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Example 7 (i) - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Aug. 20, 2021 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
Example 3 (i)
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Example 3 (iii)
Example 4
Example 5 (i)
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Example 7 (i) You are here
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Example 8 (i)
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Example 9 (i)
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Example 10 (i)
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Example 21 Important
Example 22 Important
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Example 24
Example 25 Important Deleted for CBSE Board 2022 Exams
Example 26 Deleted for CBSE Board 2022 Exams
Example 27 (i)
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Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams
Chapter 7 Class 12 Integrals (Term 2)
Serial order wise
Transcript
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)
