Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 19

Answer

$$\frac{{2{{\left( {\sin x} \right)}^{3/2}}}}{3} - \frac{{2{{\left( {\sin x} \right)}^{7/2}}}}{7} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\cos }^3}x} \sqrt {\sin x} dx \cr & {\text{split off }}\cos \theta \cr & = \int {{{\cos }^2}x} \sqrt {\sin x} \cos xdx \cr & {\text{pythagorean identity}} \cr & = \int {\left( {1 - {{\sin }^2}x} \right)} {\left( {\sin x} \right)^{1/2}}\cos xdx \cr & u = \sin x,{\text{ }}du = \cos xdx \cr & = \int {\left( {1 - {{\sin }^2}x} \right)} {\left( {\sin x} \right)^{1/2}}\cos xdx \cr & = \int {\left( {1 - {u^2}} \right){u^{1/2}}du} \cr & = \int {\left( {{u^{1/2}} - {u^{5/2}}} \right)du} \cr & {\text{evaluate the integral}} \cr & = \frac{{{u^{3/2}}}}{{3/2}} - \frac{{{u^{7/2}}}}{{7/2}} + C \cr & {\text{replace }}u = \sin x \cr & = \frac{{2{{\left( {\sin x} \right)}^{3/2}}}}{3} - \frac{{2{{\left( {\sin x} \right)}^{7/2}}}}{7} + C \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.