Integration Full Chapter Explained - Integration Class 12 - Everything you need








Last updated at Dec. 20, 2019 by Teachoo
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Example 7 Find (i) β«1βcos^2β‘π₯ ππ₯ β«1βcos^2β‘π₯ ππ₯ =β«1β((cosβ‘2π₯ + 1)/2) ππ₯ = 1/2 β«1β(cosβ‘2π₯+1) ππ₯ = 1/2 [β«1βcosβ‘2π₯ ππ₯+β«1β1 ππ₯] We know cosβ‘2π₯=2 cos^2β‘π₯β1 cosβ‘2π₯+1=2 cos^2β‘π₯ (cosβ‘2π₯ + 1)/2=cos^2β‘π₯ β«1βπππβ‘ππ π π Let 2π₯=π‘ 2 =ππ‘/ππ₯ ππ₯=1/2 ππ‘ β«1βcosβ‘π‘ . 1/2 ππ‘ =1/2 (sinβ‘π‘+πΆ1) Putting value of π‘ = 2π₯ =1/2 sinβ‘2π₯+πΆ1 β«1βπ π π =β«1βπ₯^0 ππ₯ =[π₯^(0 + 1)/(0 + 1)]+πΆ =π₯+πΆ2 Thus, β«1βcos^2β‘π₯ ππ₯=1/2 [β«1βcosβ‘2π₯ ππ₯+β«1β1 ππ₯] =1/2 [1/2 sinβ‘2π₯+πΆ1+π₯+πΆ2] =1/4 sinβ‘2π₯+π₯/2+1/2(πΆ1+πΆ2) =π/π+π/π π¬π’π§β‘ππ+πͺ ("From (1) and (2) " ) ("Let" πΆ1+πΆ2=πΆ) Example 7 Find (ii) β«1βγsinβ‘γ2π₯ γ cosβ‘3π₯ γ ππ₯ We know that 2 sinβ‘π΄ cosβ‘π΅=sinβ‘(π΄+π΅)+sinβ‘(π΄βπ΅) sinβ‘π΄ cosβ‘π΅=1/2 [sinβ‘(π΄+π΅)+sinβ‘(π΄βπ΅) ] Replace A by 2π₯ & B by 3π₯ sinβ‘2π₯ cosβ‘3π₯=1/2 [sinβ‘(2π₯+3π₯)+sinβ‘(2π₯β3π₯) ] sinβ‘2π₯ cosβ‘3π₯=1/2 [sinβ‘(5π₯)+sinβ‘(βπ₯) ] sinβ‘2π₯ cosβ‘3π₯=1/2 [sinβ‘5π₯βsinβ‘π₯ ] β«1β(sinβ‘γ2π₯ γ cosβ‘3π₯ ) ππ₯=1/2 β«1β(sinβ‘5π₯βsinβ‘π₯ ) ππ₯ =1/2 [β«1βsinβ‘5π₯ ππ₯ββ«1βsinβ‘π₯ ππ₯] β«1βπππβ‘ππ π π Let 5π₯=π‘ 5 = ππ‘/ππ₯ ππ₯=1/5 ππ‘ =β«1βsinβ‘π‘ . 1/5 ππ‘ =1/5 β«1βsinβ‘π‘ . ππ‘ =1/5 (γβcosγβ‘π‘+πΆ1) =β1/5 cosβ‘π‘+1/5 πΆ1 Putting value of π‘ =β1/5 cosβ‘5π₯+1/5 πΆ1 β«1βsinβ‘π₯ ππ₯ =βcosβ‘π₯+πΆ2 Thus, β«1βγsinβ‘γ2π₯ γ cosβ‘3π₯ γ=1/2 [β«1βγsinβ‘5π₯ ππ₯γββ«1βsinβ‘π₯ ππ₯] =1/2 [β1/5 cosβ‘5π₯+1/5 πΆ1β(βcosβ‘π₯ )+πΆ2] =1/2 [β1/5 cosβ‘5π₯+cosβ‘π₯+1/5 πΆ1βπΆ2] =1/2 [β1/5 cosβ‘5π₯+cosβ‘π₯+πΆ] =(βπ)/ππ ππ¨π¬β‘ππ+π/π ππ¨π¬β‘π+πͺ Example 7 Find (iii) β«1βsin^3β‘π₯ ππ₯ We know that π ππ 3π₯=3 π ππβ‘π₯β4 γπ ππγ^3β‘π₯ 4 γπ ππγ^3 π₯=3 π ππβ‘π₯βπ ππβ‘3π₯ γπ ππγ^3 π₯=(3 π ππβ‘π₯ β π ππβ‘3π₯)/4 β«1βsin^3β‘π₯ ππ₯=β«1β(3 sinβ‘π₯ β sinβ‘3π₯)/4 ππ₯ =1/4 β«1β(3 sinβ‘π₯βsinβ‘3π₯ ) ππ₯ =1/4 [3β«1βsinβ‘π₯ ππ₯ββ«1βsinβ‘3π₯ ππ₯] β«1βπππβ‘ππ π π Let 3π₯=π‘ 3=ππ‘/ππ₯ ππ₯=1/3 ππ‘ β«1βπππβ‘ππ π π=β«1βsinβ‘π‘ . 1/3 ππ‘ =1/3 β«1βsinβ‘π‘ . ππ‘ =1/3 (γβcosγβ‘π‘+πΆ1) =β1/3 cosβ‘π‘+1/3 πΆ1 Putting value of π‘ =β1/3 cosβ‘3π₯+1/3 πΆ1 β«1βπππβ‘π π π =βcosβ‘π₯+πΆ2 Thus, β«1βsin^3β‘π₯ ππ₯=1/4 [3β«1βγsinβ‘π₯ ππ₯γββ«1βsinβ‘3π₯ ππ₯] =1/4 [3(βcosβ‘π₯+πΆ2)β(β1/3 cosβ‘3π₯+1/3 πΆ1)] =1/4 [β3 cosβ‘π₯+3 πΆ2+ 1/3 cosβ‘3π₯+1/3 πΆ1] =1/4 [β3 cosβ‘π₯+1/3 cosβ‘3π₯+(3 πΆ2β1/3 πΆ1)] =(β3)/4 cosβ‘π₯+1/12 cosβ‘3π₯+1/4 (3 πΆ2β1/3 πΆ1) =(βπ)/π πππβ‘π+π/ππ πππβ‘ππ+πͺ ("As" 1/4 (3 πΆ2β1/3 πΆ1)=πΆ)
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