Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3

Marks : 08

Years : MAY 2016

(i) $L[\int\limits_0^t ue^{-3u}\cos ^2(2u) du] = \dfrac 1S L [e^{-3t}.t\cos^2 2t]$

$=\dfrac 1S L[t\cos^2 2t]_{S\to S+3} =\dfrac 1S[(-1) \dfrac d{dS} L[\dfrac {1+\cos 4t}2]]_{S\to S+3} \\ =\dfrac 1S [-\dfrac 12 \dfrac d{dS}[\dfrac 1S+\dfrac S{S^2+16}]]_{S\to S+3} = -\dfrac 1{2S} [-\dfrac 1{S^2}+ \dfrac {(S^2+16)-S(2S)}{(S^2+16)^2}]_{S\to S+3} \\ = -\dfrac 1{2S}[-\dfrac 1{S^2}+\dfrac {(16-S^2)}{(S^2+16)^2}]_{S\to S+3} = \dfrac 1{2S}[\dfrac 1{(S+3)^2}+\dfrac {(S+3)^2-16}{[(S+3)^2+16]^2}]$

(ii) $L[t\sqrt{1+\sin t}]=(-1) \dfrac d{ds} L[\sqrt{1+\sin t}]=-\dfrac d{ds}L[\sqrt{(\cos t/2+\sin t/2)^2}] \\ =-\dfrac d{ds}L [ \cos t/2 + \sin t/2 ]= -\dfrac d{ds}[\dfrac S {S^2+\dfrac 14 } +\dfrac {1/2}{S^2+\dfrac 14}] \\ =4[\dfrac {4S^2+4S-1 }{(4S^2+1)^2}]$