Ex 7.7, 7 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Integration by specific formulaes - Formula 8
Integration by specific formulaes - Formula 8
Last updated at April 16, 2024 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