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Ex 7.3
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Last updated at March 23, 2023 by Teachoo
Ex 7.3, 22 1/(cosβ‘(π₯ β π) cosβ‘γ(π₯ β π)γ ) β«1β1/(cosβ‘(π₯ β π) cosβ‘γ(π₯ β π)γ ) Multiply & Divide by πππβ‘(πβπ) =β«1βγsinβ‘(π β π)/sinβ‘(π β π) Γ1/(cosβ‘(π₯ β π) cosβ‘(π₯ β π) )γ ππ₯ =1/sinβ‘(π β π) β«1βsinβ‘(π β π)/(cosβ‘(π₯ β π) cosβ‘(π₯ β π) ) ππ₯ =1/sinβ‘(π β π) β«1βsinβ‘(π β π + π₯ β π₯)/(cosβ‘(π₯ β π) cosβ‘(π₯ β π) ) ππ₯ =1/sinβ‘(π β π) β«1βsinβ‘((π₯ β π) + (π β π₯))/(cosβ‘(π₯ β π) cosβ‘(π₯ β π) ) ππ₯ =1/sinβ‘(π β π) β«1βsinβ‘((π₯ β π) β (π₯ β π))/(cosβ‘(π₯ β π) cosβ‘(π₯ β π) ) ππ₯ =1/sinβ‘(π β π) β«1β(sinβ‘(π₯ β π) cosβ‘(π₯ β π) β cosβ‘(π₯ β π) sinβ‘(π₯ β π))/(cosβ‘(π₯ β π) cosβ‘(π₯ β π) ) ππ₯ =1/sinβ‘(π β π) β«1β((sinβ‘(π₯ β π) cosβ‘(π₯ β π))/(cosβ‘(π₯ β π) cosβ‘(π₯ β π) ) β (cosβ‘(π₯ β π) sinβ‘(π₯ β π))/(cosβ‘(π₯ β π) cosβ‘(π₯ β π) )) ππ₯ =1/sinβ‘(π β π) [β«1β(sinβ‘(π₯ β π)/cosβ‘(π₯ β π) β sinβ‘(π₯ β π)/cosβ‘(π₯ β π) ) ππ₯] =1/sinβ‘(π β π) [β«1βγtanβ‘(π₯βπ)βtanβ‘(π₯βπ) γ ππ₯] =1/sinβ‘(π β π) [β«1βtanβ‘(π₯βπ) ππ₯ββ«1βtanβ‘(π₯βπ) ππ₯] Using π ππβ‘(π΄βπ΅)=π ππβ‘π΄ πππ β‘π΅βπππ β‘π΄ π ππβ‘π΅ Replace A by (π₯βπ) & B by (π₯βπ) π ππβ‘((π₯βπ)β(π₯βπ)) =π ππβ‘(π₯βπ) πππ β‘(π₯βπ)βπππ β‘(π₯βπ) π ππβ‘(π₯βπ) Using β«1βtanβ‘π₯ ππ₯=βlogβ‘|cosβ‘π₯ |+πΆ =1/sinβ‘(π β π) [βlogβ‘|cosβ‘(π₯βπ) |+logβ‘|cosβ‘(π₯βπ) | ]+πΆ =1/sinβ‘(π β π) [logβ‘|cosβ‘(π₯βπ) |βlogβ‘|cosβ‘(π₯βπ) | ]+πΆ =π/πππβ‘(π β π) π₯π¨π β‘|πππβ‘(π β π)/πππβ‘(π β π) |+πͺ