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Ex 7.8

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Ex 7.8, 18 You are here

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Last updated at June 13, 2023 by Teachoo

Ex 7.8, 18 β«_0^πβ(γsππγ^2β‘γπ₯/2γβcos^2β‘γπ₯/2γ ) ππ₯ Step 1 :- Let F(π₯)=β«1β(sin^2β‘γπ₯/2βcos^2β‘γπ₯/2γ γ )ππ₯ =ββ«1β(cos^2β‘γπ₯/2βsin^2β‘γπ₯/2γ γ )ππ₯ =ββ«1βγcosβ‘π₯ ππ₯γ =βsinβ‘π₯ Hence F(π₯)=βsinβ‘π₯ Step 2 :- β«_0^πβ(π ππ^2 π/2βπππ ^2 π₯/2) ππ₯=πΉ(π)βπΉ(0) =βγπ ππ πγβ‘γ+ π ππ(0)γ =β 0+0 = π