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Integral of e^-st×sin 2t×dt |
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Answer» 1/8Step-by-step explanation:Let I=∫ 0π/4 SIN 3 2tcos2t dt and let SIN2T=u. Then, d(sin2t)=DU⇒2cos2t dt=du⇒cos2tdt= 21 duAlso, t=0⇒u=sin0=0 and t= 4π ⇒u=sin 2π =1∴I= 21 ∫ 01 u 3 du= 21 [ 4u 4 ] 01 = 81 [u 4 ] 01 = 81 (1−0)= 81 |
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