Calculus, 10th Edition (Anton)

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

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.6 Logarithmic And Other Functions Defined By Integrals - Exercises Set 6.6 - Page 461: 37

Answer

$\tan x+\cos\dfrac{1}{\sqrt 2}$

Work Step by Step

The solution of the given form is: $y(x)=y_0+\int_{x_0}^x f(t) \ dt $ We have: $y(x)=1+\int_{\pi/4}^{x} \sec^2 t-\sin t \ dt \\=1+[\tan t +\cos t]_{\pi/4}^{x}\\=1+\tan x +\cos x -1-\dfrac{1}{\sqrt 2}\\=\tan x+\cos\dfrac{1}{\sqrt 2}$
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