Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.6 The Fundamental Theorem Of Calculus - Exercises Set 4.6 - Page 320: 22

Answer

$$\frac{{{\pi ^2}}}{9} + 2\sqrt 3 $$

Work Step by Step

$$\eqalign{ & \int_{\pi /6}^{\pi /2} {\left( {x + \frac{2}{{{{\sin }^2}x}}} \right)dx} \cr & {\text{trigonometric identity }}\csc x = \frac{1}{{\sin x}} \cr & = \int_{\pi /6}^{\pi /2} {\left( {x + 2{{\csc }^2}x} \right)dx} \cr & {\text{use power rule integration formulas from table 4}}{\text{.2}}{\text{.1}} \cr & = \left( {\frac{{{x^2}}}{2} + 2\left( { - \cot x} \right)} \right)_{\pi /6}^{\pi /2} \cr & = \left( {\frac{{{x^2}}}{2} - 2\cot x} \right)_{\pi /6}^{\pi /2} \cr & {\text{part 1 of fundamental theorem of calculus}} \cr & = \left( {\frac{{{{\left( {\pi /2} \right)}^2}}}{2} - 2\cot \left( {\frac{\pi }{2}} \right)} \right) - \left( {\frac{{{{\left( {\pi /6} \right)}^2}}}{2} - 2\cot \left( {\frac{\pi }{6}} \right)} \right) \cr & = \left( {\frac{{{\pi ^2}}}{8} - 2\left( 0 \right)} \right) - \left( {\frac{{{\pi ^2}}}{{72}} - 2\left( {\sqrt 3 } \right)} \right) \cr & = \frac{{{\pi ^2}}}{8} - \frac{{{\pi ^2}}}{{72}} + 2\sqrt 3 \cr & = \frac{{{\pi ^2}}}{9} + 2\sqrt 3 \cr} $$
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