t*sin^3(t) dt.不定积分.求救!
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∫ t*sin³t dt
= ∫ t*(1-cos²t)*sint dt
= ∫ t*sint dt - ∫ t*cos²t*sint dt
= ∫ t*sint dt - (1/2)∫ t*(1+cos2t)*sint dt
= (1/2)∫ t*sint dt - (1/2)∫ t*cos2t*sint dt
= (1/2)∫ t*sint dt - (1/4)∫ t*(sint3t-sint) dt
= (1/2)∫ t*sint dt - (1/4)∫ t*sin3t dt + (1/4)∫ t*sint dt
= (3/4)∫ t*sint dt - (1/4)∫ t*sin3t dt
= -(3/4)∫ t dcost + (1/12)∫ t dcos3t
= -(3/4)tcost + (3/4)∫ cost dt + (1/12)tcos3t - (1/12)∫ cos3t dt
= -(3/4)tcost + (3/4)sint + (1/12)tcos3t - (1/36)sin3t + C
= (3/4)(sint-tcost) + (3tcos3t-sin3t)/36 + C
= ∫ t*(1-cos²t)*sint dt
= ∫ t*sint dt - ∫ t*cos²t*sint dt
= ∫ t*sint dt - (1/2)∫ t*(1+cos2t)*sint dt
= (1/2)∫ t*sint dt - (1/2)∫ t*cos2t*sint dt
= (1/2)∫ t*sint dt - (1/4)∫ t*(sint3t-sint) dt
= (1/2)∫ t*sint dt - (1/4)∫ t*sin3t dt + (1/4)∫ t*sint dt
= (3/4)∫ t*sint dt - (1/4)∫ t*sin3t dt
= -(3/4)∫ t dcost + (1/12)∫ t dcos3t
= -(3/4)tcost + (3/4)∫ cost dt + (1/12)tcos3t - (1/12)∫ cos3t dt
= -(3/4)tcost + (3/4)sint + (1/12)tcos3t - (1/36)sin3t + C
= (3/4)(sint-tcost) + (3tcos3t-sin3t)/36 + C
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