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Ex 7.8

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Last updated at June 13, 2023 by Teachoo

Ex 7.8, 17 β«_0^(π/4)β(2 sec^2β‘π₯+π₯^3+2) ππ₯ Let F(π₯)=β«1βγ(2 sec^2β‘γπ₯+π₯^3+2γ ) ππ₯γ =2β«1βγsec^2β‘γπ₯ ππ₯γ+β«1βγπ₯^3 ππ₯+β«1βγ2 ππ₯γγγ =2 tanβ‘γπ₯+π₯^4/4+2π₯γ Hence F(π₯)=2 tanβ‘γπ₯+π₯^4/4+2π₯γ Now, β«_0^(π/4)βγ(2π ππ^2 π₯+π₯^3+2)ππ₯=πΉ(π/4)βπΉ(0) γ =[2π‘ππ π/4+(π/4)^4/4+2 π/4]β[2π‘ππ(0)+(0)^4/4+2 Γ0] =2π‘ππ π/4+π^4/4^4 Γ 1/4+π/2β[2Γ0+0+0] =2 Γ1+π^4/4^5 +π/2β0 =π+π ^π/ππππ+π /π