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The three basic trigonometric substitutions are in the table below. Integral Contains Try using Substitution Va z = a sin 0, - <0< Va? +z2 * = a tan 0, z = a sec 0, 0<0< or <0 < Part 1. Using the substitution: ¤ = tan 0, – <<0 < , re-write the indefinite integral then evaluate in terms of 0. dr Note: type 'theta' for 0 and 'dtheta' for do. Part 2. Back substituting in the antiderivative you found in Part 1. above we have dz 1+x² Note: answer should be in terms of z only.

The three basic trigonometric substitutions are in the table below. Integral Contains Try using Substitution Va z = a sin 0, - <0< Va? +z2 * = a tan 0, z = a sec 0, 0<0< or <0 < Part 1. Using the substitution: ¤ = tan 0, – <<0 < , re-write the indefinite integral then evaluate in terms of 0. dr Note: type 'theta' for 0 and 'dtheta' for do. Part 2. Back substituting in the antiderivative you found in Part 1. above we have dz 1+x² Note: answer should be in terms of z only.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Trigonometric Substitutions The three basic trigonometric substitutions are in the table below. Integral Contains Try using Substitution a sin θ,-플 < θ< 폴 - <8< 플 z=asec 8, 0<0< 풀 or 풀 <8<ㅠ Va? +1 I= a tan 0, a2 Part 1. Using the substitution: z = tan e, - 5<0< 5, re-write the indefinite integral then evaluate in terms of 0. dz = V1+x? Note: type 'theta' for e and 'dtheta' for de. Part 2. Back substituting in the antiderivative you found in Part 1. above we have zp Note: answer should be in terms of r only.
Transcribed Image Text:Trigonometric Substitutions The three basic trigonometric substitutions are in the table below. Integral Contains Try using Substitution a sin θ,-플 < θ< 폴 - <8< 플 z=asec 8, 0<0< 풀 or 풀 <8<ㅠ Va? +1 I= a tan 0, a2 Part 1. Using the substitution: z = tan e, - 5<0< 5, re-write the indefinite integral then evaluate in terms of 0. dz = V1+x? Note: type 'theta' for e and 'dtheta' for de. Part 2. Back substituting in the antiderivative you found in Part 1. above we have zp Note: answer should be in terms of r only.
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