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Example 6 (ii) - Chapter 7 Class 12 Integrals (Term 2)
Last updated at March 30, 2023 by Teachoo
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Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 6 Find the following integrals (ii) β«1βsinβ‘π₯/sinβ‘(π₯ + π) ππ₯ Let π₯+π=π‘ Differentiate both sides π€.π.π‘.π₯. 1=ππ‘/ππ₯ ππ₯=ππ‘ Hence, our equation becomes β«1βsinβ‘π₯/sinβ‘(π₯ + π) ππ₯ Putting the value of (π₯β‘+ π) and ππ₯ = β«1βsinβ‘π₯/sinβ‘π‘ ππ₯" " = β«1βπππβ‘(π β π)/sinβ‘π‘ ππ‘" " = β«1β(πππβ‘π ππ¨π¬β‘π β π¬π’π§β‘π ππ¨π¬β‘π)/sinβ‘π‘ ππ‘ = β«1β(sinβ‘γπ‘ cosβ‘π γ/sinβ‘π‘ β (sinβ‘π cosβ‘π‘)/sinβ‘π‘ ) ππ‘ = β«1βsinβ‘γπ‘ cosβ‘π γ/sinβ‘π‘ . ππ‘ββ«1β(sinβ‘π cosβ‘π‘)/sinβ‘π‘ . ππ‘ = β«1βcosβ‘π . ππ‘ββ«1βγsinβ‘π coπ‘β‘π‘ γ . ππ‘ = cos πβ«1β1. ππ‘βsinβ‘π β«1βπππβ‘π . π π = cos π (π‘)βsinβ‘π πππβ‘|πππβ‘π |+ πΆ = π‘.cosβ‘πβπ ππβ‘π πππβ‘|π ππβ‘π‘ |+ πΆ Putting back value of t = x + a (Using x + a = t β΄ x = t β a) (β(ππ πππ sinβ‘(πβπ)[email protected]β‘π cosβ‘πβsinβ‘π cosβ‘π )) ( β«1βcoπ‘β‘π₯ ππ₯=logβ‘|sinβ‘π₯ | ) = β«1βπππβ‘(π β π)/sinβ‘π‘ ππ‘" " = β«1βπππβ‘γπ ππ¨π¬β‘π βπ¬π’π§β‘π ππ¨π¬β‘π γ/sinβ‘π‘ ππ‘ = β«1β(sinβ‘γπ‘ cosβ‘π γ/sinβ‘π‘ β (sinβ‘π cosβ‘π‘)/sinβ‘π‘ ) ππ‘ = β«1βsinβ‘γπ‘ cosβ‘π γ/sinβ‘π‘ . ππ‘ββ«1β(sinβ‘π cosβ‘π‘)/sinβ‘π‘ . ππ‘ = β«1βcosβ‘π . ππ‘ββ«1βγsinβ‘π coπ‘β‘π‘ γ . ππ‘ = cos πβ«1β1. ππ‘βsinβ‘π β«1βπππβ‘π . π π = cos π (π‘)βsinβ‘π [πππβ‘|πππβ‘π | ]+πΆ1 = π‘.cosβ‘πβsinβ‘π [logβ‘|sinβ‘π‘ | ]+ πΆ1 Putting back value of t = x + a = (π₯+π) cosβ‘πβsinβ‘π [logβ‘|sinβ‘(π₯+π) | ]+ πΆ1 (Using x + a = t x = t β a) (β(ππ πππ sinβ‘(πβπ)[email protected]β‘π cosβ‘πβsinβ‘π cosβ‘π )) ( β«1βcoπ‘β‘π₯ ππ₯=logβ‘|sinβ‘π₯ | ) = (π₯+π) cosβ‘πβsinβ‘π logβ‘γ |sinβ‘(π₯+π) |γ+ πΆ = π₯ cosβ‘π+γπ cosγβ‘πβsinβ‘π logβ‘|sinβ‘(π₯+π) |+πΆ = π₯ cosβ‘πβsinβ‘π logβ‘|sinβ‘(π₯+π) |+γπ πππγβ‘π+πͺ = π πππβ‘πβπ¬π’π§β‘γ πγ πππβ‘|πππβ‘(π+π) |+ππ (πβπππ πΆ1=π cosβ‘π+πΆ)
