Evaluate the integral I=∫_0^(π/4)ln(1+tanθ ) dθ
I=∫_0^(π/4)ln(1+tan(π/4-θ) ) dθ= ∫_0^(π/4)ln(1+(1-tanθ)/(1+tanθ )) dθ= ∫_0^(π/4)ln(2/(1+tanθ )) dθ=∫_0^(π/4)ln2 dθ-∫_0^(π/4)ln(1+tanθ ) dθ=ln2.π/4-I∴2I=π/4 ln2 → ∴I=π/8 ln2
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I=∫_0^(π/4)ln(1+tan(π/4-θ) ) dθ
need an explanation for this answer? contact us directly to get an explanation for this answer= ∫_0^(π/4)ln(1+(1-tanθ)/(1+tanθ )) dθ= ∫_0^(π/4)ln(2/(1+tanθ )) dθ
=∫_0^(π/4)ln2 dθ-∫_0^(π/4)ln(1+tanθ ) dθ=ln2.π/4-I
∴2I=π/4 ln2 → ∴I=π/8 ln2