Ex 7.7
Last updated at April 16, 2024 by Teachoo
Ex 7.7, 1 β(4βπ₯2) β«1βγβ(4βπ₯^2 ).ππ₯γ =β«1βγβ((π)^πβπ^π ).π πγ =1/2 π₯β((2)^2βπ₯^2 )+(2)^2/2 π ππ^(β1) π₯/2+πΆ =π/π β(πβπ^π )+π πππ^(βπ) π/π+πͺ It is of the form β«1βγβ(π^2βπ₯^2 ). ππ₯=1/2 π₯β(π^2βπ₯^2 )+π^2/2 π ππ^(β1) π₯/π+πΆγ Replacing a with 2, we get