## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 50

#### Answer

$$2$$

#### Work Step by Step

\eqalign{ & \int_{\pi /16}^{\pi /8} {8{{\csc }^2}4x} dx \cr & = \int_{\pi /16}^{\pi /8} {2{{\csc }^2}4x} \left( 4 \right)dx \cr & {\text{take out the constant 2}} \cr & = 2\int_{\pi /16}^{\pi /8} {{{\csc }^2}4x} \left( 4 \right)dx \cr & {\text{use }}\int {{{\csc }^2}udu} = - \cot u + C \cr & = 2\left( { - \cot 4x} \right)_{\pi /16}^{\pi /8} \cr & {\text{using The Fundamental Theorem}} \cr & = - 2\left( {\cot 4\left( {\frac{\pi }{8}} \right) - \cot 4\left( {\frac{\pi }{{16}}} \right)} \right) \cr & {\text{simplify}} \cr & = - 2\left( {\cot \left( {\frac{\pi }{2}} \right) - \cot \left( {\frac{\pi }{4}} \right)} \right) \cr & = - 2\left( {0 - 1} \right) \cr & = 2 \cr}

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