-6 sin 60 de = du Step 4 Rewrite the integral by substituting for cos 60. 1 5/2 (u) 5/2 sin 60 V cos 60 sin 60 de = - du Apply the power rule of integration. xn + 1 + C = xP uX n + 1 Hence, 7/2 3/2 3/2 sin- 60 V cos 66 sin 60 de = - + C. 7/2 7/2 3/2 Step 5 Rewrite the right side in terms of 0. 7/27 1 cos/2 6e sin 60 V cos 60 sin 60 de = - + C 3/2 7/2 Submit Skip (you cannot come back)
-6 sin 60 de = du Step 4 Rewrite the integral by substituting for cos 60. 1 5/2 (u) 5/2 sin 60 V cos 60 sin 60 de = - du Apply the power rule of integration. xn + 1 + C = xP uX n + 1 Hence, 7/2 3/2 3/2 sin- 60 V cos 66 sin 60 de = - + C. 7/2 7/2 3/2 Step 5 Rewrite the right side in terms of 0. 7/27 1 cos/2 6e sin 60 V cos 60 sin 60 de = - + C 3/2 7/2 Submit Skip (you cannot come back)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.2: Trigonometric Equations
Problem 69E
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Question
Rewrite the side in terms of theta
Transcribed Image Text:-6 sin 60 de = du
Step 4
Rewrite the integral by substituting for cos 60.
1
5/2
(u)
5/2
du
sin 66 Vcos 66 sin 60 de = -
Apply the power rule of integration.
xn + 1
+ C
= xP uX
n + 1
Hence,
u7/2
3/2
3/2
sin 60 V cos 66 sin 60 de = -
+ C.
7/2
7/2
3/2
Step 5
Rewrite the right side in terms of 0.
7/27
1 cos/2 66
sin 60 V cos 60 sin 60 de = -
+ C
3/2
7/2
Submit
Skip (you cannot come back)
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