1. Chapter 7 Class 12 Integrals (Term 2)
2. Concept wise
3. Integration by substitution - Trignometric - Normal

Transcript

Misc 7 Integrate the function sinβ‘π₯/sinβ‘(π₯ β π) Let I = β«1βsinβ‘π₯/sinβ‘(π₯ β π) ππ₯ Put t = π₯ β π Differentiating π€.π.π‘.π₯ ππ‘/ππ₯ = π(π₯ β π)/ππ₯ ππ‘/ππ₯ = 1 ππ₯ = ππ‘ Therefore β«1βγsin γβ‘(π‘ + π)/sinβ‘π‘ ππ‘ = β«1β(sinβ‘π‘ cosβ‘π + cosβ‘π‘ sinβ‘π)/sinβ‘π‘ ππ‘ = β«1β((sinβ‘π‘ cosβ‘π)/sinβ‘π‘ + (cosβ‘π‘ sinβ‘π)/sinβ‘π‘ ) ππ‘ = β«1βcosβ‘π ππ‘ + β«1βπππ‘β‘π‘ sinβ‘π ππ‘ = cosβ‘π β«1βππ‘ + sinβ‘π β«1βcotβ‘π‘ ππ‘ = cos a Γ t + sin a log |sinβ‘π‘ | + C Putting back t = x β a = (x β a) cos a + sin a |sinβ‘γ(π₯βπ)γ | + C sinβ‘(π΄+π΅) =sinβ‘π΄ cosβ‘π΅+cosβ‘π΄ sinβ‘π΅ (πππππ π ππβ‘π,πππ β‘π πππ ππππ π‘πππ‘π ) = sin a log |sinβ‘γ(π₯βπ)γ | + x cos a β a cos a + C = sin a log |π¬π’π§β‘γ(πβπ)γ | + x cos a + π_π (πΆ_1= β a cos a + C)