Ex 7.7, 5
Last updated at March 11, 2017 by Teachoo
Transcript
Ex7.7, 5 β(1β4π₯βπ₯2) β«1βγβ(1β4π₯βπ₯^2 ) .ππ₯γ =β«1βγβ(1β(4π₯+π₯^2 ) ) .ππ₯γ =β«1βγβ(1β(π₯^2+4π₯) ) .ππ₯γ =β«1βγβ(1β(π₯^2+2.2.π₯) ) ππ₯γ =β«1βγβ(1β[π₯^2+2.2.π₯+(2)^2β(2)^2 ] ) ππ₯γ =β«1βγβ(1β[(π₯+2)^2β4] ) .ππ₯γ =β«1βγβ(1β(π₯+2)^2+4 ) ππ₯γ =β«1βγβ(5β(π₯+2)^2 ) ππ₯γ =β«1βγβ((β5)^2β(π₯+2)^2 ) ππ₯γ = ((π₯ + 2))/2 β((β5)^2β(π₯+2)^2 )+(β5)^2/2 sin^(β1)β‘((π₯ + 2)/β5)+πΆ = ((π₯ + 2))/2 β(5βγ(π₯γ^2+4π₯+4))+ 5/2 γ sinγ^(β1)β‘((π₯ + 2)/β5)+πΆ = ((π₯ + 2))/2 β(5βπ₯^2β4π₯β4)+5/2 γ sinγ^(β1)β‘((π₯ + 2)/β5)+πΆ = π/π γ πππγ^(βπ)β‘((π + π)/βπ)+((π + π))/π β(πβπ^πβππ)+πͺ
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