cr}

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