Example 35 - Chapter 7 Class 12 Integrals
Last updated at August 11, 2026 by Teachoo
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Example 1 (i)
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Chapter 7 Class 12 Integrals
Serial order wise
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Transcript
Example 35 Evaluate β«1βcosβ‘γ6π₯ β(1+sinβ‘6π₯ )γ ππ₯ β«1βcosβ‘γ6π₯ β(1+sinβ‘6π₯ )γ ππ₯ Put π‘ = β(1+sinβ‘6π₯ ) π‘^2 = 1+sinβ‘6π₯ Differentiate π€.π.π‘.π₯ (ππ‘^2)/ππ₯=π/ππ₯ (1+sinβ‘6π₯ ) 2π‘. ππ‘/ππ₯=6 cos 6 π₯ (2π‘ ππ‘)/(6 cosβ‘6π₯ )=ππ₯ Therefore, β«1βcosβ‘γ6π₯ β(1+sinβ‘6π₯ )γ =β«1βcosβ‘γ6π₯ π‘γ . ( 2 π‘ ππ‘)/γ6 cosγβ‘6π₯ =β«1βπ‘^2/3β‘ππ‘ =1/3 β«1βπ‘^2β‘ππ‘ =1/3 π‘^(2 + 1)/(2 + 1) + πΆ =1/3 π‘^3/3 + πΆ = π‘^3/9 + πΆ Putting back π‘ = β(1+π ππβ‘6π₯ ) = (β(1 + sinβ‘6π₯ ))^3/9 + πΆ = (1 + sinβ‘6π₯ )^(1/2 Γ 3)/9 + πΆ = π/π (π + πππβ‘ππ )^(π/π)+πͺ
