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Question

π/40tanθdθ=1(2)log(21)+π2k(2).Find the value of k.

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Solution

Let I=tanθdθ=12(I1I2)
Where
I1=(tanθ+cotθ)dθ=sinθ+cosθcosθsinθdθ
Put sinθ+cosθ=u
I1=2dt1t2=2sin1t=2sin1(sinθcosθ)
And
I2=(cotθtanθ)dθ
=cosθsinθcosθsinθdθ
Put sinθ+cosθ=t2sinθcosθ=t21
I2=2dtt21=2log(1+t21)=2log(sinθ+cosθ+sin2θ)I=2sin1(sinθcosθ)+2log(sinθ+cosθ+sin2θ)
therefore
π/40Idθ=2log(2+1)+π22



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