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Question

The value of (2+1)198π3π/4π/4ϕ1+sinϕdϕ is

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Solution

Let I=3π/4π/4ϕ1+sinϕdϕ=3π/4π/4(πϕ)1+sin(πϕ)dϕ=π3π/4π/411+sinϕdϕ3π/4π/4ϕ1+sinϕdϕ
2I=π3π/4π/411+sinϕdϕI=π23π/4π/411+sinϕdϕ=π23π/4π/411+cos(π2ϕ)dϕ=π43π/4π/4sec2(π4ϕ2)dϕ=π4×2[tan(π4ϕ2)]3π/4π/4=π2[tan(π8)tan(π8)]=π(21)
Therefore
(2+1)198π3π/4π/4ϕ1+sinϕdϕ=(2+1)198π×π(21)=198

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