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Ex 7.7, 7 Important You are here
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Ex 7.7, 12 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams
Ex 7.7, 13 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams
Ex 7.7, 14 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams
Last updated at Dec. 20, 2019 by Teachoo
Ex 7.7, 7 β(1+3π₯βπ₯2) β«1βγβ(1+3π₯βπ₯^2 ) ππ₯γ =β«1βγβ(1β(β3π₯+π₯^2 ) ) ππ₯γ =β«1βγβ(1β(π₯^2β3π₯) ) ππ₯γ =β«1βγβ(1β[π₯^2β2(3/2)(π₯)] ) ππ₯γ =β«1βγβ(1β[π₯^2β2(3/2)(π₯)+(3/2)^2β(3/2)^2 ] ) ππ₯γ =β«1βγβ(1β[(π₯β3/2)^2β9/4] ) ππ₯γ =β«1βγβ(1β(π₯β3/2)^2+9/4) ππ₯γ =β«1βγβ(1+γ9/4β(π₯β3/2)γ^2 ) ππ₯γ =β«1βγβ(γ13/4β(π₯β3/2)γ^2 ) ππ₯γ =β«1βγβ(γ(β13/2)^2β(π₯β3/2)γ^2 ) ππ₯γ =(π₯ β 3/2)/2 β((β13/2)^2β(π₯β3/2)^2 )+(β13/2)^2/2 π ππ^(β1) ((π₯ β 3/2))/(β13/2)+πΆ 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 β΄ Replacing π₯ by π₯β3/2 and a by β13/2 , we get =((2π₯ β 3)/2)/2 β(13/4β(π₯β3/2)^2 )+(13/4)/2 π ππ^(β1) (((2π₯ β 3)/2))/(β13/2)+πΆ =(2π₯ β 3)/4 β(13/4β(π₯β3/2)^2 )+13/8 π ππ^(β1) (2π₯ β 3)/β13+πΆ =(2π₯ β 3)/4 β(13/4β[π₯^2+9/4β2(π₯)(3/2)] )+13/8 π ππ^(β1) (2π₯ β 3)/β13+πΆ =(2π₯ β 3)/4 β(13/4βπ₯^2β9/4+3π₯)+13/8 π ππ^(β1) (2π₯ β 3)/β13+πΆ =(2π₯ β 3)/4 β(13/4β9/4+3π₯βπ₯^2 )+13/8 π ππ^(β1) (2π₯ β 3)/β13+πΆ =(ππ β π)/π β(π+ππβπ^π )+ππ/π πππ^(βπ) ((ππ β π)/βππ)+πͺ