Ex 7.9, 10 (MCQ) - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 7.9, 10 If π(π₯)=β«_0^π₯βγπ‘ sinβ‘π‘ ππ‘γ, then πβ²(π₯) is (A) cos π₯+π₯ π ππ π₯ (B) π₯ π ππ π₯ (C) π₯ cos π₯ (D) sinβ‘π₯ π₯ cosβ‘π₯ If π(π₯)=β«_0^π₯βγπ‘ sinβ‘γπ‘ ππ‘γ γ Integrating by parts π(π₯)=β«_0^π₯βγπ‘ sinβ‘γπ‘ ππ‘γ γ By ILATE rule , π‘ is first function and sin π‘ is second function Hence β«_0^π₯βγπ‘ sinβ‘γπ‘ ππ‘=[π‘β«1βsinβ‘γπ‘ ππ‘ββ«1β(π/ππ‘ π‘Γβ«1βsinβ‘γπ‘ ππ‘γ )ππ‘γ ]_0^π₯ γ γ =[π‘Γ(βcosβ‘π‘ )ββ«1β(1Γcosβ‘π‘ )ππ‘ ]_0^π₯ =[βπ‘ cosβ‘γπ‘+sinβ‘π‘ γ ]_0^π₯ =βπ₯ cosβ‘γπ₯+sinβ‘γπ₯β(β0Γcosβ‘γ0+sinβ‘0 γ )γ γ =sinβ‘π₯βπ₯ cosβ‘π₯ Hence π(π₯)=sinβ‘γπ₯βπ₯ cosβ‘π₯ γ Therefore π^β² (π₯)=π/ππ₯ (sinβ‘γπ₯βπ₯ cosβ‘π₯ γ ) =π/ππ₯ sinβ‘γπ₯βπ/ππ₯ (π₯ cosβ‘π₯ )γ =cosβ‘γπ₯β(π₯ (π(cosβ‘π₯))/ππ₯+cosβ‘γπ₯ ππ₯/ππ₯γ )γ =cosβ‘γπ₯β(π₯(βsinβ‘π₯ )+cosβ‘π₯ )γ =cosβ‘γπ₯+π₯ sinβ‘γπ₯βcosβ‘π₯ γ γ =π₯ sinβ‘π₯ Hence, Option B is Correct
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