L5 Trig Substitution Part I - 2 ex 2 Z2 p3 x3 p16 x 2 dx 40...

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Lecture 5: Techniques for Integration (III) Trigonometric Substitution, part I Idea: Use the ’ trig ’ substitution x = sin to evaluate the integral Z 1 x 2 p 1 x 2 dx Solution: Let x = sin ; 2 2 : Then dx = cos d ; and p 1 x 2 = p 1 sin 2 = p cos 2 = cos Z 1 x 2 p 1 x 2 dx = Z cos d (sin 2 )(cos ) = Z csc 2 d = cot + c = 1
Table of Trigonometric Substitutions 1 : p a 2 x 2 x = a sin p a 2 x 2 = a cos 2 : p a 2 + x 2 x = a tan p a 2 + x 2 = a sec 3 : p x 2 a 2 x = a sec p x 2 a 2 = a tan ex. 1. Z 1 p x 2 9 dx (ln j x + p x 2 9 j ln 3 + c = ln j x + p x 2 9 j + C )

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