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Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by Teachoo
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Example 25 Important Deleted for CBSE Board 2023 Exams
Example 26 Deleted for CBSE Board 2023 Exams
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Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams You are here
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 25 (Supplementary NCERT) β«1βπ₯ β(1+π₯βπ₯^2 ) ππ₯ β«1βπ₯ β(1+π₯βπ₯^2 ) ππ₯ We can write it as:- x = A [π/ππ₯ (1+π₯βπ₯^2 )]+ B x = A [0+1β2π₯]+ B x = A [1β2π₯]+ B x = "A"β2"A" π₯+ B x = β2"A" π₯+(π΄+π΅) Comparing x and constant term Thus, we can write x = A [1β2π₯] + B x = ((β1)/2)[1β2π₯] + 1 x = (β2A) x π₯/π₯ = β2A 1 = β2A A = (β1)/2 0 = A + B B = βA B = β((β1)/2) B = 1/2 Integrating β«1βγπ₯β(1+π₯βπ₯^2 ) γ ππ₯ = β«1βγ[(β1/2)[1β2π₯]+1/2] γ β(1+π₯βπ₯^2 ) ππ₯ = β«1βγ[(β1/2)[1 β2π₯] β(1+π₯βπ₯^2 )+1/2 β(1+π₯βπ₯^2 )] γ ππ₯ = β«1βγ(β1/2)[1β2π₯] β(1+π₯βπ₯^2 ) ππ₯+γ β«1βγ1/2 β(1βπ₯βπ₯^2 )γ ππ₯ = β1/2 β«1βγ[1β2π₯] β(1+π₯βπ₯^2 ) ππ₯+γ 1/2 β«1ββ(1+π₯βπ₯^2 ) ππ₯ Solving π°_π I_1 = (β1)/2 β«1βγ[1β2π₯] β(1+π₯βπ₯^2 )γ ππ₯ Let 1 + π₯ β π₯^2 = t Diff. both sides w.r.t.x 0 + 1 β2x = ππ‘/ππ₯ (1 β 2x) dx = dt dx = ππ‘/(1 β 2π₯) Thus, our equation becomes I_1 = (β1)/2 β«1βγ[1β2π₯] β(1+π₯βπ₯^2 )γ ππ₯ Putting the value if (1+π₯βπ₯^2) and dx, we get I_1 = (β1)/2 β«1βγ[1β2π₯] βπ‘γ. ππ₯ I_1 = (β1)/2 β«1βγ[1β2π₯] βπ‘γ. ππ‘/[1 β 2π₯] πΌ_1 = (β1)/2 β«1ββπ‘. ππ‘ I_1 = (β1)/2 β«1βγ(π‘)γ^(1/2) ππ‘ I_1 = (β1)/2 γπ‘ γ^(1/2 + 1)/((1/2 + 1) )+ C_1 I_1 = (β1)/2 γπ‘ γ^(3/2 )/((3/2) )+ C_1 I_1 = (β1)/3 γπ‘ γ^(3/2 )+ C_1 I_1 = (β1)/3 γ(1βπ₯ βπ₯^2 ) γ^(3/2 )+ C_1 ("Using t = " 1βπ₯ βπ₯^2 ) Solving π°_π I_2 = 1/2 β«1ββ(1+π₯βπ₯^2 ) ππ₯ I_2 = 1/2 β«1ββ(β(π₯^2βπ₯β1)) ππ₯ I_2 = 1/2 β«1ββ(β[π₯^2β2(π₯)(1/2)β1] ) ππ₯ I_2 = 1/2 β«1βγβ(β[π₯^2β2(π₯)(1/2)+(1/2)^2β(1/2)^2β1] ) γ ππ₯ I_2 = 1/2 β«1βγβ(β[(π₯β1/2)^2β(1/2)^2β1] ) γ ππ₯ I_2 = 1/2 β«1βγβ(β[γ (π₯β1/2)γ^2+(β1 β4)/4] ) γ ππ₯ I_2 = 1/2 β«1βγβ(β[γ (π₯β1/2)γ^2 β5/4] ) γ ππ₯ I_2 = 1/2 β«1βγβ(5/4 γ β(π₯β1/2)γ^2 ) γ ππ₯ I_2 = 1/2 β«1βγβ((β5/4)^2 γ β(π₯β1/2)γ^2 ) γ ππ₯ I_2 = 1/2 ((π₯ β 1/2)/2 β((β5/4)^2 γ β(π₯β1/2)γ^2 )+(β5/2)^2/2 γπ ππγ^(β1) ((π₯ + 1/2)/(β5/2))+ C_2 ) I_2 = 1/2 (((2π₯ β1)/2)/2 β(5/4β[π₯^2 +1/4β2π₯(1/2)] ) +(5/4)/2 γπ ππγ^(β1) (((2π₯ β 1)/2)/(β5/2))+ C_2 ) It is of form β(π^2βπ₯^2 ) ππ₯=1/2 π₯β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1) π₯/π+ C_2 Replacing x by (x β 1/2) and a by β5/2 , we get I_2 = 1/2 ((2π₯ β1)/4 β(5/4β[π₯^2 +1/4 βπ₯] ) +5/8 γπ ππγ^(β1) ((2π₯ β 1)/β5)+ C_2 ) I_2 " = " (2π₯ β1)/8 β(1+π₯+π₯^2 ) +5/16 γπ ππγ^(β1) ((2π₯ β 1)/β5)+C_3 Putting the value of I_1 and I_2 in (1) β«1βπ₯ β(1+ π₯βπ₯^2 ) dπ₯ = (β1)/2 β«1βγ[1β2π₯] β(1+π₯βπ₯^2 )γ ππ₯+β«1ββ(1+π₯βπ₯^2 ) ππ₯ = (β1)/3 γ(1+π₯βπ₯^2)γ^(3/2) + C_1 + ((2π₯ β1))/8 β(1+π₯βπ₯^2 )+5/16 γπ ππγ^(β1) ((2π₯ β 1)/β5)+ C_3 = (βπ)/π γ(π+πβπ^π)γ^(π/π) +π/π (ππ βπ) β(π+πβπ^π )+ π/ππ γπππγ^(βπ) ((ππ β π)/βπ)+ πͺ
