求1╱(3cos2x+4sin2x)的不定积分
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3cos2x+4sin2x
= 5[ (3/5)cos2x + (4/5)sin2x]
=5cos(2x- arccos(3/5))
∫dx/(3cos2x+4sin2x)
=(1/5)∫dx/cos(2x- arccos(3/5))
=(1/5)∫sec(2x- arccos(3/5)) dx
=(1/10)ln|sec(2x- arccos(3/5)) + tan(2x- arccos(3/5))| + C
= 5[ (3/5)cos2x + (4/5)sin2x]
=5cos(2x- arccos(3/5))
∫dx/(3cos2x+4sin2x)
=(1/5)∫dx/cos(2x- arccos(3/5))
=(1/5)∫sec(2x- arccos(3/5)) dx
=(1/10)ln|sec(2x- arccos(3/5)) + tan(2x- arccos(3/5))| + C
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