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x2n+1 The Maclaurin Series of sin x is sin x = for all x E R. (2n + 1)! n=0 1. Express the integral x sin x dx as a series of constant terms. 2. Find the 5th degree Maclaurin polynomial for sin(2x) and use it to approximate the value of sin(0.02). (Do not simplify.)

x2n+1 The Maclaurin Series of sin x is sin x = for all x E R. (2n + 1)! n=0 1. Express the integral x sin x dx as a series of constant terms. 2. Find the 5th degree Maclaurin polynomial for sin(2x) and use it to approximate the value of sin(0.02). (Do not simplify.)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I've been trying to answer these two for hours. Pls help me with both since they are connected and actually under 1 item. I also included reference for more convenience on your part. Thank you and will upvote!

x2n+1 E(-1)". (2n + 1)! The Maclaurin Series of sin x is sin x = for all x E R. n=0 1. Express the integral x° sin x dx as a series of constant terms. 2. Find the 5th degree Maclaurin polynomial for sin(2x) and use it to approximate the value of sin(0.02). (Do not simplify.)
Transcribed Image Text:x2n+1 E(-1)". (2n + 1)! The Maclaurin Series of sin x is sin x = for all x E R. n=0 1. Express the integral x° sin x dx as a series of constant terms. 2. Find the 5th degree Maclaurin polynomial for sin(2x) and use it to approximate the value of sin(0.02). (Do not simplify.)
x" = 1 +x + x² + x³ + • ·. R= 1 e Σ 1 + + 2! 1! R=00 ... 3! x2a+1 sin x= E(-1)ª. R= 0 = X - 3! + 7! - (2n + 1)! 5! n-0 x* cos x= E(-1)-. 1 2! R= 0 %D ... (2n)! 4! 6! n-0 19 x2a+1 tan 'x= E(-1)“- = X R=1 %3D 2n+1 x2 In(1 + x) = E(-1)ª-1 R=1 = X- - + +... - 2 3 4 (1 + x)* = E (* k(k – 1) -x² + 2! k(k – 1)(k – 2) 3 +... x" = 1 + kx + R= 1 3! +
Transcribed Image Text:x" = 1 +x + x² + x³ + • ·. R= 1 e Σ 1 + + 2! 1! R=00 ... 3! x2a+1 sin x= E(-1)ª. R= 0 = X - 3! + 7! - (2n + 1)! 5! n-0 x* cos x= E(-1)-. 1 2! R= 0 %D ... (2n)! 4! 6! n-0 19 x2a+1 tan 'x= E(-1)“- = X R=1 %3D 2n+1 x2 In(1 + x) = E(-1)ª-1 R=1 = X- - + +... - 2 3 4 (1 + x)* = E (* k(k – 1) -x² + 2! k(k – 1)(k – 2) 3 +... x" = 1 + kx + R= 1 3! +
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