## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 13 - The Trigonometric Functions - 13.3 Integrals of Trigonometric Functions - 13.3 Exercises - Page 697: 12

#### Answer

$$2\sqrt {\sin x} + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{\cos x}}{{\sqrt {\sin x} }}} dx \cr & {\text{or we can write }}\sqrt {\sin x} {\text{ as}} \cr & = \int {\frac{{\cos x}}{{{{\left( {\sin x} \right)}^{1/2}}}}} 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 {\frac{{\cos x}}{{{{\left( {\sin x} \right)}^{1/2}}}}} dx = \int {\frac{{\cos x}}{{{u^{1/2}}}}} \left( {\frac{{du}}{{\cos x}}} \right) \cr & = \int {\frac{1}{{{u^{1/2}}}}} du \cr & = \int {{u^{ - 1/2}}} du \cr & {\text{integrate by using the power rule for integration}} \cr & = \frac{{{u^{1/2}}}}{{1/2}} + C \cr & {\text{simplifying}} \cr & = 2{u^{1/2}} + C \cr & = 2\sqrt u + C \cr & {\text{write in terms of }}x \cr & = 2\sqrt {\sin x} + C \cr}

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