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Evaluate Solution √25 Let x = 5 sin(8), where 플 플 √25 - x² = √√√25 - 25 sin²(0) = √√25 cos² (8) = 5|cos(0)| = 5 cos(0). (Note that cos(0) ≥ 0 because sos ..) Thus, the inverse substitution rule gives √25-x² dx. dx = sos 5 cos(0) 25 sin²(0) cos² (8) Then dx = = [cot²(8) de = √ (csc²(8) − 1) de de + C. de de and

Evaluate Solution √25 Let x = 5 sin(8), where 플 플 √25 - x² = √√√25 - 25 sin²(0) = √√25 cos² (8) = 5|cos(0)| = 5 cos(0). (Note that cos(0) ≥ 0 because sos ..) Thus, the inverse substitution rule gives √25-x² dx. dx = sos 5 cos(0) 25 sin²(0) cos² (8) Then dx = = [cot²(8) de = √ (csc²(8) − 1) de de + C. de de and

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.6: The Matrix Of A Linear Transformation
Problem 30EQ
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Evaluate [v Solution J 25 - x² +² TT Let x = 5 sin(0), where < # < 2 √25-x² 25x² +² (Note that cos(0) ≥ 0 because dx. dx = -TT 2 = = 1₂ = = ✓ 25 - 25 sin²(0) = √ 25 cos² (0) = 5|cos(0)| = 5 cos(0). 5 cos(0) Then dx = ≤0 ≤.) Thus, the inverse substitution rule gives 2 2 25 sin²(0) cos²(0) cot² (0) de = [(csc²(0) — 1) de de de and + C. de Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot(0) in terms of sin(0) = U|X
Transcribed Image Text:Evaluate [v Solution J 25 - x² +² TT Let x = 5 sin(0), where < # < 2 √25-x² 25x² +² (Note that cos(0) ≥ 0 because dx. dx = -TT 2 = = 1₂ = = ✓ 25 - 25 sin²(0) = √ 25 cos² (0) = 5|cos(0)| = 5 cos(0). 5 cos(0) Then dx = ≤0 ≤.) Thus, the inverse substitution rule gives 2 2 25 sin²(0) cos²(0) cot² (0) de = [(csc²(0) — 1) de de de and + C. de Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot(0) in terms of sin(0) = U|X
Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot(0) in terms of sin(0) or by drawing a diagram, as in the figure below, where is interpreted as an angle of a right triangle. cot (8) X Since sin(0) = ¹ we label the opposite side and the hypotenuse as having lengths x and 5. Then the Pythagorean theorem gives the length of the adjacent side as 25 x2, so we can simply read the value of cot(0) from the figure. A Since sin(8) = ₁ 25- +² X +2 (Although > 0 in the diagram, this expression for cot(0) is valid even when 0 <0.) 5 25 dx = we have 0 = sin¯¹ X + C. and so OneDrive Screenshot saved X5 X
Transcribed Image Text:Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot(0) in terms of sin(0) or by drawing a diagram, as in the figure below, where is interpreted as an angle of a right triangle. cot (8) X Since sin(0) = ¹ we label the opposite side and the hypotenuse as having lengths x and 5. Then the Pythagorean theorem gives the length of the adjacent side as 25 x2, so we can simply read the value of cot(0) from the figure. A Since sin(8) = ₁ 25- +² X +2 (Although > 0 in the diagram, this expression for cot(0) is valid even when 0 <0.) 5 25 dx = we have 0 = sin¯¹ X + C. and so OneDrive Screenshot saved X5 X
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