Solve the indefinite integral sin(1.5piX+.25pi)
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Solve the indefinite integral sin(1.5piX+.25pi)

[From: ] [author: ] [Date: 12-02-21] [Hit: ]
25pi)y =- cos (1.5pix +0.25pi)/1.5pi +c-No.......
Let u = 1.5piX+.25pi
du = 1.5pi dx
dx = (1/1.5pi) = 2/3pi

∫ sin(1.5piX+.25pi) dx = (2/3pi) ∫ sin(u) du = (-2/3pi) cos(u) + C

=(-2/3pi) cos(1.5piX+.25pi) +C

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http://answers.yahoo.com/question/index;_ylt=Au3VJxTY5FGcwuPSVZDxEWHty6IX;_ylv=3?qid=20120220090757AA7dwM9

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dy/dx = sin(1.5pix+0.25pi) y = - cos (1.5pix +0.25pi)/1.5pi +c

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No.
1
keywords: indefinite,1.5,Solve,integral,25,pi,sin,the,piX,Solve the indefinite integral sin(1.5piX+.25pi)
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