Example 7 (iii) - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Examples
Example 1 (i)
You are here
You are here
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
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)
