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Ex 7.8
Last updated at April 16, 2024 by Teachoo
Ex 7.8, 12 β«_0^(π/2)βγπππ 2 π₯ ππ₯γ Step 1 :- Let F(π₯)=β«1βγπππ ^2 π₯ ππ₯γ = β«1β(cosβ‘2π₯ + 1)/2 ππ₯ =1/2 β«1βγπππ 2π₯ ππ₯+1/2 β«1βππ₯γ =1/2 Γ (π ππ 2π₯)/2+π₯/2 =1/4 π ππ 2π₯+ π₯/2 Hence , F(π₯)=1/4 π ππ 2π₯+ π₯/2 Step 2 :- β«_0^(π/2)βγπππ ^2 π₯=πΉ(π/2)βπΉ(0) γ =1/4 π ππ(2 Γπ/2)+((π/2))/2β1/4 π ππ(2Γ0/2)β0/2 =1/4 π ππ(π)+π/4β1/4 γ sinγβ‘γ0 β0γ =1/4 Γ0+ π/4β 1/4 Γ0β0 = π /π