求(6)的不定积分的具体步骤
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(4)
∫ dx/(sinxcosx)
=2∫ dx/sin2x
=2∫ csc2x dx
= ln|csc2x-cot2x| + C
(6)
∫xdx/√(2-3x)
= -(2/3)∫xd√(2-3x)
=-(2/3)x√(2-3x) + (2/3) ∫√(2-3x) dx
=-(2/3)x√(2-3x) - (2/9) ∫√(2-3x) d(2-3x)
=-(2/3)x√(2-3x) - (4/27) [(2-3x)]^(3/2) + C
∫ dx/(sinxcosx)
=2∫ dx/sin2x
=2∫ csc2x dx
= ln|csc2x-cot2x| + C
(6)
∫xdx/√(2-3x)
= -(2/3)∫xd√(2-3x)
=-(2/3)x√(2-3x) + (2/3) ∫√(2-3x) dx
=-(2/3)x√(2-3x) - (2/9) ∫√(2-3x) d(2-3x)
=-(2/3)x√(2-3x) - (4/27) [(2-3x)]^(3/2) + C
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