Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.7 Exercises - Page 380: 30

Answer

$$\frac{\pi }{4}$$

Work Step by Step

$$\eqalign{ & \int_0^{\pi /2} {\frac{{\cos x}}{{1 + {{\sin }^2}x}}} dx \cr & {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx,{\text{ }}dx = \frac{1}{{\cos x}}du \cr & {\text{The new limits of integration are}} \cr & x = \pi ,{\text{ }}u = \sin \left( {\frac{\pi }{2}} \right) = 1 \cr & x = 0,{\text{ }}u = \sin 0 = 0 \cr & {\text{Substituting}} \cr & \int_0^{\pi /2} {\frac{{\cos x}}{{1 + {{\sin }^2}x}}} dx = \int_0^1 {\frac{{\cos x}}{{1 + {u^2}}}} \left( {\frac{1}{{\cos x}}} \right)du \cr & = \int_0^1 {\frac{1}{{1 + {u^2}}}} du \cr & {\text{Integrate using basic integration rules}} \cr & = \left[ {{{\tan }^{ - 1}}u} \right]_0^1 \cr & = {\tan ^{ - 1}}\left( 1 \right) - {\tan ^{ - 1}}\left( 0 \right) \cr & = \frac{\pi }{4} \cr} $$
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.