Ex 7.7, 2 - 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, 2 (Method 1) β(1β4π₯2) β«1βγβ(1β4π₯^2 ).ππ₯γ =β«1βγβ(4(1/4βπ₯^2 ) ).ππ₯γ =β«1βγβ4 β(1/4βπ₯^2 ).ππ₯γ =2β«1βγβ((1/2)^2βπ₯^2 ).ππ₯" " γ 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+πΆ1γ Replacing a with 2 and 1/2 , we get =2[1/2 π₯β((1/2)^2βπ₯^2 ) +(1/2)^2.1/2 π ππ^(β1) π₯/(1/2)+πΆ1] =2[1/2 π₯β(1/4βπ₯^2 ) +1/4 . 1/2 π ππ^(β1) 2π₯+πΆ1] =π₯β(1/4βπ₯^2 )+1/4 π ππ^(β1) 2π₯+2πΆ1 =1/4 π ππ^(β1) 2π₯+π₯β((1 β 4π₯^2)/4)+πΆ =π/π πππ^(βπ) ππ+π/π πβ(π β ππ^π )++πͺ [where πΆ=2πΆ1] Ex 7.7, 2 (Method 2) β(1β4π₯2) Let 2π₯=π‘ Differentiating both sides w.r.t. π₯ 2=ππ‘/ππ₯ ππ₯=ππ‘/2 Integrating the function β«1βγβ(1β4π₯^2 ) ππ₯γ =β«1βγβ(1β(2π₯)^2 ) ππ₯γ Putting value of t = 2π₯ and ππ₯ = ππ‘/2 =β«1βγβ(1βπ‘^2 ) .ππ‘/2γ =1/2 β«1βγβ(1βπ‘^2 ) ππ‘γ =1/2 β«1βγβ((1)^2β(π‘)^2 ) ππ‘γ =1/2 [π‘/2 β((1)^2βπ‘^2 )+(1)^2/2 π ππ^(β1) π‘/((1) )+πΆ1] =π‘/4 β(1βπ‘^2 )+1/4 π ππ^(β1) (π‘)+πΆ1" " /2 =π‘/4 β(1βπ‘^2 )+1/4 π ππ^(β1) (π‘)+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 +πΆ1γ Replacing a by 1 and π₯ by π‘ , we get Putting back t = 2x =2π₯/4 β(1β(2π₯)^2 )+1/4 π ππ^(β1) 2π₯+πΆ =π/π πππ^(βπ) ππ+π/π β(πβππ^π )+πͺ